Math book

Hi,
This book is based on some quicker methods
for different numerical problems, hence useful for Bank
Exams.
MULTIPLICATION TABELE (10x10)
1 2 3 4 5 6 7 8 9 10
1 1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 12 14 16 18 20
3 9 12 15 18 21 24 27 30
4 16 20 24 28 32 36 40
5 25 30 35 40 45 50
6 Think
As
8*4
4 *8,..
36 42 48 54 60
7 49 56 63 70
8 68 72 80
9 81 90
10 100
MULTIPLICATION TABLE (25x10)
21 21 42 63 84 105 126 147 168 189 210
22 22 44 66 88 110 132 154 176 198 220
23 23 46 69 92 115 138 161 184 207 230
24 24 48 72 96 120 144 168 192 216 240
25 25 50 75 100 125 150 175 200 225 250
1 2 3 4 5 6 7 8 9 10
11 11 22 33 44 55 66 77 88 99 110
12 12 24 36 48 60 72 84 96 108 120
13 13 26 39 52 65 78 91 104 117 130
14 14 28 42 56 70 84 98 112 126 140
15 15 30 45 60 75 90 105 120 135 150
16 16 32 48 64 80 96 112 128 144 160
17 17 34 51 68 85 102 119 136 153 170
18 18 36 54 72 90 108 126 144 162 180
19 19 38 57 76 95 114 133 152 171 190
20 20 40 60 80 100 120 140 160 180 200
MULTIPLICATION TABLE (11-20 x11-20)
11 12 13 14 15 16 17 18 19 20

11 121 132 143 154 165 176 187 198 209 220
12 144 156 168 180 192 204 216 228 240
13 169 182 195 208 221 234 247 260
14 196 210 224 238 252 266 280
15 225 240 255 270 285 300
16 256 272 288 304 320
17 289 306 323 340
18 324 342 360
19 361 380
20 400

For quickest math-
1. Many diagrams are given in almost every example which would be the reminder of the
process when u
completely learn the techniques.
2. Try to understand the colors, every color is explaining the steps and other important things.
Edited by
RAKES PRASAD
First we learn how to find square, square root and cube roots with multiplication techniques.
Rule no 1. You are to remember squares of 1 to 32. They are given as follows
No Square No Sq No Sq No Sq
no sq
1 1 9 81 17 289 25 625 33 1089
2 4 10 100 18 324 26 676 34 1156
3 9 11 121 19 361 27 729 35 1225
4 16 12 144 20 400 28 784 36 1296
5 25 13 169 21 441 29 841 38 1444
6 36 14 196 22 484 30 900 41 1681
7 49 15 225 23 529 31 961
8 64 16 256 24 576 32 1024
N.B. The square of 38 contains 444 and the sq of an odd prime no 41 contains two perfect
squares as 4
2
= 16 and 9
2
= 81.
Rule No 2. Squares of numbers ending in 5 :we are showing by giving
an
example.
For the number 25, the last digit is 5 and the 'previous' digit is 2. 'to multiply the previous digit 2
by one
more , that is, by 3'. It becomes the L.H.S of the result, that is, 2 X 3 = 6. The R.H.S of the result
is

5
2
, that is, 25. Thus 25
2
= 2 X 3 / 25 = 625. In the same way, 105
2
= 10 X 11/25 = 11025; 135
2
= 13 X
14/25 = 18225; see the figure below…
….
Now we are extending the formula as
Rule no 3: If sum of the last two digits give 10 then you can use the formula but
L.H.S should
be same. Check the examples
Ex 1 : 47 X 43 .See the end digits sum 7 + 3 = 10 ; then 47 x 43 = ( 4
+ 1 ) x 4 / 7 x 3
= 20 / 21 = 2021.
Example 2: 62 x 68
2 + 8 = 10, L.H.S. portion remains the same i.e.,, 6.
Next of 6 gives 7 and 62 x 68 = ( 6 x 7 ) / ( 2 x 8 ) = 42 / 16 =
4216.
Example 3: 127 x 123
127 x 123 = 12 x 13 / 7 x 3 = 156 / 21 = 15621. (see the fig )
Example 4. 395
2
395
2
= 395 x 395 = 39 x 40 / 5 x 5 = 1560 / 25 = 156025.
You can rem the formula as 125
2
=(12
2
+12)25=(144+12)25=15625. Both
are same things but it can use in bigger case easily.
Rule 4: Also we can extend the formula where the sum of last two digits
being 5 as
Ex 1. 82 x 83=(8
2
+8/2) /3 x 2=(64+4) / 06=6806 i.e.( n
2
+n/2) n is even but note that
the R.H.S term should be of two digits as 06 .( in this ex.)

Ex 2.181 x 184=(18
2
+9) / 04=(324+9) / 04=33304.
But we know what are u thinking of? Yes, if the number be odd then what
happen?
And ur ans is as follows:
Ex 3: 91 x 94= (9
2
+9/2) / 04=(81+4) / 54. Surprised? We just pass ½ as 5 in the next
row since ½ belongs to hundred’s place so 1/2 x 100=50. Just pass 5 in such
cases.
See the fig.
Ex 4 . 51 x 54= (25+2) / 54 =2754 Ex 5. 171 x 174= (289+8) /
54=29754
Rule no 5 . Now we are discussing the general method of finding squares of
any
number. See the chart first.
See 5
2
=25
And 10
2
= 100 , so on.
Now u can understand the tricks.
base range Trick alternative
50 26-74 25(±)( read as 25 plus
minus)
100 76-126 100(±)2(±) Given no(±)
150 126-174 225(±)3(±)
200 176-224 400(±)4(±)
250 226-274 625(±)5(±)
300 276-224 900(±)6(±)
350 326-374 1225(±)7(±)
400 376-424 1600(±)8(±)
450 426-474 2025(±)9(±)
500 476-524 2500(±)10(±)
……… ………….. ……………………….
1000 976-1024 100,00(±)20(±)
1500 1476-1524 225,00,00(±)30(±)
2000 1976-2024 4,000,000(±)40(±)
For base 50: Now we are explaining one by one through some examples.
Ex 1. 43
2

step 1. Find base –no as 50-43=07
Step 2. Sq the no as 07
2
=49.write it on the right side.
Step 3. Subtract the no from 25 as 25-7=18.write it on extreme left.
Step 4. Now this is Ur ans. 1849
Ex 2. 57
2
step 1. 57-50=7
Step2. 7
2
=49
Step3. 25+7=32
step4. Ans is 3249.
Ex 3. 69
2
step1. 69-60=19 , 19
2
=361
Step2. carry over 3 and write 61 in right side.
Step3 . 25+19=44, now add 44 with 3 (carry number) and write
it with extreme left.
i.e. (25+19+3)
Step 4. ur ans is 4761
Ex4 . 38 2 step1. 50-38=12 , 12
2
=144 ,carry 1 and write 44 at right side .
Step2. now (25-12)+1=14 i.e. Carry number always adds (it
never subtract). So ur ans is 1444.
So, 1. At first see the given no is it more or less than Ur base. If it is more
then addition rule and if it be less, the n subtraction rule will be
applied. That’s why (±) sign is given
2. And always add the carried no.
3. range is chosen making (±) 24 from base .
4. Trick is done by squaring the base excluding a 0 (zero) such for base 50 :
5 2 =25.for base 350 , 35 2 = 1225
5. for(±) divide the base without zero by 5 as for base 100: 10/5=2,so trick
is
100 (±)2(±) and for base 350: 35/5=7 ,and the trick is 1225(±)7(±) and
so on.

Now we are giving more examples to clear the fact, a regular practice can make
u
master of the art.ok, see
The following examples.
For base 100
Ex 5. 89
2
Step1. 100-89=11 ,11
2
=121 ,carry 1
Step2. 100-(2 x 11 )+1=79, so the ans is 7921
Also u can use the alternative formula here,”given no (±)” such as
Step 1. 100-89=11,sq is 121,carry 1
Step2. 89-11+1=79 ,ass is 7921
(I think this process is better for numbers whose base is near 100)
Ex 6. 117²
Step1. 117-17=17, 17
2
=289 ,carry 2
Step2. 117+17=134, 134+2=136 ,ans is 13689
For base 150
Ex 6. 139²
1. 150-139=11 , sq is 121 ,carry 1
2. 225-3×11=225-33=192, 192+1=193 ,ans is 19, 321
Ex 7. 164 ² 1. 164-150=14, 14 ² =196, carry 1
2. 225+3(14)=225+42=267 ,267+1=268 ,ans is 26,896
Ex 8. 512 ² 1. 12 ² =144 ,carry 1, 2500+10(12)+1=2621, ans is 262144
Try yourself to various numbers to learn the technique quickly. now we learn
to
find the square root .before starting note that…
Rule no 6:
unit digit of perfect
square
1 4569
unit digit of square root 1,9
(1+9)=10

2,8
2+8=10

5
5+5=10

4,6
4+6=10

3,7
3+7=10

Notes : 1 . any no end with 2,3,7,8 can’t be a perfect square.
2. any no ends with odd no zero’s can’t be a perfect square.
Now check the given boxes…
range sq range sq range sq range
1²=1 1-3 9²=81 81-99 17²=289 289-323 25²=625 625-675
2²=4 4-8 10²=100 100-120 18²=324 324-360 26²=676 676-728
3²=9 9-15 11²=121 121-143 19²=361 361-399 27²=729 729-783
4²=16 16-24 12²=144 144-168 20²=400 400-440 28²=784 784-840
5²=25 25-35 13²=169 169-195 21²=441 441-483 29²=841 841-899
6²=36 36-48 14²=196 196-224 22²=484 484-528 30²=900 900-960
7²=49 49-63 15²=225 225-255 23²=529 529-575 31²=961 961-
1023
8²=64 64-80 16²=256 256-288 24²=576 576-624 32²=1024 1024-
1088
Rule: 1. separate the two numbers of extreme right.
2. find the range of the square of the rest, write it on left side.
3. Multiply the range with its preceding number and see that the separated
number is more or less than the multiplied number.
4. if more, then choose the large no of unit digit as the R.H.S.
Ex1. √ (2601)
1. 26 / 01
2. 26 falls on the range of 5 ,write 5 in the left.
3. now 5×6=30 ,26 is smaller than 30.
4. so choose 1 as the right side no. ans is 51
Ex2. √ (6241)
1. 62 / 41
2. 62 falls in range of 7 ,write 7 in left.
3. 7×8=56 i.e.,62 is more than 56,so we choose 9 as the right
side
Ex3. √ 2704
1. 27 / 04
2. 27 falls in range of 5, so 5×6=30 ,but 27 is less than 30, so
We choose 2 between 2 & 8 (allocated for 4)
3.so the ans is 52
But it is easy if the number ends with 5 such as
Ex 4. √ (99225)
1. 992 /25
2. Here u can insert 5 as right side because there is only 5 allocated
for 5 and no choice need here.

3. clearly 992 falls in the range of 31 and the ans is 315
Exercise 1. √ 34225 2. √ 105625 3. √ (0.00126025) 4. √ 2209
5. √ ___________________2916 6. √ 2116 7. √ 15129 8. √ 16129
9. √ 55696 10. √ 66564 11. √ 8.8804 12. √ 0.00101124
13. √ 0.125316
Rule no 7: let us know how to find cube root quickly. See the box first.
unit digit of
perfect cube
1 2 3 4 5 6 7 8 9
unit digit of
cube root
1 8
2+8=10
7
3+7=10
4 5 6 3
7+3=10
2
8+2=10
9
n.b. 1, 4 , 5, 6, 9 has no change as they were in the sq formula and 2,3,7,8
has there
compliment with 10, i.e., 2+8=10, 3+7=10 etc.
Ex1. (46656)
1/3
1. 46 / 656
2. write 6 for 6 on right side
3. see 46 falls in the range of 3 3
4.so the ans is 36
cube range cube range cube range
1 =1 1-7 6 =216 216-342 11 =1331 1331-1727
2 =8 8-26 7 =343 343-511 12 =1728 1728-2196
3 =27 27-63 8 =512 512-728 13 =2197 2197-2743
4 =64 64-124 9 =729 729-999 14 =2744 2744-3374
5 =125 125-215 10 =1000 1000-1330 15 =3375 3375-4095
21=9261
25=15625
Ex2 . cube of 0.0020048383
1. 0.0020048 / 383
2. write 7 for 3 , and see 2048 falls in the range of 12

3. so the ns be 0 .123
n.b. (se 3 places to shift 1 decemal place. )
Rule no 8: Let us know to find the cube of any two digits number.
Ex 1. (18) 3
Step1. Find the ratio of the numbers such as 1:8 in this example.
Step 2. Write cube of first digit and then write three successive terms in
horizontal line which are multiplied by the ratio. in this ex as
Step 3. Make double of second and third terms and write them just below
their own positions
Step 4. Add successive terms (carry over if more than one digit) and u will
get the required result .Lets see the methods …
1. 18 , that is 1:8
4 24 51
2. 1 3 =1 / 1x8=8 / 8 x 8=64 / 64 x 8 =512
3. 8 x 2=16 64 x 2=128
4. ------------------------------------------------------------
( 4 +1) ( 24 +8+16) ( 51 +64+128) 51 2
5 = 4 8 = 24 3
5. So the ans is 5832.
Ex 2. (33)
3
Rakes Prasad , 2008
It contains some vedic methds for multiplication of two,three
and
more digits,finding h.c.f. ofr two polynomial equations.
Edited by
Rakes Prasad, 2008
Rule no (1) If R.H.S. contains less number of digits than the number of
zeros in the base, the remaining digits are filled up by giving zero or zeroes
on the
left side of the R.
Note: If the number of digits are more than the number of zeroes in he
base, the excess digit or digits are to be added to L.H.S of the
answer.
Case 1: Let N1 and N2 be two numbers near to a given base in
powers of 10, and D1 and D2 are their respective
deviations(difference)
from the base. Then N1 X N2 can be represented as
Ex. 1: Find 97 X 94. Here base is 100.

.
Ex. 2: 98 X 97 Base is 100.
ans: 95/06=9506
Ex. 3: 75X95. Base is 100. .
Ex. 4: 986 X 989. Base is 1000
Ex. 5: 994X988. Base is 1000 .
Ex. 6: 750X995
Case 2:
Ex. 7: 13X12. Base is 10
.Ex. 8: 18X14. Base is 10
Ex. 9: 104X102. Base is 100.
104 04
102 02
¯¯¯¯¯¯¯¯¯¯¯¯
Ans is 106 / 4x2 = 10608 ( rule - f ) .
Ex. 10: 1275X1004. Base is 1000.
1275 275
1004 004
...............................................
1279 /275x4 =1279 / 1 / 100 (rule f) =1280100
Case ( iii ): One number is more and the other is less than the base.
Ex.11: 13X7. Base is 10
Ex. 12: 108 X 94. Base is 100.
Ex. 13: 998 X 1025. Base is 1000.
Find the following products by the formula.
1) 7 X 4 2) 93 X 85 3) 875 X 994
4) 1234 X 1002 5) 1003 X 997 6) 11112 X 9998
7) 1234 X 1002 8) 118 X 105
Rules no 2,3,4 of chapter no 1 can also be extended as
Eg. 1: consider 292 x 208. Here 92 + 08 = 100, L.H.S portion
is same i.e. 2
292 x 208 = ( 2 x 3 ) / 92 x 8 / =736 ( for 100 raise the
L.H.S. product by 0 ) = 60736.
Eg. 2: 848 X 852
Here 48 + 52 = 100, L.H.S portion is 8 and its next number is 9.
848 x 852 = 8 x 9 / 48 x 52
720 = 2496
= 722496.
[Since L.H.S product is to be multiplied by 10 and 2 to be carried
over as the base is 100].
Eg. 3: 693 x 607
693 x 607 = 6 x 7 / 93 x 7 = 420 / 651 = 420651.
Find the following products .

1. 318 x 312 2. 425 x 475 3. 796 x 744
4. 902 x 998 5. 397 x 393 6. 551 x 549
(2) 'One less than the previous'
1) The use of this sutra in case of multiplication by 9,99,999.. is as
follows
a) The left hand side digit (digits) is ( are) obtained by deduction 1
from
the left side digit (digits) .
b) The right hand side digit is the complement or difference between
the
multiplier and the left hand side digit (digits)
c) The two numbers give the answer
Example 1: 8 x 9
Step ( a ) gives 8 – 1 = 7 ( L.H.S. Digit )
Step ( b ) gives 9 – 7 = 2 ( R.H.S. Digit )
Step ( c ) gives the answer 72
Example 2: 15 x 99
Step ( a ) : 15 – 1 = 14 Step ( b ) : 99 – 14 = 85 ( or 100 – 15 )
Step ( c ) : 15 x 99 = 1485
Example 3: 24 x 99 Answer :
Example 5: 878 x 9999 Answer :
find out the products
64 x 99 723 x 999 3251 x 9999
43 x 999 256 x 9999 1857 x 99999
(3) Multiplication of two 2 digit numbers.
Ex.1: Find the product 14 X 12
The symbols are operated from right to left .
Step i) : Step ii) :
Step iii) :
Ex.4: 32 X 24
Step (i) : 2 X 4 = 8
Step (ii) : 3 X 4 = 12; 2 X 2 = 4; 12 + 4 = 16.
Here 6 is to be retained. 1 is to be carried out to left side.
Step (iii) : 3 X 2 = 6. Now the carried over digit 1 of 16 is to be
added. i.e., 6 + 1 = 7. Thus 32 X 24 = 768
Note that the carried over digit from the result (3X4) + (2X2)
= 12+4 = 16 i.e., 1 is placed under the previous digit 3 X 2 = 6 and
added.
After sufficient practice, you feel no necessity of writing in this way
and
simply operate or perform mentally.
(4) Consider the multiplication of two 3 digit numbers.

Ex 1. 124 X 132 =16368.
Proceeding from right to left
i) 4 X 2 = 8. First digit = 8
ii) (2 X 2) + (3 X 4) = 4 + 12 = 16. The digit 6 is retained and 1 is carried
over to left side. Second digit = 6.
iii) (1 X 2) + (2 X 3) + (1 X 4) = 2 + 6 + 4 =12. The carried over 1 of
above
step is added i.e., 12 + 1 = 13. Now 3 is retained and 1 is carried over
to left side. Thus third digit = 3.
iv) ( 1X 3 ) + ( 2 X 1 ) = 3 + 2 = 5. the carried over 1 of above step is
added i.e., 5 + 1 = 6 . It is retained. Thus fourth digit = 6
v) ( 1 X 1 ) = 1. As there is no carried over number from the previous
step it is retained. Thus fifth digit = 1
(5) Cubing of Numbers:
Example : Find the cube of the number 106. We proceed as follows:
i) For 106, Base is 100. The surplus is 6. Here we add double of the
surplus i.e. 106+12 = 118. (Recall in squaring, we directly add
the surplus) This makes the left-hand -most part of the answer.
i.e. answer proceeds like 118 / - - - - -
ii) Put down the new surplus i.e. 118-100=18 multiplied
by the initial surplus i.e. 6=108.
Since base is 100, we write 108 in carried over form 108 i.e. .
As this is middle portion of the answer, the answer proceeds
like 118 / 108 /....
iii) Write down the cube of initial surplus i.e. 63 = 216 as the last
portion
i.e. right hand side last portion of the answer. Since base is 100,
write 216 as 216 as 2 is to be carried over. Answer is 118 / 108 / 216
Now proceeding from right to left and adjusting the carried over,
we get the answer 119 / 10 / 16 = 1191016.
Eg.(1): 1023 = (102 + 4) / 6 X 2 / 23 = 106/ 12 / 08 = 1061208.
Observe initial surplus = 2, next surplus =6 and base = 100.
Eg.(2): 943
Observe that the nearest base = 100.
i) Deficit = -6. Twice of it -6 X 2 = -12 add it to the number = 94 -12
=82.
ii) New deficit is -18. Product of new deficit x initial deficit = -18 x -6 =
108
iii) deficit3 = (-6)3 = -216.
__ Hence the answer is 82 / 108 / -216
Since 100 is base 1 and -2 are the carried over. Adjusting the carried
over

in order, we get the answer ( 82 + 1 ) / ( 08 – 03 ) / ( 100 – 16 )
= 83 / 05 / 84 = 830584 . 16 becomes 84 after taking1 from middle
most portion i.e. 100. (100-16=84). _ Now 08 - 01 = 07 remains in the
middle portion, and 2 or 2 carried to it makes the middle
as 07 - 02 = 05. Thus we get the above result.
Eg.(3): 9983 Base = 1000; initial deficit = - 2.
9983 = (998 – 2 x 2) / (- 6 x – 2) / (- 2)3 = 994 / 012 / -008
= 994 / 011 / 1000 - 008 = 994 / 011 / 992 = 994011992.
Find the cubes of the following numbers using yavadunam sutra.
1. 105 2. 114 3. 1003 4. 10007 5. 92
6. 96 7. 993 8. 9991 9. 1000008 10. 999992.
(6) Highest common factor:
Example 1: Find the H.C.F. of x2 + 5x + 4 and x2 + 7x + 6.
1. Factorization method:
x2 + 5x + 4 = (x + 4) (x + 1)
x2 + 7x + 6 = (x + 6) (x + 1)
H.C.F. is ( x + 1 ).
2. Continuous division process.
x2 + 5x + 4 ) x2 + 7x + 6 ( 1
x2 + 5x + 4
___________
2x + 2 ) x2 + 5x + 4 ( ½x
x2 + x
__________
4x + 4 ) 2x + 2 ( ½
2x + 2
______
0
Thus 4x + 4 i.e., ( x + 1 ) is H.C.F.
OUR PROCESS
i.e.,, (x + 1) is H.C.F
Example 2: Find H.C.F. of 2x2 – x – 3 and 2x2 + x – 6
Example 3: x3 – 7x – 6 and x3 + 8x2 + 17x + 10.
Example 4: x3 + 6x2 + 5x – 12 and x3 + 8x2 + 19x + 12.
(or)
Example 5: 2x3 + x2 – 9 and x4 + 2x2 + 9
Add: (2x3 + x2 – 9) + (x4 + 2x2 + 9) = x4 + 2x3 + 3x2.
÷ x2 gives x2 + 2x + 3 ------ (i)
Subtract after multiplying the first by x and the second by 2.
Thus (2x4 + x3 – 9x) - (2x4 + 4x2 + 18) = x3 - 4x2 – 9x – 18 ------ ( ii )
Multiply (i) by x and subtract from (ii)
x3 – 4x2 – 9x – 18 – (x3 + 2x2 + 3x) = - 6x2 – 12x – 18

÷ - 6 gives x2 + 2x + 3.
Thus ( x2 + 2x + 3 ) is the H.C.F. of the given expressions.
Find the H.C.F. in each of the following cases using Vedic sutras:
1 x2 + 2x – 8, x2 – 6x + 8
2 x3 – 3x2 – 4x + 12, x3 – 7x2 + 16x - 12
3 x3 + 6x2 + 11x + 6, x3 – x2 - 10x - 8
4 6x4 – 11x3 + 16x2 – 22x + 8, 6x4 – 11x3 – 8x2 + 22x – 8.
Edited by
Rakes Prasad , 2008
Contains: (1) Use the formula ALL FROM 9 AND THE LAST
FROM 10 to perform instant subtractions.
(2) Multiplying numbers just over 100.
(3) The easy way to add and subtract fractions.
(4) Multiplying a number by 11.
(5) Method for diving
(6) SOME BASIC RELATIONS AND REVIEW OF SQUARE
AND CUBE FORMULA:
(7) SQUARE ROOTS AND MULTIPLICATION FORMULA :
(8) TWO INTERESTING FACTS:
(9) ROMAN NUMERALS
(10) NUMER REPRESENTATION:
(11) DECIMAL REPRENTATION AND PREFIXS:
(12) POWER OF INTIGERS:
(13) SOME PROOFS BUT WITHOUT WORDS:
(14) THE LARGEST PRIME NUMBERS DISCOVERED SO
FAR:
Edited by
Rakes Prasad, 2008
(1) Use the formula ALL FROM 9 AND THE LAST FROM 10 to
perform
instant subtractions.
● For example 1000 - 357 = 643 We simply take each figure in 357 from 9 and the
last figure
from 10.
So the answer is 1000 - 357 = 643
And thats all there is to it!
This always works for subtractions from numbers consisting of a 1 followed by
noughts: 100;
1000; 10,000 etc.

Similarly 10,000 - 1049 = 8951
For 1000 - 83, in which we have more zeros than figures in the numbers being
subtracted, we
simply suppose 83 is 083. So 1000 - 83 becomes 1000 - 083 = 917
Try some yourself:
1) 1000 - 777 2) 1000 - 283 3) 1000 - 505 4) 10,000 - 2345 5) 10000 - 9876 6)
10,000 -
1101
(2) Multiplying numbers just over 100.
● 103 x 104 = 10712
The answer is in two parts: 107 and 12, 107 is just 103 + 4 (or 104 + 3), and 12 is
just 3 x 4.
● Similarly 107 x 106 = 11342 107 + 6 = 113 and 7 x 6 = 42
Again, just for mental arithmetic Try a few:
1) 102 x 107 = 2) 106 x 103 = 1) 104 x 104 = 4) 109 x 108 =
(3) The easy way to add and subtract fractions.
Use VERTICALLY AND CROSSWISE to write the answer straight down!
Multiply crosswise and add to get the top of the answer: 2 x 5 = 10 and 1 x 3 = 3.
Then 10 + 3 =
13. The bottom of the fraction is just 3 x 5 = 15.
.
You multiply the bottom number together.
Subtracting is just as easy: multiply crosswise as before, but the subtract:
Try a few:
(4) Multiplying a number by 11.
To multiply any 2-figure number by 11 we just put the total of
the two figures between the 2 figures.
● 26 x 11 = 286
Notice that the outer
figures in 286 are the
26 being multiplied.
And the middle figure is just 2 and 6 added up.
● So 72 x 11 = 792
Multiply by 11:
1) 43 = ) 81 = 3) 15 = 4) 44 = 5) 11 =
● 77 x 11 = 847
This involves a carry figure because 7 + 7 = 14 we get 77 x 11 = 7147 = 847.
Multiply by 11:
1) 88 = 2) 84 = 3) 48 = 4) 73 = 5) 56 =
● 234 x 11 = 2574

We put the 2 and the 4 at the ends. We add the first pair 2 + 3 = 5. and we add the
last pair: 3 +
4 = 7.
Multiply by 11:
1) 151 = 2) 527 = 3) 333 = 4) 714 = 5) 909 =
(5) Method for diving by 9.
23 / 9 = 2 remainder 5
The first figure of 23 is 2, and this is the answer. The remainder is just 2 and 3
added up!
● 43 / 9 = 4 remainder 7
The first figure 4 is the answer and 4 + 3 = 7 is the remainder - could it be easier?
Divide by 9:
1) 61 = remainder 2) 33 = remainder 3) 44 = remainder 4) 53 = remainder
● 134 / 9 = 14 remainder 8
The answer consists of 1,4 and 8.1 is just the first figure of 134.4 is the total of the
first two
figures 1+ 3 = 4,and 8 is the total of all three figures 1+ 3 + 4 = 8.
Divide by 9:
6) 232 = 7) 151 = 8) 303 = 9) 212 = remainder remainder remainder remainder
● 842 / 9 = 812 remainder 14 = 92 remainder 14
Actually a remainder of 9 or more is not usually permitted because we are trying to
find how
many 9's there are in 842.
Since the remainder, 14 has one more 9 with 5 left over the final answer will be 93
remainder
5
Divide these by 9:
1) 771 2) 942 3) 565 4) 555 5) 777 6) 2382
(6) SOME BASIC RELATIONS AND REVIEW OF SQUARE AND CUBE
FORMULA:
Relations :
If a = b , b = c then a= c
If a > b , b > c then a> c
If a < b , b < c then a < c
If a b< b c then a < c
If ab > bc then b > c
If a > b , c> d then a + c> b+ d
If a > b , c< d then a -c> b –d
If a < b , c< d then a -c< b -d
If a > b & a ,b both positive then 1/a < 1/b
x = n then x = n or -n
2 2

(a + b) = a + 2ab + b
2 2 2

(a – b) = a –2ab + b 2 2 2

(a + b) = a + 3a b + 3ab + b 3 3 2 2 3

(a - b) = a -3a b + 3ab -b 3 3 2 2 3

(a + b+ c) = a + b + c + 2ab + 2bc+ 2ac 2 2 2 2

Factorization
a –b = ( a + b )(a –b)
2 2

a + b = (a + b) ( a -ab + b )
3 3 2 2

a -b = (a -b ) (a + ab+ b )
3 3 2 2

Identities
(a+ b) + (a –b) = 2(a + b ) 2 2 2 2

(a+ b) -(a–b) = 4ab 2 2

In dices
a
m
xa
n
=a
(m+n)
a
m
/a
n
=a
(m-n)
a=1 0

a =1/a
-m

m
a = m√a
1/m

(a *b ) = amx b m m

(a /b ) = a m

m
/b
m
(a ) = a m n m*n

Logarithms
a = n then log a n= x
x

log a (mn )= log am + log an
log a (m/n) = log a m -log an
log a (m) = n log am n

log b n= log an/log ab
Surd
( √a ) = a
n n

n√ a * √ b = √ab n n

n√ a / √ b = √a /b n n

m√ √ a = √a = √ √ a
n mn n m

m√ √ a = √a = √a
p p mp p m

Angle Measurement
Total of Interior Angles in Degrees
Triangle 180
Rectangle 360
Square 360
Pentagon 540
Circle 360
Some Facts
In a triangle, interior opposite angle is always less than exterior angle.

Sum of 2 interior opposite angles of a triangle is always equal to exterior angle.
Triangle can have at one most obtuse angle.
Angle made by altitude of a triangle with side on which it is drawn is equal to 90 degrees.
in parallelogram opposite angles are equal.
Squares
Number Square Number Square Number Square
1 1 11 121 21 441
2 4 12 144 22 484
3 9 13 169 23 529
4 16 14 196 24 576
5 25 15 225 25 625
6 36 16 256 26 676
7 49 17 289 27 729
8 64 18 324 28 784
9 81 19 361 29 841
10 100 20 400 30 900
Cubes & Other Powers
Number ( X) X3 X4 X5
1111
2 8 16 32
3 27 81 243
4 64 256 1024
5 125 625 3125
6 216 1296 7776
7 343 2401 16807
8 512 4096 32768
9 729 6561 59049
10 1000 10000 100000
(7) SQUARE ROOTS AND MULTIPLICATION FORMULA :
No Roots No Roots No Roots No Roots
1 1 5 2.23 9 3 36 6
2 1.41 6 2.44 10 3.16 49 7
3 1.73 7 2.64 16 4 64 8
4 2 8 2.82 25 5 81 9
1 to 10 :
1 2 3 4 5 6 7 8 9 10
1 1 2 3 4 5 6 7 8 9 10
2 2 4 6 8 10 12 14 16 18 20
3 3 6 9 12 15 18 21 24 27 30
4 4 8 12 16 20 24 28 32 36 40
5 5 10 15 20 25 30 35 40 45 50
6 6 12 18 24 30 36 42 48 54 60
7 7 14 21 28 35 42 49 56 63 70
8 8 16 24 32 40 48 56 64 72 80
9 9 18 27 36 45 54 63 72 81 90
10 10 20 30 40 50 60 70 80 90 100
11to 20

11 12 13 14 15 16 17 18 19 20
1 11 12 13 14 15 16 17 18 19 20
2 22 24 26 28 30 32 34 36 38 40
3 33 36 39 42 45 48 51 54 57 60
4 44 48 52 56 60 64 68 72 76 80
5 55 60 65 70 75 80 85 90 95 100
6 66 72 78 84 90 96 102 108 114 120
7 77 84 91 98 105 112 119 126 133 140
8 88 96 104 112 120 128 136 144 152 160
9 99 108 117 126 135 144 153 162 171 180
10 110 120 130 140 150 160 170 180 190 200
(8) TWO INTERESTING FACTS:
(9) ROMAN NUMERALS
(10) NUMER REPRESENTATION:
(11) DECIMAL REPRENTATION AND PREFIXS:
(12) POWER OF INTIGERS:
(13) SOME PROOFS BUT WITHOUT WORDS:

(14) THE LARGEST PRIME NUMBERS DISCOVERED SO FAR:
THE END
Edited by
Rakes Prasad ,2008
Contains: 1. Some different techniques of squaring
ending with 1,2,3,4,6,7,8,9,etc .
2. adding different number sequence
3. finding the difference of squares.
4. dividing a 6-digit/3-digit number by
13,7,37037,11,41,15873 etc
5. dividing numbers by 75,125,625,12 2/3 etc..
6. know why divisibility rules works?
7. finding percentage quickly.
8. various types of multiplication by
99,72,84,………
9. squaring of some special
numbers
Rakes Prasad
2008.
Squaring a 2-digit number ending in 1

.Take a 2-digit number ending in 1. Subtract 1 from the
number. Square
the difference. Add the difference twice to its square. Add 1.
Example:
If the number is 41, subtract 1: 41 - 1 = 40. 40 x 40 = 1600
(square the
difference). 1600 + 40 + 40 = 1680 (add the difference twice to
its square).
1680 + 1 = 1681 (add 1). So 41 x 41 = 1681.
See the pattern?
For 71 x 71, subtract 1: 71 - 1 = 70.
70 x 70 = 4900 (square the difference).
4900 + 70 + 70 = 5040 (add the difference twice to its
square). . So 71 x 71
= 5041.
I Take a 2-digit number ending in 2. f The last digit will be _ _
_ 4. t
Multiply the first digit by 4: the 2nd number will be h the next
to the last
digit: _ _ X 4. e n Square the first digit and add the number
carried from u
the previous step: X X _ _. m b e Example: r i
s 52, the last digit is _ _ _ 4. 4 x 5 = 20 (four times the first
digit): _ _ 0 4. 5
x 5 = 25 (square the first digit), 25 + 2 = 27 (add carry): 2 7 0 4.
For 82 x
82, the last digit is _ _ _ 4. 4 x 8 = 32 (four times the first
digit): _ _ 2 4. 8 x
8 = 64 (square the first digit), 64 + 3 = 67 (add carry): 6 7 2 4.
Squaring a 2-digit number ending in 3 Take a 2-digit
number
ending in 3. 2 The last digit will be _ _ _ 9. 3 Multiply the first
digit by 6:

the 2nd number will be the next to the last digit: _ _ X 9. 4
Square the first
digit and add the number carried from the previous step: X X
_ _.
Example:
1 If the number is 43, the last digit is _ _ _ 9. 2 6 x 4 = 24 (six
times the
first digit): _ _ 4 9. 3 4 x 4 = 16 (square the first digit), 16 + 2 =
18 (add
carry): 1 8 4 9.
4 So 43 x 43 = 1849.
See the pattern?
1 For 83 x 83, the last digit is _ _ _ 9. 2 6 x 8 = 48 (six times
the first digit):
_ _ 8 9.
38 x 8 = 6 4 ( s q u a r e t he first digit), 64 + 4 = 68 (add carry):
6 8 8 9. 4
So 83 x 83 = 6889.
1 Take a 2-digit number ending in 4. 2 Square the 4; the last
digit is 6: _ _
_ 6 (keep carry, 1.)
3 Multiply the first digit by 8 and add the carry (1); the 2nd
number will be
the next to the last digit: _ _ X 6 (keep carry). 4 Square the
first digit and
add the carry: X X _ _.
Example:
1 If the number is 34, 4 x 4 = 16 (keep carry, 1); the last digit
is _ _ _ 6. 2 8
x 3 = 24 (multiply the first digit by 8), 24 + 1 = 25 (add the
carry): the next
digit is 5: _ _ 5 6. (Keep carry, 2.) 3 Square the first digit and
add the

carry, 2: 1 1 5 6. 4 So 34 x 34 = 1156.
See the pattern?
1 For 84 x 84, 4 x 4 = 16 (keep carry, 1);
the last digit is _ _ _ 6.
2 8 x 8 = 64 (multiply the first digit by 8),
64 + 1 = 65 (add the carry):
the next digit is 5: _ _ 5 6. (Keep carry, 6.)
Square the first digit and add the carry, 6: 7 0 5 6.
So 84 x 84 = 7056.
See previous chapters…
1 Choose a 2-digit number ending in 6. 2 Square the second
digit (keep
the carry): the last digit of the answer is always 6: _ _ _ 6
3 Multiply the first digit by 2 and add the carry (keep the
carry): _ _ X _
4 Multiply the first digit by the next consecutive number and
add the
carry: the product is the first two digits: XX _ _.
Example:
1 If the number is 46, square the second digit : 6 x 6 = 36; the
last digit of
the answer is 6 (keep carry 3): _ _ _ 6 2 Multiply the first digit
(4) by 2 and
add the carry (keep the carry): 2 x 4 = 8, 8 + 3 = 11; the next
digit of the
answer is 1: _ _ 1 6
3 Multiply the first digit (4) by the next number (5) and add
the carry: 4 x
5 = 20, 20 + 1 = 21 (the first two digits): 2 1 _ _
4 So 46 x 46 = 2116.
See the pattern?
1 For 76 x 76, square 6 and keep the carry (3): 6 x 6 = 36; the
last digit of
the answer is 6: _ _ _ 6 2 Multiply the first digit (7) by 2 and
add the carry:

2 x 7 = 14, 14 + 3 = 17; the next digit of the answer is 7 (keep
carry 1): _ _
7 6 3 Multiply the first digit (7) by the next number (8) and add
the carry:
7 x 8 = 56, 56 + 1 = 57 (the first two digits: 5 7 _ _ 4 So 76 x 76
= 5776.
1 Choose a 2-digit number ending in 7. 2 The last digit of the
answer is
always 9: _ _ _ 9 3 Multiply the first digit by 4 and add 4 (keep
the carry):
__X_
add the
Example:
1 If the number is 47: 2 The last digit of the answer is 9: _ _ _
9 3 Multiply
the first digit (4) by 4 and add 4 (keep the carry): 4 x 4 = 16,
16 + 4 = 20;
the next digit of the answer is 0 (keep carry 2): _ _ 0 9
4 Multiply the first digit (4) by the next number (5) and add
the carry (2): 4
x 5 = 20, 20 + 2 = 22 (the first two digits): 2 2 _ _
5 So 47 x 47 = 2209.
See the pattern?
1 For 67 x 67 2 The last digit of the answer is 9: _ _ _ 9 3
Multiply the first
digit (6) by 4 and add 4 (keep the carry): 4 x 6 = 24, 24 + 4 =
28; the next
digit of the answer is 0 (keep carry 2): _ _ 8 9 4 Multiply the
first digit (6)
by the next number (7) and add
5 the carry (2): 6 x 7 = 42, 42 + 2 = 44 (the first two digits): 4 6
So 67 x 67
= 4489.

answer is
always 4: _ _ _ 4 3 Multiply the first digit by 6 and add 6 (keep
the carry):
__X_
add the
Example:
4 3 Multiply
the first digit (7) by 6 and add 6 (keep the carry): 7 x 6 = 42,
42 + 6 = 48;
the next digit of the answer is 8 (keep carry 4): _ _ 8 4 4
Multiply the first
digit (7) by the next number (8) and add the carry (4): 7 x 8 =
56, 56 + 4 =
60 (the first two digits): 6 0 _ _ 5 So 78 x 78 = 6084.
See the pattern?
1 For 38 x 38
The last digit of the answer is 4: _ _ _ 4
Multiply the first digit (3) by 6 and add 6 (keep the
carry): 3 x 6 = 18, 18 + 6 = 24; the next digit of the
answer is 4 (keep carry 2): _ _ 4 4
Multiply the first digit (3) by the next number (4)
and add the carry (2):
3 x 4 = 12, 12 + 2 = 14 (the first two digits): 1 4 _ _
So 38 x 38 = 1444
answer
is always
1. Multiply the first digit by 8 and add 8 (keep the carry): _ _ X
_3

Multiply the first digit by the next consecutive number and
add the
Example:
13
Multiply the first digit (3) by 8 and add 8 (keep the carry): 8 x
3=
24, 24 + 8 = 32; the next digit of the answer is 2 (keep carry
3): _ _
2 1 4 Multiply the first digit (3) by the next number (4) and add
the
carry (3): 3 x 4 = 12, 12 + 3 = 15 (the first two digits): 1 5 _ _
5 So 39 x 39 = 1521. rn?
1 For 79 x 79
2 The last digit of the answer is 1: _ _ _ 1
Multiply the first digit (7) by 8 and add 8 (keep the carry): 8 x
7 = 56,
56 + 8 = 64; the next digit of the answer is 4 (keep carry 6): _
_41
Multiply the first digit (7) by the next number (8) and add the
carry (6)
: 7 x 8 = 56, 56 + 6 = 62 (the first two digits): 6 2 _ _
So 79 x 79 = 6241.
..
Choose two 2-digit numbers less than 20 (no limits for
experts). Add all
the numbers between them:
1 Add the numbers; 2 Subtract the numbers and add 1; 3
Multiply half
the sum by this difference + 1, OR Multiply the sum by half
the difference
+ 1.
Example:

1 If the two numbers selected are 6 and 19: 2 Add the
numbers: 6 + 19 =
25. 3 Subtract the numbers: 19 - 6 = 13. Add 1: 13 + 1 = 14. 4
Multiply 25
by half of 14: 25 x 7 = 175. 5 So the sum of the numbers from
6 through 19
is 175.
Therefore 6+7+8+9+10+11+12+13+14+15+16+17+18+19=175
See the pattern?
1 If the two numbers selected are 4 and 18: 2 Add the
numbers: 4 + 18 =
22. 3 Subtract the numbers: 18 - 4 = 14. Add 1: 14 + 1 = 15.
Multiply half of 22 by 15: 11 x 15 = 165 (10 x 15 + 15). So the
sum of the
numbers from 4 through 18 is 165.
1 Choose a 2-digit odd number. Add all the odd numbers
starting with
one through this 2-digit number: 2 Add one to the 2-digit
number. 3
Divide this sum by 2 (take half of it). 4 Square this number.
This is the
sum of all odd numbers from 1 through the 2-digit number
chosen.
Example:
1 If the 2-digit odd number selected is 35: 2 35+1 = 36 (add 1).
3 36/2 = 18
(divide by 2) or 1/2 x 36 = 18 (multiply by 1/2). 4 18 x 18 = 324
(square 18):
18 x 18 = (20 - 2)(18) = (20 x 18) - (2 x 18) = 360 - 36 = 360 -30 -
6 = 324. 5
So the sum of all the odd numbers from one through 35 is
324.
See the pattern?
1 If the 2-digit odd number selected is 79: 2 79+1 = 80 (add 1).
3 80/2 = 40

(divide by 2) or 1/2 x 80 = 40 (multiply by 1/2). 4 40 x 40 = 1600
(square
40).
5. So the sum of all the odd numbers from one through 79 is
1600.
1 Have a friend choose a a single digit number. (No
restrictions for
experts.) 2 Ask your friend to jot down a series of doubles
(where the
next term is always double the preceding one), and tell you
the last term.
3 Ask your friend to add up all these terms. 4 You will give
the answer
before he or she can finish: The sum of all the terms of this
series will be
two times the last term minus the first term.
Example:
if the number selected is 9: 1 The series jotted down is: 9, 18,
36, 72, 144.
2 Two times the last term (144) minus the first (9): 2 x 144 =
288; 288 - 9 =
279.
3 So the sum of the doubles from 9 through 144 is 279.
See the pattern? Here's one for the experts:
1 The number selected is 32: 2 The series jotted down is: 64,
128, 256,
512. 3 Two times the last term (512) minus the first (64): 2 x
512 = 1024;
1024 - 32 = 1024 - 30 - 2 = 994 - 2 = 992.
4 So the sum of the doubles from 32 through 512 is 992.
Remember to subtract in steps from left to right. With
practice you will be
expert in summing series.
restrictions for experts.) Ask your friend to jot down a series
of

quadruples (where the next term is always four times the
preceding one),
and tell you only the last term. Ask your friend to add up all
these terms.
You will give the answer before he or she can finish: The
sum of all the
terms of this series will be four times the last term minus the
first term,
divided by 3. Example:
If the number selected is 5:
1 The series jotted down is: 5, 20, 80, 320, 1280. 2 Four times
the last term
(1280) minus the first (5): 4000 + 800 + 320 - 5 = 5120 - 5 =
5115 Divide by
3: 5115/3 = 1705
3 So the susum of the quadruples from 5 through 1280 is
1705.
See the pattern? Here's one for the experts:
1 The number selected is 32: 2 The series jotted down is: 32,
128, 512,
2048. 3 Four times the last term (2048) minus the first (32):
8000 + 160 +
32 - 32 = 8,160
Divide by 3: 8160/3 = 2720.
4. So the sum of the quadruples from 32 through 2048 is
2720.
Practice multiplying from left to right and dividing by 3. With
practice you
will be an
expert quad adder.
Add a sequence from one to a selected 2-digit number
1 Choose a 2-digit number. 2 Multiply the 2-digit number by
half the next
number, orMultiply half the 2-digit number by the next
number.

Example:
1 If the 2-digit even number selected is 51: 2 The next
number is 52.
Multiply 51 times half of 52. 3 51 x 26: (50 x 20) + (50 x 6) + 1 x
26) = 1000
+ 300 + 26 = 1326
4 So the sum of all numbers from 1 through 51 is 1326.
See the pattern?
1 If the 2-digit even number selected is 34: 2 The next
number is 35.
Multiply half of 34 x 35. 3 17 x 35: (10 x 35) + (7 x 30) + (7 x 5)
= 350 + 210
+ 35 = 560 + 35 = 595
4 So the sum of all numbers from 1 through 34 is 595.
With some multiplication practice you will be able to find
these sums of
sequential numbers easily and faster than someone using a
calculator!
1 Choose a 1-digit number. 2 Square it.
Example:
1 If the 1-digit number selected is 7: 2 To add 1 + 2 + 3 + 4 + 5
+6+7+6+
5 + 4 + 3 + 2 + 1 3 Square 7: 49 4 So the sum of all numbers
from 1
through 7 and back is 49.
See the pattern?
1 If the 1-digit number selected is 9: 2 To add 1 + 2 + 3 + 4 + 5
+6+7+8+
9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 3 Square 9: 81 4 So the sum of
all numbers
from 1 through 9 and back is 81.
Add sequences of numbers in the 10's
1 Choose a 2-digit number in the 10's. To add all the 10's
from 10 up

through this number and down from it: 2 Square the 2nd digit
of the
number (keep the carry) _ _ X 3 The number of terms is 2 x
the 2nd digit +
1. 4 The first digits = the number of terms (+ carry). nbsp; X X
_
Example:
1 If the 2-digit number in the 10's selected is 16: (10 + 11 +
12 ... 16 + 15 +
14 ... 10) 2 Square the 2nd digit of the number: 6 x 6 = 36
(keep carry 3) _ _
6 3 No. of terms = 2 x 2nd digit + 1: 2 x 6 + 1 = 13
4 No. of terms (+ carry): 13 + 3 = 16 1 6 _ 5 So the sum of the
sequence is
166.
See the pattern?
1. If the 2-digit number in the 10's selected is 18:
(10 + 11 + 12 + ... 18 + 17 + 16 + 15 ... 10)
1 Square the 2nd digit of the number: 8 x 8 = 64 (keep carry
6) _ _ 4 2 No.
of terms = 2 x 2nd digit + 1: 2 x 8 + 1 = 17 3 No. of terms (+
carry): 17 + 6 =
23 2 3 _ 4 So the sum of the sequence is 234.
from 20 up
of the
the 2nd digit +
1. 4 Multiply the number of terms by 2 (+ carry). X X _
Example:
1 If the 2-digit number in the 20's selected is 23: (20 + 21 + 22
+ 23 + 22 +
21 + 20) 2 Square the 2nd digit of the number: 3 x 3 = 9 _ _ 9 3
No. of

terms = 2 x 2nd digit + 1: 2 x 3 + 1 = 7 4 2 x no. of terms: 2 x 7
= 14 1 4 _ 5
So the sum of the sequence is 149.
See the pattern?
22 ... + 28 + 27
+ ... 22 + 21 + 20) 2 Square the 2nd digit of the number: 8 x 8
= 64 (keep
carry 6) _ _ 4 3 No. of terms = 2 x 2nd digit + 1: 2 x 8 + 1 = 17
4 2 x no. of
terms (+ carry): 2 x 17 + 6 = 40 4 0 _ 5 So the sum of the
sequence is 404.
2 S 1 Choose a 2-digit number in the 30's. To add all the 30's
qfrom 30 up
through this number and down from it: u
are the 2nd digit of the number (keep the carry) _ _ X 3 The
number of
terms is 2 x the 2nd digit + 1. 4 Multiply the number of terms
by 3 (+
carry) X X _
Example:
+ 33 + 34 +
33 + 32 + 31 + 30) 2 Square the 2nd digit of the number: 4 x 4
= 16 (keep
4 3 x no. of terms: 3 x 9 + 1 = 28 2 8 _ 5 So the sum of the
sequence is
286.
See the pattern?
32 + ... + 38 +
37 + ... 32 + 31 + 30) 3 Square the 2nd digit of the number: 8 x
8 = 64 (keep
carry 6) _ _ 4

No. of terms = 2 x 2nd digit + 1: 2 x 8 + 1 = 17
3 x no. of terms: 3 x 17 + 6 = 51 + 6 = 57 5 7 _ So the sum of
the sequence
is 574.
from 40 up through this number and down from it:
2 Square the 2nd digit of the number
(keep the carry) _ _
X
3 The number of terms is 2 x the 2nd digit + 1.
4 Multiply the number of terms by 4 (+ carry) X X _
e 2-digit number in the 40's selected is 43: (40 + 41 + 42 + 43
+ 42 + 41 +
40) 1 Square the 2nd digit of the number: 3 x 3 = 9 _ _ 9 2 No.
of terms = 2
x 2nd digit + 1: 2 x 3 + 1 = 7
3 4 x no. of terms: 4 x 7 = 28 2 8 _ 4 So the sum of the
sequence is 289.
See the pattern?
+ ... + 48 +
8 = 64 (keep
4 4 x no. of
terms: 4 x 17 + 6 = 40 + 28 + 6 = 68 + 6 = 74 7 4 _
5 So the sum of the sequence is 744.
from 50 up
of the
the 2nd digit +
1. 4 Multiply the number of terms by 5. X X _
Example:

+ 53 + 52 +
No. of
= 35 3 5 _ 5
See the pattern?
2 ... + 57 + 56 + ... 52 + 51 + 50) 2 Square the 2nd digit of the
number: 7 x 7
= 49 (keep carry 4) _ _ 9
3 No. of terms = 2 x 2nd digit + 1: 2 x 7 + 1 = 15
5 x no. of terms (+ carry): 5 x 15 + 4 = 75 + 4 = 79 7 9 _
Add sequences of numbers in the 60's
from 60 up through
this number and down from it: 2 Square the 2nd digit of the
number (keep
the carry) _ _ X 3 The number of terms is 2 x the 2nd digit + 1.
4 Multiply the
number of terms by 6 (+ carry) X X _
Example:
+ 63 + 62 + 61 +
60) 2 Square the 2nd digit of the number: 3 x 3 = 9 _ _ 9 3 No.
of terms = 2 x
2nd digit + 1: 2 x 3 + 1 = 7
4
5 6 x no. of terms: 6 x 7 = 42 4 2 _ So the sum of the
sequence is 429.
See the pattern?
62 + ... + 68 + 67 + ... 62 + 61 + 60) 2 Square the 2nd digit of

the number: 8 x 8 = 64 (keep carry 6) _ _ 4 3 No. of terms =
2 x 2nd digit + 1: 2 x 8 + 1 = 17 4 6 x no. of terms: 6 x 17 +
6 = 42 + 1 = 102 + 6 = 108 1 0 8 _
5oS
the sum of the sequence is 1084.
from 70 up
of the
number (keep the carry) _ _ X
3 The number of terms is 2 x the 2nd digit + 1. 4 Multiply the
number of
terms by 7 (+ carry) X X _
Example:
+ 73 + 72 +
No. of
terms = 2 x 2nd digit + 1: 2 x 3 + 1 = 7
4 7 x no. of terms: 7 x 7 = 49 4 9 _ 5 So the sum of the
sequence is 499.
See the pattern?
+ ... + 78 +
8 = 64 (keep
4 7 x no. of
terms: 7 x 17 + 6 = 49 + 6 = 119 + 6 = 125 1 2 5 _
5 So the sum of the sequence is 1254.
1 Choose a 2-digit number in the 80's. To add all the80's
from 80 up through this number and down from it:
2 Square the 2nd digit of the number
(keep the carry) _ _

umber of terms is 2 x the 2nd digit + 1 4 Multiply the number
of terms by
8 (add the carry) X X _
Example:
+ 83 + 82 +
No. of
= 56 5 6 _ 5
See the pattern?
82 ... + 88 + 87
+ ... 83 + 82 + 81 + 80) 2 Square the 2nd digit of the number: 8
x 8 = 64
(keep carry 6) _ _ 4 3 No. of terms = 2 x 2nd digit + 1: 2 x 8 + 1
= 17
8 x no. of terms (+ carry): 8 x 17 + 6 = 80 + 56 + 6 = 136 + 6 =
42 4 2 _
1 Have a friend choose and write down a single-digit number.
(Two
digits for experts.) 2 Ask your friend to name and note a third
number
by adding the first two. 3 Name a fourth by adding the
second and third.
Continue in this way, announcing each number, through ten
numbers. 4
Ask your friend to add up the ten numbers. You will give the
answer
before he or she can finish:
The sum of all the terms of this series will be the seventh
number
multiplied by 11.

Example:
1 If the numbers selected are 7 and 4:
e series jotted down is: 4, 7, 11, 18, 29, 47, 76, 123, 199, 322. 3
The
seventh number is 76. 11 x 76 = 836 (use the shortcut for 11:
7 is the first
digit, 6 is the third digit; the middle digit will be 7 + 6, and
carry the 1:
836).
4 So the sum of the ten numbers is 836.
Here are some of the calculations that you can do mentally
after
practicing the exercises. See how many you can do. Check
your answers
with a calculator. Your mental math powers should be
impressive!
15 x 15 89 x 89 394 x 101 32 x 38 51 x 59
101 x
101 x 94 93 x 93 228 x 101 25 x 25
147 56 x 56 94 x 96 101 x 448 83 x 87 687 x 101 64 x 66 52 x 52
101 x 654
35 x 35 11 x 19 101 x
61 x 69 89 x 101 61 x 61 34 x 36
206 101 x 48 52 x 52 45 x 45 456 x 101 54 x 54 41 x 41 101 x 41
32 x 32 83 x
101 29 x 29 55 x 55 63 x 67 479 x 101 82 x 82 882 x 101 14 x 16
319 x 101
73 x 77 65 x 65 101 x 859 13 x 17 101 x 149 41 x 49 82 x 101 75
x 75 101 x
71 x 79 69 x 101 22 x 22 993 x 101
144 51 x 51 85 x 85 101 x 101 129 x 101 69 x 69 31 x 39 738 x
101 21 x 29
71 x 71 77 x 101 95 x 95 265 x 101 74 x 76 53 x 53 101 x 409 51
x 51 98 x
101 94 x 94 42 x 42 53 x 57 339 x

62 x 68 101 x 88 19 x 19 238 x 101
101
43 x 47 96 x 96 21 x 21 101 x 279 81 x 89
648 x
62 x 62 101 x 57 24 x 26 101 x 668
101 99 x 99 23 x 27 22 x 28 515 x 101 97 x 97 79 x 79 101 x 826
245 x 101
31 x 31 37 x 33 101 x
82 x 88 39 x 39 126 x 101 91 x 99
218
68 x 101 51 x 101 598 x 101 98 x 98 72 x 72
91 x 91 54 x 56 101 x 29 57 x 57 771 x 101 81 x 81 559 x 101 52
x 58 59 x 59
84 x 86 46 x 44 101 x 25 12 x 18 101 x 697 92 x 92 349 x
93 x 97 101 x 188 42 x 48 49 x 49
101
58 x 58 92 x 92 101 x 78 37 x 101
1 Select two consecutive 2-digit numbers. 2 Add the two 2-
digit numbers!
Examples:
1 24 + 25 = 49. (Try it on a calculator and see, or if you're
really sharp, do
it mentally: 24 x 24 = 576, 25 x 25 = 625, 625 - 576 = 49.) 2 If 63
and 64 are
selected, then 63 + 64 = 127. (For larger number addition, do
it in steps:
63 + 64 = 63 + 60 + 4 = 123 + 4 = 127.)
1 Select two consecutive 2-digit numbers, one not more than
10 larger
than the other (experts need not use this limitation). 2
Subtract the
smaller number from the larger. 3 Add the two numbers. 4
Multiply the
first answer by the second.
Examples:

1 If 71 and 64 are selected: 2 71 - 64 = 7. 3 71 + 64 =
Add left to right: 71 + 64 = 71 + 60 + 4 = 131 + 4 = 135)
4 Multiply these results: 7 x 135 = 945 (Multiply left to right: 7
x 135 = 7 x
(100+30+5) = 700 + 210 + 35 = 910 + 35 = 945)
5 So the difference of the squares of 71 and 64 is 945.
See the pattern?
1 If 27 and 36 are selected: 3 2 36 - 27 = 9.
36 + 27 = 63 (Think: 27 + 30 + 6 = 57 + 6 = 63)
Multiply these results: 9 x 63 = 567 (Think: 9 x (60+3) = 540 +
27 = 567)
So the difference of the squares of 27 and 36 is 567.
Why do the divisibility 'rules' work?
From: cms@dragon.com (Cindy Smith) To: Dr. Math
<dr.math@forum.
swarthmore.edu> Subject: Factoring Tricks
In reading a popular math book, I came across several
arithmetic
factoring tricks. Essentially, if the last digit of a number is
zero,
then the entire number is divisible by 10. If the last number is
even, then the entire number is divisible by 2. If the last two
digits are divisible by 4, then the whole number is. If the last
three digits divide by 8, then the whole number does. If the
last
four digits divide by 16, then the whole number does, etc. If
the
last digit is 5, then the whole number divides by 5.
Now for the tricky ones.
If you add the digits in a number and the sum is divisible by
3,
then the whole number is. Similarly, fif you add the digits in
a number and the sum is divisible by 9, then the whole
number is. For example, take the number 1233: 1 + 2 + 3 + 3 =
9.

Therefore, the whole number is divisible by 9 and the
quotient is
137. The number is also divisible by 3 and the quotient is
411.
It works for extremely large numbers too (I checked on my
calculator). Now here's a really tricky trick. You add up the
alternate digits of a number and then add up the other set
of alternate digits. If the sums of the alternate digits equal
each other then the whole number is divisible by 11. Also,
if the difference of the alternate digits is 11 or a multiple of
11, then the whole number is divisible by 11. For example,
123,456,322. 1 + 3 + 5 + 3 + 2 = 14 and 2 + 4 + 6 + 2 = 14.
Therefore, the whole number is divisible by 11 and the
quotient
is 11,223,302. Also, if a number is divisible by both 3 and 2,
then
the whole number is divisible by 6.
The only single digit number for which there is no trick listed
is 7.
I find these rules interesting and useful, especially when
factoring
large numbers in algebraic expressions. However, I'm not
sure
why all these rules work. Can you explain to me why these
math
tricks work? If I could understand why they work, I think it
would
improve my math skills. Thanks in advance for your help.
Cindy Smith cms@dragon.com
From: Dr. Math <dr.math@forum.swarthmore.edu> To:
cms@dragon.
com (Cindy Smith) Subject: Re: Factoring Tricks
Thank you for this long and very well-written question. I will
try to
write as clearly as you as I answer. All these digital tests for

divisibility are based on the fact that our system of numerals
is
written using the base of 10.
The digits in a string of digits making up a numeral are
actually the
coefficients of a polynomial with 10 substituted for the
variable. For
example, 1233 = 1*10^3 + 2*10^2 + 3*10^1 + 3*10^0 which is
gotten
from the polynomial 1*x^3 + 2*x^2 + 3*x^1 + 3*x^0 = x^3 +
2*x^2 +
3*x + 3 by substituting 10 for x. We can explain each of these
tricks
in terms of that fact: 10 - numerals ending in 0 represent
numbers
divisible by 10:
Since the last digit is zero, and all other terms in the
polynomial form
are divisible by 10, the number is divisible by 10. Similarly, if
the
number is divisible by 10, since all the terms except the last
one
are automatically divisible by 10 no matter what the
coefficients or
digits are, the number will be divisible by 10 only if the last
digit is.
Since all the digits are smaller than 10, the last digit has to be
0 to
be a multiple of 10.2 - numerals ending in an even digit
represent
numbers divisible by 2: f o Same argument as above about all
the
terms except the last one being divisible by 2. The last digit
is

divisible by 2 (even) if and only if the whole number is. R 4 - numerals ending
with
a two-digit multiple of 4 represent numbers divisible by 4:o f
Similar to the above, but since 10 is not a multiple of 4, but
10^2 is,
we have to look at the last two digits instead of just the last
digit.
a n 8 - numerals ending with a three-digit multiple of 8
represent
numbers divisible by 8: u Similar to 4, but now 10^2 is not a
multiple
of 8, but 10^3 is, so we have to look at the last three mdigits.
er
a
16 - You figure this one!
l
5 - You figure this one, too!
ook
3 - numerals whose sum of digits is divisible by 3 represent
numbers
s divisible by 3: This one is different, because 3 does not
divide any
power of 10 evenly. That means we will have to consider the
effect of
all the l digits. Here we use this fact: 10^k - 1 = (10 - 1)*(10^(k-
1) = ... +
10^2 + i 10 + 1) This is a fancy way of saying 9999...999 =
9*1111...111
. We use k this to rewrite our powers of 10 as 10^k = 9*a[k] +
1. Now the
polynomial e this: d[k]*10^k + d[k-1]*10^(k-1) + ... + d[1]*10 +
d[0] = d[k
]*(9*a[k] + 1) + ... + d[1]*(9*a[1] + 1) + d[0] = 9*(d[k]*a[k] + ... +
d[1]*a[1])

+ d[k] + ... + d[1] + d[0] Now notice that 3 divides the first
part, so the
whole number is divisible by 3 if and only if the sum of the digits is.
9 - numerals whose sum of digits is divisible by 9 represent
numbers
divisible by 9: Use the same equation as the previous case.
You figure
the rest!
11 - numerals whose alternating sum of digits is divisible by
11 represent
numbers divisible by 11:
Here the phrase "alternating sum" means we alternate the
signs from
positive to negative to positive to negative, and so on. We
use this fact:
10 to an odd power plus 1 is divisible by 11, and 10 to an
even power
minus 1 is divisible by 11. The first part is a fancy way of
writing 10000.
..0001 = 11*9090...9091 (where there are an even number of
0's on the
left-hand side). The second part is a fancy way of writing
99999...9999
= 11*9090...0909 (where there are an even number of 9's in
the left
hand side). We write 10^(2*k) = 11*b[2*k] + 1 and 10^(2*k+1) =
11*b
[2*k+1] - 1. Here k is any nonnegative integer. We substitute
that
into the polynomial form, so:d[2*k]*10^(2*k) + d[2*k-
1]*10^(2*k-1) + .
.. + d[1]*10 + d[0]
= d[2*k]*(11*b[2*k] + 1) + d[2*k-1]*(11*b[2*k-1] - 1) + ...

� + d[1]*(11*a[1] - 1) + d[0] = 11*(d[2*k]*b[2*k] + ... + d[1]*a[1])
+ d[2*k]
- d[2*k-1]
� + ... - d[1] + d[0]
The first part is divisible by 11 no matter what the digits are,
so the
whole number is divisible by 11 if and only if the last part,
which is
the alternating sum of the digits, is
divisible by 11. If you prefer, you can write
d[2*k] - d[2*k-1] + ... - d[1] + d[0] = (d[0] + d[2] + ... + d[2*k]) -
(d[1] +
d[3] + ... + d[2*k-1]), so that you add up every other digit,
starting from
the units digit, and then add up the remaining digits, and
subtract the
two sums. This will compute the same result as the
alternating sum of
the digits. 7 - There is a trick for 7 which is not as well known
as the
others. It makes use of the fact that 10^(6*k) - 1 is divisible by
7, and
10^(6*k - 3) + 1 is divisible by 7. It goes like this: Mark off the
digits in
groups of threes, just as you do when you put commas in
large numbers.
Starting from the right, compute the alternating sum of the
groups as
three-digit numbers. If the result is negative, ignore the sign.
If the
result is greater than 1000, do the same thing to the resulting
number
until you have a result between 0 and 1000 inclusive. That 3-
digit

number is divisible by 7 if and only if the original number is
too.
Example:
123471023473 = 123,471,023,473, so make the sum
473 - 23 + 471 - 123 = 450 + 348 = 798.
798 = 7*114, so 798 is divisible by 7, and 123471023472 is,
too.
An extra trick is to replace every digit of 7 by a 0, every 8 by a
1, and
every 9 by a 2, before, during, or after the sum, and the fact
remains.
The sum could also have been computed as
473 - 23 + 471 - 123 --> 403 - 23 + 401 - 123 = 380 + 278 --> 310
+ 201 =
511 = 7*73. You can figure out why this "casting out 7's" part
works.
There is another way of testing for 7 which uses the fact that
7 divides
2*10 + 1 = 21. Start with the numeral for the number you want
to test.
Chop off the last digit, double it, and subtract that from the
rest of the
number. Continue this until you get stuck. The result is 7, 0,
or -7, if
and only if the original number is a multiple of 7.
Example:
123471023473 --> 12347102347 - 2*3 = 12347102341 -->
1234710234
- 2*1 = 1234710232 --> 123471023 - 2*2 = 123471019 -->
12347101 - 2*9
= 12347083 --> 1234708 - 2*3 = 1234702
--> 123470 - 2*2 = 123466 --> 12346 - 2*6 = 12334 --> 1233 - 2*4
= 1225
--> 122 - 2*5 = 112 --> 11 - 2*2 = 7.

13 - The same trick that works for 7 works for 13; that is, 13
divides
10^(6*k) - 1 and 10^(6*k - 3) + 1, so the alternating sum of
three-digit
groups works here, too. 17 - This is harder. You would have
to
use alternating sums of 8-digit groups!
1 Select a 3-digit number. 2 Repeat these digits to make a 6-
digit
number. 3 Divide these 6 digits by 7, then by 13. 4 The
answer i
s 11 times the first three digits!
Example:
1 If the 3-digit number selected is 234: 2 The 6-digit number
is 234234.
3. Divide by 7, then by 13: multiply by 11
-to multiply 234 by 11, work right to left: last digit on right = _
__4
next digit to left = 3 + 4 = 7: _ _ 7 _ next digit to left = 2 + 3 =
5: _ 5 _
_ last digit on left = 2 _ _ _
3 So 234234 divided by 7, then 13 is 2574.
See the pattern?
is 461461.
3. Divide by 7, then by 13: multiply by 11
-to multiply 461 by 11, work right to left: last digit on right = _
__1
nextdigit to left = 4 + 6 = 10: _ 0 _ _ last digit on left = 4 + 1
(carry) = 5: 5 _ _ _
Practice multiplying by 11 - this process works for
multiplying any
number by 11.

1 Select a 3-digit number. 2 Repeat these digits to make a 6-
digit
number. 3 Divide these 6 digits by 13, then by 11. 4 The
answer
is 7 times the first three digits!
Example:
is 231231.
3 Divide by 13, then by 11:
7 x 231 = 1400 + 210 + 7 = 1617.
See the pattern?
is 412412.
Reme
mber
to
multiply left to right and add in increments. Then you will
be able to give these answers quickly and accurately.
1 Select a 6-digit number repeating number. 2 Repeat these
digits to
make a 6-digit number. 3 Multiply a single digit by 3, then by
5.
Example:
1 If the 6-digit repeating number selected is 333333: 2
Multiply 3 x 3: 9 3
Multiply 9 x 5: 45 4 So 333333 divided by 37037 and
multiplied by 5 is 45.
See the pattern?
You can expand this exercise by using a different number in
the final step.
Example: multiply by 4:
1 If the 6-digit repeating number selected is 555555: 2
Multiply 3 x 5: 15. 3

Multiply 15 x 4: 60 4 So 555555 divided by 37037 and
multiplied by 4 is 60.
By changing the last step you can generate many extensions
of this
3 dsivide by 13, then by 11:
7 x 412 = 2800 + 70 + 14 = 2884.
exercise. Be inventive and create some impressive
calculations.
Dividing a repeating 6-digit number by 7, 11, 13; subtract 101
Select a 3-digit number.
Repeat these digits to make a 6-digit number.
Divide this 6-digit by 7, then 11, then 13.
Subtract 101.
The answer is the original number minus 101!
Example:
If the 3-digit number selected is 289:
The 6-digit number is 289289.
Divide by 7, then by 11, then by 13:
the answer is 289.
Subtract 101: 289 - 101 = 188
So 289289 divided by 7, then 11, then 13 minus 101 is 188.
See the pattern?
If the 3-digit number selected is 983:
The 6-digit number is 983983.
Divide by 7, then by 11, then by 13:
the answer is 983.
1 Subtract 101: 983 - 101 = 882
So 983983 divided by 7, then 11, then 13 minus 101 is 882.
1 Select a repeating 3-digit number . 2 The answer is 3 times
one
of the digits plus 41!
Example:
1 Select 999. 2 Multiply one digit by 3: 9 x 3 = 27. 3 Add 41: 27
+ 41

= 68. 4 So (999 / 37) + 41 = 68.
Change the last step to add other numbers, and thus
produce many
new exercises.
1 Select a repeating 6-digit number . 2 The answer is 7 times
the first
digit of the number!
Example:
1 777777 / 15873 = 7 x 7 = 49. 2 555555 / 15873 = 7 x 5 = 35. 3
999999 /
15873 = 7 x 9 = 63. Not very demanding mental math, but
good for a
quick challenge or two.
Dividing mixed numbers by 2
1 Select a mixed number (a whole number and a fraction).
� 2. If the whole number is even, divide by 2 - this is the
whole number
of the answer.
�� The numerator of the fraction stays the same; multiply
the
denominator by 2.
� 3. If the whole number is odd, subtract 1 and divide by 2 -
this
is the whole number of the answer.
� Add the numerator and the denominator of the fraction -
this
will be the new numerator of the fraction;
� Multiply the denominator by 2.
Even whole number example:
If the first number selected is 8 3/4: Divide the whole number
(8) by

2: 8/2 = 4 (whole number) Use the same numerator: 3 Multiply
the
denominator by 2: 4 x 2 = 8 (denominator) So 8 3/4 divided by
2 = 4 3/8.
Odd whole number example:
1 If the first number selected is 13 2/5: 2 3 Subtract 1 from the
whole
number and divide by 2: 13 - 1 = 12, 12/2 = 6.
Add the numerator and the denominator: 2 + 5 = 7. This is the
numerator of the fraction. Multiply the denominator by 2: 5 x
2 = 10
. So 13 2/5 divided by 2 = 6 7/10.
file:///
1 Select a 2-digit number. 2 Multiply it by 8 (or by 2 three
times).
3 Move the decimal point 2 places to the left.
Example:
1 The 2-digit number chosen to multiply by 12 1/2 is 78. 2
Multiply
by 2 three times: 2 x 78 = 156 2 x 156 = 312 2 x 312 = 624
3 Move the decimal point 2 places to the left: 6.24 4 So 78
divided
by 12 1/2 = 6.24.
See the pattern?
1 If the 2-digit number chosen to multiply by 12 1/2 is 91: 2
Double three times: 182, 364, 728. 3 Move the decimal point
2 places to the left: 7.28 4 So 91 divided by 12 1/2 = 7.28.5
ividing a 2-digit number by 15
1 Select a 2-digit number. 2 Multiply it by 2. 3 Divide the
result by 3.
4 Move the decimal point 1 place to the left.
Example:
1 The 2-digit number chosen to multiply by 15 is 68. 2
Multiply by 2:

2 x 68 = 120 + 16 = 136 3 Divide the result by 3: 136/3 = 45 1/3
4
Move the decimal point 1 place to the left: 4.5 1/3 5 So 68
divided by 15 = 4.5
1/3.
See the pattern?
Multiply
by 2: 2 x 96 = 180 + 12 = 192 3 Divide the result by 3: 192/3 =
64
4 Move the decimal point 1 place to the left: 6.4 5 So 96/15 =
6.4.
With this method you will be able to divide numbers by 15
with two quick
1 Select a 2-digit number. (Choose larger numbers when you
feel sure
about the method.) 2 Multiply by 4 (or by 2 twice). 3 Move the
decimal
l point two places to the left.
Example:
1 The 2-digit number chosen to divide by 25 is 38. 2 Multiply
by 4: 4 x 38
= 4 x 30 = 120 + 32 = 152. 3 Move the decimal point 2 places
to the left:
1.52 4 So 38 divided by 25 = 1.52.
See the pattern?
by 2 twice:
2 x 64 = 1282. 2 x 1282 = 2400 + 164 = 2564.
3 Move the decimal point 2 places to the left: 25.64. 4 So 641
divided by
25 = 25.64.
1 Select a 2-digit number (progress to larger ones). 2 Multiply
it by 3. 3
Move the decimal point 2 places to the left.

Example:
Multiply by 3:
3 x 46 = 3(40 + 6) = 120 + 18 = 138 3 Move the decimal point 2
places to
the left: 1.38 4 So 46 divided by 33 1/3 = 1.38. (If you divide by
33.3
using a calculator, you will not get the exact answer.)
See the pattern?
1 If the 3-digit number chosen to multiply by 33 1/3 is 650: 2
Multiply
by 3: 3 x (600 + 50) = 1800 + 150 = 1950 3 Move the decimal
point 2
places to the left: 19.50 4 So 650 divided by 33 1/3 = 19.5.
Practice multiplying left to right and this procedure will
become an
easy one - and you will get exact answers, too.
Dividing a 2- or 3-digit number by 35
feel sure
about the method.) 2 Multiply by 2. 3 Divide the resulting
number by 7.
4 Move the decimal point 1 place to the left.
Example:
If the number chosen to divide by 35 is 61: Multiply by 2: 2 x
61 = 122.
Divide by 7: 122/7 = 17 3/7 Move the decimal point 1 place to
the left:
1.7 3/7 So 61 divided by 35 = 1.7 3/7.
See the pattern?
1 If the number chosen to divide by 35 is 44: 2 Multiply by 2:
2 x 44 =
88 3 Divide by 7: 88/7 = 12 4/7 4 Move the decimal point 1
place to the
left: 1.2 4/7 5 So 44 divided by 35 = 1.2 4/7.

Division done by calculator will give repeating decimals
(unless the
original number is a multiple of 7), truncated by the limits of
the display.
The exact answer must be expressed as a mixed number.
1 Select a 2-digit number. 2 Multiply by 8. 3 Divide the
product by 3. 4
Example:
1 The 2-digit number chosen to divide by 37 1/2 is 32. 2
Multiply by 8:
8 x 32 = 240 + 16 = 256 3 Divide by 3: 256/3 = 85 1/3 4 Move
the decimal
point two places to the left: .85 1/3 5 So 32 divided by 37 1/2 =
.85 1/3.
See the pattern?
Multiply by 8:
8 x 51 = 408 3 Divide by 3: 408/3 = 136 4 Move the decimal
point two
places to the left: 1.36 5 So 51 divided by 37 1/2 = 1.36.
Dividing a 2-digit number by 45
1 Select a 2-digit number. 2 Divide by 5. 3 Divide the resulting
number
by 9.Example:
1 f the number chosen to divide by 45 is 32: 2 Divide by 5:
32/5 = 6.4 3 Divide the result by 9: 6.4/9 = .71 1/9 4 So 32
divided
by 45 = .71 1/9.
See the pattern?
1 If the number chosen to divide by 45 is 61: 2 Divide by 5:
61/5 =
12.2 3 Divide the result by 9: 12.2/9 = 1.35 5/9 4 So 61 divided
by 45
1.35 5/9.

feel
sure about the method.) 2 Multiply by 4 (or by 2 twice). 3
Move the
decimal point two places to the left. 4 Divide by 3 (express
remainder
as a fraction).
Example:
by 4:
4 x 82 = 328. 3 Move the decimal point 2 places to the left:
3.28 4
Divide by 3: 3.28/3 = 1.09 1/3 5 So 82 divided by 75 = 1.09 1/3.
See the pattern?
by 4
(multiply left to right): 4 x 631 = 2400 + 120 + 4 = 2520 + 4 =
2524.
3 Move the decimal point 2 places to the left: 25.24. 4 Divide
by 3:
25.24/3 = 8.41 1/3 5 So 631 divided by 75 = 8.41 1/3.
Select a number.
2 Multiply it by 2 3 Divide the result by 3.
Example:
1 The number chosen to divide by 1 1/2 is 72. 2 Multiply by 2:
2 x 72 =
144 3 Divide by 3: 144 / 3 = 48 4 So 72 divided by 1 1/2 = 48.
See the pattern?
2 x 83 =
166 3 Divide by 3: 166 / 3 = 55 1/3 4 So 83 divided by 1 1/2 =
55 1/3.
1 Select a 2-digit number. 2 Multiply by 3. 3 Divide by 4.
Example:

Multiply by 3:
3 x 47 = 120 + 21 = 141 3 Divide by 4: 141/4 = 35 1/4 4 So 47
divided
by 1 1/3 = 35 1/4.
See the pattern?
Multiply by 3:
3 x 82 = 246 3 Divide by 4: 246/4 = 61 1/2 4 So 82 divided by
1 1/3 = 61 1/2.
With this pattern you will be able to give answers quickly, but
most
importantly, your answers will be exact. If a calculator user
divides
by 1.3, the answer will NOT be correct.
1 Select a 2-digit number. 2 Multiply by 8. 3 Move the decimal
point
one place to the left.
Example:
Multiply by 8:
8 x 32 = 240 + 16 = 256 3 Move the decimal point one place to
the left
: 25.6 4 So 32 divided by 1 1/4 = 25.6.
See the pattern?
Multiply by 8:
the left:
51.2 4 So 64 divided by 1 1/4 = 51.2.
Multiply from left to right for ease and accuracy. You will
soon be
doing this division by a mixed number quickly.
1 Select a number. 2 Multiply it by 5 3 Divide the result by 6.
Example:

5 x 24
= 120 3 Divide by 6: 120/6 = 20 4 So 24 divided by 1 1/5 = 20.
See the pattern?
5 x 76
= 350 + 30 = 380 3 Divide by 6: 380/6 = 63 2/6 4 So 76 divided
by 1 1/5
= 63 1/3.
Dividing a 2-digit number by 1 2/3
1 Select a 2-digit number. 2 Multiply by 6 (or by 2 and 3). 3
Move the decimal point one place to the left.
Example:
Multiply
by 3: 3 x 78 = 210 + 24 = 234 3 Multiply by 2: 2 x 234 = 468 4
Move
the decimal point one place to the left: 46.8 5 So 78 divided
by
1 2/3 = 46.8.
See the pattern?
Multiply by 3:
3 x 32 = 96 3 Multiply by 2: 2 x 96 = 180 + 12 = 192 4 Move the
decimal
point one place to the left: 19.2 5 So 32 divided by 1 2/3 =
19.2.
Practice multiplying from left to right and you will become
adept at
mentally dividing a number by 1 2/3.
1 Select a 2-digit number. 2 Multiply by 8. 3 Move the decimal
point
3 places to the left.
Example:

by 8:
8 x 72 = 560 + 16 = 576. 3 Move the decimal point 3 places to
the left
: .576 4 So 72 divided by 125 = .576.
See the pattern?
by 8:
8 x 42 = 320 + 16 = 336.
3 ove the decimal point 3 places to the left: .336 4 So 42
divided by
125 = .336.
product by 7.
Example:
Multiply by 4:
4 x 34 = 136 3 Divide the product by 7: 137/6 = 19 3/7 (If you
use a
calculator, you will get a long, inexact decimal number.)
4 So 34 divided by 1 3/4 = 19 3/7.
See the pattern?
Multiply by 4:
4 x 56 = 224 3 Divide the product by 7: 224/7 = 32 4 So 56
divided
by 1 3/4 = 32.
Notice that numbers divisible by 7 will produce whole-
number
quotients. For numbers not divisible by 7, your calculator will
give
you long decimal results that are not exact.
1 Select a 2-digit number. 2 Multiply it by 7. 3 Move the
decimal
point 1 place to the left.

Example:
1 The number chosen to multiply by 1 3/7 is 36. 2 Multiply by
7:
7 x 36 = 210 + 42 = 252 3 Move the decimal point 1 place to
the
left: 25.2 4 So 36 divided by 1 3/7 = 25.2.
See the pattern?
1 The number chosen to multiply by 1 3/7 is 51. 2 Multiply by
7:
7 x 51 = 357 3 Move the decimal point 1 place to the left: 35.7
4
So 36 divided by 1 3/7 = 35.7.
1 Select a 2- or 3-digit number. 2 Multiply by 4 (or by 2 twice).
3 Move the decimal point one place to the left.
Example:
Multiply
by 4: 4 x 80 + 4 x 6 = 320 + 24 = 344 3 Move the decimal point
one place to the left: 34.4 4 So 86 divided by 2 1/2 = 34.4.
See the pattern?
Multiply
by 2: 2 x 624 = 1248 3 Multiply by 2: 2 x 1248 = 2400 + 96 =
2496
4 Move the decimal point one place to the left: 249.6 5 So 624
divided by 2 1/2 = 249.6.
Multiply by 4 when this is easy; otherwise use two steps and
multiply by 2 twice.
1 Select a 2-digit number. 2 Multiply by 3. 3 Divide the result
by 7.
Example:
1 he 2-digit number chosen to divide by 2 1/3 is 42. 2 Multiply
by 3:

3 x 42 = 126 3 Divide by 7: 126/7 = 18 4 So 42 divided by 2 1/3
= 18.
See the pattern?
Multiply by 3:
3 x 73 = 219 3 Divide by 7: 219/7 = 31 2/7 4 So 73 divided by 2
1/3 = 31 2/7.
If the number chosen is divisible by 7, the quotient will be a
whole
number. If the number is not divisible by 7, a calculator user
will
get a long, inexact decimal, while your answer will be exact.
1 Select a 2-digit number. 2 Multiply by 3. 3 Divide by 8.
Example:
Multiply by 3:
3 x 32 = 96 3 Divide by 8: 96/8 = 12 4 So 32 divided by 2 2/3 =
12.
See the pattern?
Multiply by 3:
2/3 = 22 7/8.
Using this method, your answers will be exact. Those using
calculators will only get approximations.
1 Select a 2-digit number. 2 Mltmultiiply the number by 2. 3
Divide the
product by 7.
Example:
Multiply by 2: 2 x
42 = 84 3 Divide by 7: 84/7 = 12 4 So 42 divided by 3 1/2 = 12.
See the pattern?
Multiply by 2:

1/2 = 17 3/7.
If the number chosen is divisible by 7, the answer will be a
whole number.
For numbers not divisible by 7, a calculator will get a
repeating decimal,
but your fractional answer will be exact.
1 Select a 2- or 3-digit number. 2 Multiply by 3. 3 Move the
decimal point
one place to the left.
Example:
Multiply by 3:
72 x 3 = 216 3 Move the decimal point one place to the left:
21.6 4 So
72 divided by 3 1/3 = 21.6.
See the pattern?
Multiply by 3:
the left:
14.4 4 So 48 divided by 3 1/3 = 14.4.
After practicing, choose larger numbers. Insist on exact
answers (you
won't get an exact answer if you divide by 3.3 using a
calculator).
Multiply from left to right in steps and impress your friends
with
your mental powers.
Dividing a 2-digit number by 375
product by
3 (express remainder as a fraction). 4 Move the decimal point
three places to the left.
Example:

1 The number chosen to divide by 375 is 32. 2 Multiply by 8:
8 x 32 = 240 + 16 = 256 3 Divide by 3: 256/3 = 85.3 1/3 4 Move
the decimal point 3 places to the left: .0853 1/3 5 So 32
divided by 375 = .0853 1/3.
See the pattern?
1 The number chosen to divide by 375 is 61. 2 Multiply by 8:
8 x 61 = 480 + 8 = 488 3 Divide by 3: 488/3 = 162 2/3 4 Move
the decimal point 3 places to the left: .162 2/3 5 So 61 divided
by 375 = .162 2/3.
72 3 Divide by 9:
72/9 = 8 4 So 36
divided by 4 1/2 =
8.
For numbers not
divisible by 9, your
calculator will get
a
repeating decimal,
but your fractional
answer will be
exact.
Dividing a 2-
digit number by
625
product
by 5. 4 Move the decimal point 3 places to the left.
Example:
by
8: 8 x 65 = 480 + 40 = 520 3 Divide by 5: 520/5 = 104 4 Move
the
decimal point 3 places to the left: .104 5 So 65 divided by 625
= .104.
See the pattern?

by
8: 8 x 32 = 240 + 16 = 256 3 Divide by 5: 256/5 = 51.2 4 Move
the
decimal point 3 places to the left: .0512 5 So 32 divided by
625 =
.0512.
1 Select a 2-digit number. 2 Multiply it by 4 (or by 2 twice). 3
Divide
by 3. 4 Move the decimal point 1 place to the left.
2M u lt i p l
Example: y b 1 The 2-digit number chosen to multiply by 7
1/2 is 42. y
4: 42 x 4 = 168 3 Divide by 3: 168/3 = 56 4 Move the decimal
point
1 place to the left: 5.6 5 So 42 divided by 7 1/2 = 5.6.
See the pattern?
Multiply by
2
1 Select a 2-digit number.
2 Multiply the number by 2.
3 Divide the product by 9.
Example:
1 The 2-digit number chosen to divide by 4 1/2 is 62.
2
3
Multiply by 2: 2 x 62 = 124 Divide by 9: 124/9 = 13 7/9
4 So 62 divideby 4 1/2 = 13 7/9.
See the pattern?
1 The 2-digit number chosen to divide by 4 1/2 is 36.
4: 93 x 4 = 360 + 12 = 372 3 Divide by 3: 372/3 = 124 4 Move
the

decimal point 1 place to the left: 12.4 5 So 93 divided by 7 1/2
= 12.4.
Practice and you will soon be cranking out these quotients
with
speed and accuracy.
Dividing a 2- or 3-digit number by 16 2/3
feel
sure about the method.) 2 Multiply by 6 (or by 3 and then 2). 3
Move the decimal point two places to the left.
Example:
Multiply
by 3: 3 x 72 = 216 3 Multiply by 2: 2 x 216 = 432 4 Move the
decimal
point 2 places to the left: 4.32 5 So 72 divided by 16 2/3 =
4.32.
See the pattern?
Multiply by
3: 3 x 212 = 636 3 Multiply by 2: 2 x 636 = 1200 + 72 = 1272 4
Move the
decimal point 2 places to the left: 12.72 5 So 212 divided by
16 2/3 = 12.72.
Practice multiplying by 3, then by 2, and you will be able to
do these
problems quickly.
Divisibility Rules
Why do these 'rules' work? - Dr. Rob
Divisibilidad por 13 y por números primos (13,17,19...)
-en español, de la lista SNARK
From the Archives of the Math Forum's Internet project Ask
Dr. Mathour
thanks to Ethan 'Dr. Math' Magness, Steven 'Dr. Math'
Sinnott,

nd, for the explanation of why these rules work, Robert L.
Ward (Dr. Rob).
Dividing by 3
Add up the digits: if the sum is divisible by three, then the
number is
as well. Examples: 1 111111: the digits add to 6 so the whole
number
is divisible by three. 2 87687687. The digits add up to 57, and
5 plus
seven is 12, so the original number is divisible by three.
Why does the 'divisibility by 3' rule work?
From: "Dr. Math" To: keving@ecentral.com (Kevin Gallagher)
Subject:
Re: Divisibility of a number by 3
As Kevin Gallagher wrote to Dr. Math
On 5/11/96 at 21:35:40 (Eastern Time),
>I'm looking for a SIMPLE way to explain to several very
bright 2nd
>graders why the divisibility by 3 rule works, i.e. add up all
the >digits;
if the sum is evenly divisible by 3, then the number is as well.
>Thanks!
>Kevin Gallagher
The only way that I can think of to explain this would be as
follows:
Look at a 2 digit number: 10a+b=9a+(a+b). We know that 9a is
divisible by 3, so 10a+b will be divisible by 3 if and only if a+b
is.
Similarly, 100a+10b+c=99a+9b+(a+b+c), and 99a+9b is
divisible
by 3, so the total will be iff a+b+c is.
This explanation also works to prove the divisibility by 9 test.
It
clearly originates from modular arithmetic ideas, and I'm not
sure

if it's simple enough, but it's the only explanation I can think
of.
Doctor Darren, The Math Forum
Check out our web site -http://forum.swarthmore.edu/dr.math/
Dividing by 4
ook at the last two digits. If they are divisible by 4, the
number is
as well. Examples:
1 100 is divisible by 4. 2 1732782989264864826421834612 is
divisible by four also, because 12 is divisible by four.
Dividing by 5
If the last digit is a five or a zero, then the number is divisible
by 5.
Dividing by 6
Check 3 and 2. If the number is divisible by both 3 and 2, it is
divisible by 6 as well. Robert Rusher writes in:
Another easy way to tell if a [multi-digit] number is
divisible by six . . .
is to look at its [ones digit]: if it is even, and the sum of
the [digits] is
a multiple of 3, then the number is divisible by 6.
Dividing by 7
To find out if a number is divisible by seven, take the last
digit,
double it, and subtract it from the rest of the number.
Example:
If you had 203, you would double the last digit to get six, and
ubtract that from 20 to get 14. If you get an answer divisible
by 7
(including zero), then the original number is divisible by
seven.
If you don't know the new number's divisibility, you can
apply
the rule again.
Dividing by 8

Check the last three digits. Since 1000 is divisible by 8, if the
last three digits of a number are divisible by 8, then so is the
whole number. Example: 33333888is divisible by 8; 33333886
isn't. How can you tell whether the last three digits are
divisible by 8? Phillip McReynolds answers:
If the first digit is even, the number is divisible by 8 if the
last two digits are. If the first digit is odd, subtract 4 from
the last two digits; the number will be divisible by 8 if the r
esulting last two digits are. So, to continue the last example,
33333888 is divisible by 8 because the digit in the hundreds
place is an even number, and the last two digits are 88, which
is divisible by 8. 33333886 is not divisible by 8 because the
digit in the hundreds place is an even number, but the last
two digits are 86, which is not divisible by 8.
Dividing by 9
Add the digits. If they are divisible by nine, then the number
is as well. This holds for any power of three.
Dividing by 10
If the number ends in 0, it is divisible by 10.
Dividing by 11Let's look at 352, which is divisible by 11;
the answer is 32. 3+2 is 5; another way to say this
is that 35 -2 is 33. Now look at 3531, which is also
divisible by 11. It is not a coincidence that 353-1 is
352 and 11 x 321 is 3531. Here is a generalization
of this system. Let's look at the number 94186565.
First we want to find whether it is divisible by 11, but on the
way we are going to save the numbers that we use: in every
step we will subtract the last digit from the other digits, then
save the subtracted amount in order. Start with
9418656 - 5 = 9418651 SAVE 5 Then 941865 - 1 = 941864
SAVE 1
Then
94186 - 4 = 94182 SAVE 4 Then 9418 - 2 = 9416 SAVE 2 Then
941
- 6 = 935

SAVE 6 Then 93 - 5 = 88 SAVE 5 Then 8 - 8 = 0 SAVE 8
Now write the numbers we saved in reverse order, and we
have
8562415, which multiplied by 11 is 94186565.
Here's an even easier method, contributed by Chis
Foren:Take
any number, such as 365167484.Add the 1,3,5,7,..,digits....
.3 + 5 + 6 + 4 + 4 = 22Add the 2,4,6,8,..,digits.....6 + 1 + 7 + 8 =
22If the difference, including 0, is divisible by 11, then so is
the
number. 22 - 22 = 0 so 365167484 is evenly divisible by
11.See also
Divisibility by 11 in the Dr. Math archives.
Dividing by 12
Check for divisibility by 3 and 4.
Dividing by 13
Here's a straightforward method supplied by Scott Fellows:
D e lete the last digit from the given number. Then subtract
nine
times the deleted digit from the remaining number. If what is
left
is divisible by 13, then so is the original number.
And here's a more complex method that can be extended to
other
formulas:
1 = 1 (mod 13)10 = -3 (mod 13) (i.e., 10 - -3 is divisible by
13)100
= -4 (mod 13) (i.e., 100 - -4 is divisible by 13)1000 = -1 (mod
13)
(i.e., 1000 - -1 is divisible by 13)10000 = 3 (mod 13)100000 = 4
(mod 13)1000000 = 1 (mod 13)
Call the ones digit a, the tens digit b, the hundreds digit c, .....
and you get:
a - 3*b - 4*c - d + 3*e + 4*f + g - .....

If this number is divisible by 13, then so is the original
number.
You can keep using this technique to get other formulas for
divisibility for prime numbers. For composite numbers just
check for divisibility by divisors.
Finding 2 1/2 percent of a number
1 Choose a number (start with 2 digits and advance to 3 with
practice).
2 Divide by 4 (or divide twice by 2). 3 Move the decimal point
one place
to the left.
Example:
1 If the number selected is 86: 2 Divide 86 by 4: 86/4 = 21.5 3
Move the
decimal point one place to the left.: 2.15 4 So 2 1/2% of 86 =
2.15.
See the pattern?
1 If the number selected is 648: 2 Divide 648 by 2 twice: 648/2
= 324,
324/2 = 162 3 Move the decimal point one place to the left.:
16.2 4 So
2 1/2% of 648 = 16.2.
Practice dividing by 4, or by 2 twice, and you will be able to
find these
answers faster than with a calculator.
Finding 5 percent of a number
1 Choose a large number (or sum of money). 2 Move the
decimal
point one place to the left. 3 Divide by 2 (take half of it).
Example:
1 If the amount of money selected is $850: 2 Move the
decimal point
one place to the left.: 85 3 Divide by 2: 85/2 = 42.50 4 So 5%
of $850 = $42.50.
See the pattern?

1 If the amount of money selected is $4500: 2 Move the
decimal point
one place to the left.: 450 3 Divide by 2: 450/2 = 225 4 So 5%
of $4500 =
$225.
1 Choose a 2-digit number. 2 Multiply the number by 3. 3
Divide by 2
. 4 Move the decimal point one place to the left.
Example:
1 If the number selected is 43: 2 Multiply by 3: 3 x 43 = 129 3
Divide
by 2: 129/2 = 64.5 4 Move the decimal point one place to the
left: 6.45
5 So 15% of 43 = 6.45.
See the pattern?
1 If the number selected is 72: 2 Multiply by 3: 3 x 72 = 216 3
Divide by
2: 216/2 = 108 4 Move the decimal point one place to the left:
10.8
5S o
15%
of72=10.8
.
1 Choose a 2-digit number. 2 Multiply the number by 9. 3
Divide by 2.
4 Move the decimal point one place to the left.
Example:
1 If the number selected is 36: 2 Multiply by 9: 9 x 36 = 270 +
54 = 324 3
Divide by 2: 324/2 = 163 4 Move the decimal point one place
to the left
: 16.2 5 So 45% of 36 = 16.2.
See the pattern?

If the number selected is 52: Multiply by 9: 9 x 52 = 450 + 18 =
468 Divide
by 2: 468/2 = 234 Move the decimal point one place to the
left: 23.4 So
45% of 52 = 23.4.
n 1 Choose a 2-digit number. e 2 Multiply the number by 11.
(Add digits
from right to left - see examples). x 3 Divide by 2. t 4 Move the
decimal
point one place to the left. d
i Example: g
it 1 If the number selected is 81: t 2 Multiply by 11: 11 x 81 =
891 o right
digit is 1 l eft is 1 + 8 = 9 last digit to left is 8
3 Divide by 2: 891/2 = 445.5 4 Move the decimal point one
place to the
left: 44.55 5 So 55% of 81 = 44.55.
See the pattern?
1 If the number selected is 59:
2. Multiply by 11: 11 x 59 = 649 right digit is 9 next digit to left
is 9 + 5
= 14 (use the 4 and carry 1)
last digit to left is 5 + 1 = 6
2 Divide by 2: 649/2 = 324.5
3 Move the decimal point one place to the left: 32.45
4 So 55% of 59 = 32.45.
Why do the divisibility "rules" work?
The Power of Modulo Arithmetic
Modulo arithmetic is a powerful tool that can be used to test
for
divisibility by any number. The great disadvantage of using
modulo arithmetic to test for divisibility is the fact that it is
usually
a slow method. Under some circumstances, however, the

application of modulo arithmetic leads to divisibility rules
that can
be used. These rules are important to mathematics because
they
save us lots of time and effort.
In fact, many of the divisibility rules that are commonly used
(rules for 3, 9, 11) have their roots in modulo arithmetic. This
page will show you how some of these common divisibility
rules are connected to modulo arithmetic.
Applying the Rules of Modulo Arithmetic
The rules of modulo arithmetic state that the number N is
divisible
by some number (P) if the above expression is also divisible
by P
after the base (10) is replaced by the remainder of 10 divided
by P.
In compact notation, this remainder is denoted by (10 mod
P). T
hus, N is divisible by P if the following expression is also
divisible by P:
[Dn * (10 mod P)^(n-1)] + ... + [D2 * (10 mod P)^1] + [D1 * (10
mod P)^0].
Origin of Divisibility Rules for 3, 9, and 11
D We can now see how the divisibility rules for P = 3, 9, and
11 are
rooted in modulo arithmetic. n
Consider the case where P = 3. .
Because 10 mod 3 is equal to 1, any number is divisible by 3
if the
following expression is
. also divisible by 3: .
-
[Dn * (1)^n-1] + ... + [D2 * (1)^1] + [D1* D
2
(1)^0]. Because 1 raised to any power is equal to 1, the above

+
expression can be simplified as: Dn + ... + D2 + D1 + D0,
D
which is equal to the sum of digits in the number. Thus
modulo
arithmetic allows us to 1
state that any number is divisible by 3 if the sum of its digits is
Because 10 mod 9 is also 1, any number is divisible by 9 is
the
sum of its digits are also
divisible by 9. Thus, the divisibility rules for 3 and 9 come
directly
from modulo n
arithmetic.
n)
Finally, consider the case where P = 11. a
Because 10 mod 11 is equal to -1, any number is divisible by
11
if the n following expression is also divisible by 11: d
D[Dn * (-1)^n-1] + ... + [D2 * (-1)^1] + [D1* (-1)^0].
n
Because -1 raised to any even power is equal to 1 and -1
raised to
any odd +power is equal to -1, the above expression becomes
a summation of the digits involving alternated signs: .
. . - D2 + D1 (for odd n).
This "alternating" summation rule for P = 11 is well known
and has two versions. The first version states a number is
divisible by 11 if the alternating sum of the digits is also
divisilbe
by 11 (i.e., the alternating sum is a muliple of 11). The second
version states that a number is divisible by 11 if the sum of
every
other digit starting with the rightmost digit is equal to the
sum of

every other digit starting with the second digit from the right
(i.e.,
the alternating sum is 0). Both of these versions come
directly
from the rules of modulo arithmetic.
Bases Other Than 10
v Up to this point, we have only considered numbers written
in
base 10. A number can, however, be i written in any base (B).
Such a number can be expressed by the following
summation: g
it N = [Dn * B^n-1] + [Dn-1 * B^(n-2)] + ... + [D2 * B^1] + [D1*
B^0]. s
"
The rules of modulo arithmetic apply no matter what base is
used
. These rules tell us that N is divisible by P if the following
expression is also divisible by P: [Dn * (B mod P)^n-1] + ...
+ [D2 * (B mod P)^1] + [D1* (B mod P)^0]. Most bases other
than 10 are difficult to use, but bases
that are powers of 10 can easily be
used to construct divisibility rules using modulo arithmetic.
Consider writing the base-10 number N = 1233457 in base-
100.
This can be done by starting with the rightmost digit and
grouping the digits in pairs of two. bEach grouping of
two digits is considered a single "digit" when the number
is ywritten in base-100. Written in base-100, the number is
1 23 34 57 where 57, 34, 123, and 1 are considered single
"digits." In a similar fashion, this number written 1in
base-1000 is 1 233 457 where 457, 233, and 1 are considered
single "digits." .Once the base-10 digits are grouped to form
the "digits," the above expression Bcan be used to test for
divisibility. Consider P = 7.
Use the base, B = 1000. Because 1000 mod 7 = -1, the

alternating summation rule a (used for P = 11 in base-10) can
be applied. Before applying this rule, the base-10 pdigits
must
be properly grouped to form base-1000 "digits." p
e Stryker explains it this way: When you are multiplying by 9,
on
your fingers (starting with your thumb) count the number you
are
multiplying by and hold down that finger. The number of
fingers
before the finger held down is the
first digit of the answer and the number of finger after the
finger
held down is the second digit of the answer.
Example: 2 x 9. your index finder is held down, your thumb is
before, representing 1, and there are eight fingers after your
index
finger, representing 18.
● Polly Norris suggests: When you multiply a number times
9, count
back one from that number to get the beginning of your
product. (
5 x 9: one less than 5 is 4). To get the rest of your answer,
just think
of the add fact families for 9:
1+8=92+7=93+6=94+5=9
8+1=97+2=96+3=95+4=9
5 x 9 = 4_. Just think to yourself: 4 + _ = 9 because the digits
in your
product always add up to 9 when one of the factors is 9.
Therefore,
4 + 5 = 9 and your answer is 45! I use this method to teach
the "nines
in multiplication to my third graders and they learn them in
one lesson!

Tamzo explains this a little differently:
1 Take the number you are multiplying 9 by and subtract one.
That
number is the first number in the solution. 2 Then subtract
that
number from nine. That number is the second number of the
solution.
Examples:
4 * 9 = 36
1 4-1=3
2 9-3=6
3 solution = 36
8 * 9 = 72
1 8-1=7 2 9-7=2 3 solution = 72
5 * 9 = 45
1 5-1=4 2 9-4=5 3 solution = 45
● Sergey writes in: Take the one-digit number you are
multipling by
nine, and insert a zero to its right. Then subtract the original
number
from it.
For example: if the problem is 9 * 6, insert a zero to the right
of the six,
then subtract six: 9 * 6 = 60 - 6 = 54
Multiplying a 2-digit number by 11
● A tip sent in by Bill Eldridge: Simply add the first and
second
digits and place the result between them.
Here's an example using 24 as the 2-digit number to be
multiplied
by 11: 2 + 4 = 6 so 24 x 11 = 264.
This can be done using any 2-digit number. (If the sum is 10
or greater
, don't forget to carry the one.)
Multiplying any number by 11

● Lonnie Dennis II writes in:
Let's say, for example, you wanted to multiply 54321 by 11.
First, let's
look at the problem the long way...
54321
x 11
54321 + 543210
=
597531
Now let's look at the easy way...
11 x 54321
4432
=51
+5 +3 +2 +1
531
Do you see the pattern? In a way, you're simply adding the
digit to
whatever comes before it. But you must work from right to
left.
The reason I work from right to left is that if the numbers,
when
added together, sum to more than 9, then you have
something
to carry over. Let's look at another example... 11 x 9527136
Well,
we know that 6 will be the last number in the answer. So the
answer
now is
???????6.
Calculate the tens
place: 6+3=9, so now
we know that the
product has the form
??????96. 3+1=4, so
now we know that the

product has the form
?????496. 1+7=8, so
????8496. 7+2=9, so
???98496. 2+5=7, so
??798496. 5+9=14. Here's where carrying digits comes in: we
fill in the hundred thousands place with the ones digit of the
sum
5+9, and our product has the form ?4798496. We will carry
the extra
10 over to the next (and final) place. 9+0=9, but we need to
add the
one carried from the previous sum: 9+0+1=10. So the product
is
104798496. Calculation Tips & Tricks
Multiplication Tips
Multiplying by five
● Jenny Logwood writes: Here is an easy way to find an
answer to
a 5 times question. If you are multiplying 5 times an even
number:
halve the number you are multiplying by and place a zero
after the
number. Example: 5 x 6, half of 6 is 3, add a zero for an
answer of
30. Another example: 5 x 8, half of 8 is 4, add a zero for an
answer
of 40.
If you are multiplying 5 times an odd number: subtract one
from
the number you are multiplying, then halve that number and
place
a 5 after the resulting number. Example: 5 x 7: -1 from 7 is 6,
half of
6 is 3, place a 5 at the end of the resulting number to produce
the

number 35. Another example: 5 x 3: -1 from 3 is 2, half of 2 is
1,
place a 5 at the end of this number to produce 15.
● Doug Elliott adds: To square a number that ends in 5,
multiply
the tens digit by (itself+1), then append 25. For example: to
calculate 25 x 25, first do 2 x 3 = 6, then append 25 to this
result;
the answer is 625. Other examples: 55 x 55; 5 x 6 = 30,
answer
is 3025. You can also square three digit numbers this way, by
starting with the the first two digits: 995 x 995; 99 x 100 =
9900,
answer is 990025.
Multiplying by nine
� Diana Grinwis says: To multiply by nine on your fingers,
hold
up ten fingers - if the problem is 9 x 8 you just put down your
8 finger
and there's your answer: 72. (If the problem is 9 x 7 just put
down
your 7 finger: 63.)
� Laurie Stryker explains it this way: When you are
multiplying by
9, on your fingers (starting with your thumb) count the
number you
are multiplying by and hold down that finger. The number of
fingers
before the finger held down is the
first digit of the answer and the number of finger after the
finger
held down is the second digit of the answer.
Example: 2 x 9. your index finder is held down, your thumb is

before, representing 1, and there are eight fingers after your
index
finger, representing 18.
Norris suggests: When you multiply a number times 9, count
back
one from that H number to get the beginning of your product.
(5 x 9:
one less than 5 is 4). e r To get the rest of your answer, just
think
of the add fact families for 9: e '
1 + 8 = 9 2 + 7 = 9 3 + 6 = 9 4 + 5 = 9 s8 + 1 = 9 7 + 2 = 9 6 + 3 =
9
5+4=9
5 x 9 = 4_. Just think to yourself: 4 + _ = 9 because the digits
in
your product n always add up to 9 when one of the factors is
9.
Therefore, 4 + 5 = 9 and your answer is 45! I use this method
to
each the "nines" in multiplication to my third e graders and
they
learn them in one lesson!
Tamzo explains this a little differently:
1. Take the number you are multiplying 9 by and subtract
one.
That number is the first e number in the solution. 2 Then
subtract
t that number from nine. That number is the second number
of the
solution.
Examples:
n 4 * 9 = 36 g1 4-1=3 2 9-3=6 2 3 solution = 36 4
8 * 9 = 72
a
s 1 8-1=7 2 9-7=2 t 3 solution = 72 h

e
5 * 9 = 45
2 -d 1 5-1=4 i 2 9-4=5 g3 solution = 45 i
t
n
● A tip sent in by Bill Eldridge: Simply add the first and
second
digits and place the result u
between them. m b er to be multiplied by 11: 2 + 4 = 6 so 24 x
11 = 264.
This can be done using any 2-digit number. (If the sum is 10
or g
reater, don't forget to carry the one.)
Multiplying any number by 11
● Lonnie Dennis II writes in:
Let's say, for example, you wanted to multiply 54321 by 11.
First,
let's look at the problem the long way...
54321
x 11
54321 + 543210
= 597531 Now let's look at the easy way... 11 x 54321
=51
+5 +3 +2 +1
Do you see the pattern? In a way, you're simply adding the
digit
to whatever comes before it. But you must work from right to
left.
The reason I work from right to left is that if the numbers,
when
added together, sum to more than 9, then you have
something
to carry over.

Let's look at another example... 11 x 9527136 Well, we know
that
6 will be the last number in the answer. So the answer now is
6. Calculate the tens place: 6+3=9, so now we know that the
product has the form ??????96. 3+1=4, so now we know that
the product has the form?????496. 1+7=8, so ????8496.
7+2=9, so ???98496. 2+5=7, so ??798496. 5+9=14. Here's
where c
arrying digits comes in: we fill in the hundred thousands
place
with the ones digit of the sum 5+9, and our product has the
form ?
4798496.
We will carry the extra 10 over to the next (and final) place.
9+0=9, but we need to add the one carried from the previous
sum:
9+0+1=10.
So the product is 104798496.
1 Select a 3-digit number. 2 Subtract the 1st digit plus 1 from
the number. X X X _ _
3 Subtract the last two digits of the number from 100. _ _ _ X
X
Example:
Subtract
the 1st digit + 1 from the number: 274 - 3 = 271 : 2 7 1 _ _
3 Subtract the last two digits from 100: 100 - 74 = 26: _ _ _ 2
6.
4 So 274 x 99 = 27126.
See the pattern?
Subtrac
t the 1st digit + 1 from the number: 924 - 10 = 914 : 9 1 4 _ _
3 Subtract the last two digits from 100:

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