2. The amount of coconuts after on monkey has finished will equal b = (6/7)*(x-1) where b is the number of coconuts.

3. Use the same equation for the remaining monkeys

4. So the equation set would be

5. b = (6/7)*(x-1) = amount after 1 monkey,

6. c = (6/7)*(b-1) = amount after 2 monkeys,

7. d = (6/7)*(c-1) = amount after 3 monkeys,

8. e = (6/7)*(d-1) = amount after 4 monkeys,

9. f = (6/7)*(e-1) = amount after 5 monkeys,

10. g = (6/7)*(f-1) = amount after 6 monkeys,

11. h = (6/7)*(g-1) = amount after 7 monkeys,

12. i = (1/7)*(h-1) = amount each got at the end.

13. This works as the original number of coconuts must be able to be subtracted from 6/7 when one is subtracted from the original number and form an integer

15. b = (6/7)*(x-1) Multiply each side by 7

16. 7B=6*(x-1) Distribute out the 6

17. 7B=6x-6 Put the six on one side of the equation and the variables on the other

18. 6=7B+6x

19. Repeat this process with the remaining equations; you get

20. 6a - 7b = 6,

21. 6b - 7c = 6,

22. 6c - 7d = 6,

23. 6d - 7e = 6,

24. 6e - 7f = 6,

25. 6f - 7g = 6,

26. 6g - 7h = 6,

27. h - 7i = 1

29. 6^1x - 7^1b = 6*(7^1 - 6^1)

30. 6^2x - 7^2c = 6*(7^2 - 6^2)

31. 6^3x - 7^3d = 6*(7^3 - 6^3)

32. 6^4x - 7^4e = 6*(7^4 - 6^4)

33. 6^5x - 7^5f = 6*(7^5 - 6^5)

34. 6^6x - 7^6g = 6*(7^6 - 6^6)

35. 6^7x - 7^7h = 6*(7^7 - 6^7)

36. 6^7x - 7^8i = 7^8 – 6^8

37. In the last problem X: Is going to be -6(match the 6^7X on the left side to the -6^8 on the right. You need a negative and one more six to make the two terms equal)

38. In the last problem;I is going to equal -1(again match the -7^8i on the left to the 7^8 on the right)

40. X=-6 + 7^8*T

41. And

42. I=-1 + 6^7*T

43. T is any integer

44. Set T equal to one, as that is the lowest positive integer

45. X=-6+7^8(1)

46. X=5,764,795

47. I=-1+6^7(1)

48. I=279,935