## RolyPoly 2 years ago Why is the last two digits of $$91^{71}$$ the same as 91mod100? Similarly, why is it true that the last two digits of $$81^{81}$$ the same as 81mod100, and that of $$71^{91}$$ the same as 71mod100?

1. RolyPoly

Maybe I should edit my question a bit. It should be "how do I know if the last two digits of $$91^{71}$$ is the same as 91mod100? What if I need to find last two digits of$$91^{70}$$?

2. Jack17

there the same thing, how much modular arithmetic do you know

3. RolyPoly

I only know how to find x^y mod (n) using the trick that express y in binary number first, then take mod.

4. RolyPoly

Why are you deleting your comment?

5. Jack17

oh your refering to the binary exponentiation algorithm

6. RolyPoly
7. Jack17

$$91^{71}\equiv 91 \text{ mod 100}$$ is what you are trying to show

8. RolyPoly

Wait!

9. Jack17

ok

10. RolyPoly

Suppose the question is find the last two digits of $$91^{71}$$. How would you start?

11. Jack17

Do you know what the mod operator does?

12. Jack17

if I were to write 91^71 in base 10

13. Jack17

it would be $$91^{71}=a+b10+c10^2+d10^3...$$

14. Jack17

now we can reduce that modulo 100, ok

15. RolyPoly

Ehh!!! So, finding the last two digits of n is the same as finding n mod 100?

16. Jack17

so $$91^{71}\text{ mod 100}=a+10b$$ Where a is are first digit and b are second digit

17. Jack17

yes pretty much

18. RolyPoly

And finding the last n digits of m is the same as finding m mod (10^n)?

19. Jack17

yes

20. Jack17

Which you can compute using your exponentiation by squaring algorithm or using a variety of other techniques

21. Jack17

though for small numbers like yours, I wouldn't use that technique

22. RolyPoly

What technique would you use?

23. RolyPoly

Don't tell me calculator ~_~

24. Jack17

well for $$91^{71}\text{ mod 100}$$

25. Jack17

I would reduce the 91 modulo 100, so I get,

26. Jack17

$$91^{71} \text{ mod 100} = (-9)^{71} \text{ mod 100}= -9^{71} \text{ mod 100}$$

27. RolyPoly

"reduce the 91 modulo 100"?

28. Jack17

Yes, I think maybe you should read up a little on modular arithmetic,

29. Jack17

$$91\equiv -9 \text{ mod 100}$$

30. Jack17

therefore

31. Jack17

$$91^{71}\equiv (-9)^{71}\text{ mod 100}$$

32. Jack17

so now we just have to compute -9 to the 71st power modulo 100

33. Jack17

but $$(-1)^{71}=-1$$

34. Jack17

so its just computing $$-9^{71}\text{ mod 100}$$

35. Jack17

Now 9 is coprime to 100

36. Jack17

and $$\phi(9)=6$$

37. Jack17

$$9^{6}\equiv 1 \text{ mod 100}$$

38. Jack17

raising both sides to the 12th power, gives

39. Jack17

$$9^{72}\equiv 1 \text{ mod 100 }$$

40. Jack17

$$-9^{72}\equiv -1 \text{ mod 100}$$

41. Jack17

but wqe want $$-9^{71} \text{ mod 100}$$

42. Jack17

if we let $$-9^{71}=x$$

43. Jack17

we have $$9x\equiv -1 \text{ mod 100}$$

44. Jack17

now all we have to do is solve for x modulo 100, and we have are solution, I know this looked like it took a while but its because I had to type over this chat, also there is many ways to solve somthing like this, if your just using a computer an algorithm like yours might be better.

45. Jack17

But if your doing it by hand, its helpful to be familear with some of this stuff, because sometimes there are 'easyier' ways to do things.

46. RolyPoly

I'll tag you if I have questions. I need to work it out myself to see if I really understand it. Sleep well :)

47. Jack17

alright later

48. RolyPoly

@Jack17 Problem problem problem "Now 9 is coprime to 100" "and ϕ(9)=6" " $$9^6≡1 mod 100$$" But isn't it $$a^{\phi(n)} ≡ 1 (\mod n)$$ ? Why would we find ϕ(9)???

49. Jack17

nice catch, this changes how you would proceed

50. Jack17

We want $$\phi(100)$$

51. Jack17

which is 40

52. RolyPoly

Don't be so lazy... :(

53. RolyPoly

Yay!!! Thanks!! :)

54. Jack17

$$9^{10}\equiv 1 \text{ mod 100 }$$

55. Jack17

Which you can get by exponentiation by squaring

56. Jack17

nvm im being stupid the original problem was $$-9^{71} \text{ mod 100}$$

57. Jack17

this is the same as $$-3^{142} \text{ mod 100}$$

58. Jack17

ahh maybe you should use your algorithm, i am to lazy to figure this out

59. RolyPoly

142 is a LARGE number!!!

60. Jack17

61. Jack17

use the fact $$9^{10}\equiv 1 \text{ mod 100}$$

62. Jack17

so $$9^{70}\equiv 1 \text{ mod 100}$$

63. Jack17

$$9^{71}\equiv 9\text{ mod 100}$$

64. RolyPoly

9^(40) ≡1 mod 100 since phi(100) is 40 o_o

65. Jack17

$$-9^{71}\equiv -9 \text{ mod 100}$$

66. Jack17

$$-9^{71}\equiv 91 \text{ mod 100}$$

67. Jack17

$$91^{71}\equiv 91 \text{ mod 100}$$

68. Jack17

the first two digits of $$91^{71}$$ are 1 and 9

69. RolyPoly

9^(10) ≡1 mod 100 <- How do you work this out?

70. Jack17

$$9^2\equiv -19 \text{ mod 100}$$

71. Jack17

$$9^4\equiv 19^2 \text{ mod 100}$$

72. Jack17

$$9^4\equiv -39 \text{ mod 100}$$

73. Jack17

$$9^8 \equiv 39^2 \text{ mod 100}$$

74. Jack17

$$9^8\equiv 21 \text{ mod 100}$$

75. Jack17

$$9^{10}\equiv 21*9^2 \text{ mod 100}$$

76. Jack17

$$9^{10}\equiv 1 \text{ mod 100}$$

77. RolyPoly

Got it! Have never thought that it works in this way... Another problem is that how you can draw the conclusion "so $$9^{70}≡1 \mod 100$$" from $$9^{10}≡1 \mod 100$$ quickly?

78. Jack17

I raised both sides to the 7th power

79. RolyPoly

Ok, I didn't notice that, sorry!!

80. Jack17

If $$a\equiv b \text{ mod m}$$ and $$c\equiv d \text{ mod m}$$ then $$ac\equiv bd \text{ mod m}$$

81. RolyPoly

Thanks!!! It looks pretty good now :) Btw, if you want to type LaTeX in the same line as the words, use \( instead of \[ . Thanks again for your help :)

82. Jack17

ye no problem