Posted by **JOHN** on Wednesday, July 9, 2008 at 3:30am.

I have three questions:

1. What is the remainder when 27 to the power of 1001 is divided by 13?

2. What is the remainder of when 38 to the power of 101 is divided by 13?

3. How do you show that

70 x 27 (to the power of 1001) + 31 x 38(to the power of 101)

can be divisible by 13?

Thanks

- MATHS -
**drwls**, Wednesday, July 9, 2008 at 11:06am
1. 27/13 = 2 with a remainder of 1

27^2/13 = 56 with a remainder of 1

27^3/13 = 1514 with a remainder of 1

27^4/13 = 40,880 with a remainder of 1

...

So 27 to any integer power, divided by 13, has a remainder of 1.

- MATHS -
**Count Iblis**, Wednesday, July 9, 2008 at 6:09pm
Define x Mod(13) as the remainder of x if you divide x by 13. You can iterchange taking Mod with taking powers.

Since:

27 Mod(13) = 1 --------->

27^n Mod(13) = 1^n = 1

2)

Let's omit writing Mod(13) every time, equality simply means both sides are the same Mod (13).

We can use that 38 = 12 = -1

So 38^101 = (-1)^101 = -1 = 12

70 x 27^(1001) + 31 x 38^(101) =

70 - 31 = 39 = 0

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