Question

What is the remainder when \(2^{64}\) is divided by \(7\)?

Answer 1

I get a pattern!

Answer 2

2^2 (which equals 4) divided by 7 leaves remainder 4.
2^3 (which equals 8) divided by 7 leaves remainder 1.
2^4 (which equals 16) divided by 7 leaves remainder 2.
2^5 (which equals 32) divided by 7 leaves remainder 4.
2^6 (which equals 64) divided by 7 leaves remainder 1.

Answer 3

2mod7 = 2
2^2 mod 7 = 4
2^3 mod 7 = 1
and it is repeating!

Answer 4

i think it is : 1 ..

Answer 5

64/7 =remainder = 1 ..

Answer 6

For every 2 powers, it gets 2 as the remainder.

Answer 7

Why are you dividing the exponent? :P

Answer 8

For every four*

Answer 9

And 64 is divisible by 4, so it's 2?

Answer 10

Hmm, it isn't. :/

Answer 11

2^8 mod 7 = 4
So that statement was completely false.

Answer 12

Umm I don't know about a proof but you have a repeating pattern with a period of 3, predicting that the remainder will be the same as for 2^4, that is, 2
Python:

>>> 2**64
18446744073709551616L
>>> 2**64 % 7
2L

Seems right.

Answer 13

yes the answer is 4.

Answer 14

oops sorry i meant the answer is 2

Answer 15

Scared me.

Answer 16

:D ..

Answer 17

Hmm, but that thing I said didn't satisfy for \(2^8\).

Answer 18

2^8/7 = remainder = 2

Answer 19

>>> 2**8 % 7
4

Answer 20

yes .. remainder = 4 ..
but why m i getting 2 by that way?

Answer 21

256/7 = remainder = 4

Answer 22

The answers follow this pattern:
2 divided by 7 leaves remainder 2.
2^2 (which equals 4) divided by 7 leaves remainder 4.
2^3 (which equals 8) divided by 7 leaves remainder 1.
2^4 (which equals 16) divided by 7 leaves remainder 2.
2^5 (which equals 32) divided by 7 leaves remainder 4.
2^6 (which equals 64) divided by 7 leaves remainder 1.

We should stop as soon as we notice that the cycle will repeat itself forever in this pattern: [2, 4, 1]. Every third remainder is the same. (From here on out, "remainder" always means "remainder after we divide by 7.") Since every third remainder is the same, we should look at the remainder when the power is a multiple of 3. The remainders of 2^3 and 2^6 are 1. Thus, the remainder of 2 raised to a power that is any multiple of 3 is 1.