Yet another absolutely challenging question to all, of course just for fun?

Question

What is the remainder when 2^1344452457 is divided by 11?
There is a method so don't give any more ridiculous answers please.

Answer 1

mod 11 (remainder) for powers of two go in this cycle:

2 - 4 - 8 - 5 - 10 - 9 - 7 - 3 - 6 - 1 (and repeat)

This cycle is 10 long, so the exponent (1344452457) mod 10 is all that matters.

It's the same remainder as 2^7, which is 7.

Answer 2

I’ve always liked Fermat’s Little Theorem.

X ^(p-1)≡ 1(mod p) where p is prime.

It is useful in solving this problem:
As X^(p-1) x X^(p-1) ≡ 1(mod p)
We have X^[ 2(p-1)] ≡ 1(mod p)
It follows that X^[ k(p-1)] ≡ 1(mod p) for any integer k *

Here X = 2 and p = 11,
hence p -1 = 10

The problem remains to find the remainder when 2^1344452457 is divided by 11.
Using *
Let X = 2, p = 11 and k = 134445245, then as p – 1 = 10, the problem can be written:

2^1344452457 = 2^[134445245x10] x 2^7 ≡ 1 x 2^7 (mod p)
≡ 2^7 (mod p)
≡ 2^7 (mod p)
≡ 128 (mod p)
≡ 7 (mod p)
Hence the remainder is 7.

Answer 3

2^1(mod 11) = 2
2^2(mod 11) = 4
2^3(mod 11) = 8
2^4(mod 11) = 5
2^5(mod 11) = 10
2^6(mod 11) = 9
2^7(mod 11) = 7
2^8(mod 11) = 3
2^9(mod 11) = 6
2^10(mod 11) = 1
2^11(mod 11) = 2, and the pattern will repeat from there.

Since 1344452457 ends in 7, 2 raised to this power will be congruent to 2^7(mod 11). The remainder is 7.

Answer 4

Per Fermat's little theorem, x^(n-1) ≡ 1 mod n. So:

2^1344452457 = 2^1344452450 * 2^7 ≡ 2^7 = 128 ≡ 7 mod 11

So the remainder is 7.

Answer 5

I don't know that of the top of my head.

Answer 6

None... 2^1344452457 is prime.

Answer 7

Isn't "Fun" and "Mathematics" a contradiction in terms :o)