2 power 56 when divided by 17 leaves a remainder of?

Question

anonymous · Answer

$$2^{56}/17 = 4.239*10^{15}$$ does that answer work for you or would you like more steps?

anonymous · Answer

This is same as 56*56 = 3136, when divided by 17 = 184 remainder 8

anonymous · Answer

2^16 congruent to 1 mod 17 so 2^56 = 2^8 mod17...

anonymous · Answer

56*56 nowhere near 2^56!!

anonymous · Answer

@DevinBlade  remainder is a number less than 17

anonymous · Answer

oh so its 2^(56/17)? aha my bad

anonymous · Answer

$$\frac{2^{56}}{17}$$

anonymous · Answer

ohh the remainder x_x its been so long since i did old school division totally forgot about remainders

anonymous · Answer

I guess the answer is one, follow my next post to find why.

anonymous · Answer

@alsamixer  Correct.

anonymous · Answer

$$2^{56} = 2^{8\times7} = (2^4)^{14} = (2^4 + 1 -1)^{14} = (a-1)^{14}$$
where $$a = 2^{4}+1 = 17 $$
Now expand the first equation using binomial equation. And all terms till the last term is divisible by 17, what remains is $$1^{14} = 1$$ , hence the remainder is 1.

anonymous · Answer

If you don't know it you have  a^(p-1) = 1 mod p (Fermat's Little Theorem).

anonymous · Answer

But does it apply to this problem? We wanted something like:
$$(a-1)^p = 1 \mod a$$

anonymous · Answer

I posted it above already

2^16 congruent to 1 mod 17 so 2^56 = 2^8 mod17...

2^16 three times is 1 mod 17 so goes away and 2^8/17 is simple..