Ok. So I do know that 2^2^2^2 is 2^16 and not 2^8
But in one of the older videos it was shown that 32^32^32 = 2^5120
How is this possible? Or am i missing something here?
The specific sum was to find the remainder when 32^32^32 is divided by 7. While the answer was correct, the strategy is not making sense because it says that 32^32^32 is equal to 2^5120!
Please help!
This is not true, so yes getting the right answer was just a coincidence
Do you need help with this?
Would love to see an easy to do solution, if there is one.
I did find a way of doing it finally…but needed to go through a lot more stuff than would generally need for GRE (such as totient function, fermats theorem, etc)
Almost went crazy for a couple of days
I mean if you know totient then isn’t it trivial?
Let n = 32^{32} and thus our problem is to find 32^n \pmod 7 \equiv 4^n \pmod 7
From euler’s totient, you have that since \varphi(7) = 6 so:
n = 32^{32} \pmod 6 \equiv 2^{32} \pmod 6 \equiv 2^2 \pmod 6
Thus, 32^n \pmod 7 \equiv 4^4 \pmod 7 \equiv \boxed{4 \pmod 7}
I had to learn totient for this.
Didn’t know it before this question popped up.
But would have liked to know if there is sth easier …
Uhhh i guess you can use cyclicity.
The remainder of \frac{32^n}{7} is in the order [4,2,1] (and then it repeats in this fashion) for n \geq 1.
We have that n = 32^{32} and we have to find what position (in the above group of 3) this comes (first, second, or third) in.
\frac{32^n}{3} has remainder in the [ 2,1] for n \geq 1. So 32^{32} (or rather n) leaves a remainder of 1 when divided by 3.
Thus it boils down to finding the remainder of \frac{32^1} {7}, which from our table was 4.
Thank you so much!