A Test To All Minds

Number Theory Level 3

What is the remainder when you divide $2^{32}$ by 641?

The answer is 640.

1 solution

Adhiraj Mandal
Jan 3, 2014

641=[(5^4)+(2^4)]=[5 (2^7)+1] Clearly [(5^4)+(2^4)] divides [(5^4)+(2^4)] 2^28 Hence 641 divides [2^32 + (5^4)*2^28]

Since (a+b) divides [(a^n)-(b^n)] , hence [5 (2^7)+1] divides [{5 (2^7)}^4-1^4] Therefore, 641 divides [(5^4*2^28)-1]

There 641 divides [2^32 + (5^4) 2^28] -[(5^4 2^28)-1] Hence 641 divides (2^32+1). Therefore (2^32) leaves a remainder 640 when divided by 641.[HENCE PROVED]

Sorry friends for not using latex.

It is known that $F_5=2^{2^5}+1$ is the fifth Fermat number and is divisible by 641.So our answer is $\boxed{640}$ .

Bogdan Simeonov - 7 years, 5 months ago

Good solution.But how do you know it?Is it a theorem that you can directly apply? No. The proof is provided above. You can check it.

Adhiraj Mandal - 7 years, 5 months ago

A Test To All Minds

The answer is 640.

1 solution

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