Remainder Headache!

Number Theory Level 3

What is the remainder when $(3^{2^n})-1$ is divided by $2^{n+3}$ ?

$n$ is any positive integer.

2^(n-1) 2^(n+1) 2^(n+2) 2^(n-2)

2 solutions

A Former Brilliant Member
Aug 29, 2014

just put any no. and do hit and trial......for the convenience put n=2 and check the ans.

Kunal Mandil
Aug 14, 2014

3^2^n - 1

=[(1+2)^2^n] -1

open it with the help of binomial theorem......

let 2^n = m

therefore we have to open the expansion of [(1+2)^m] - 1

{ [mC0(1^m)(2^0)] + [mC1(1^(m-1))(2^1)] + [mC2(1^(m-2))(2^2)] + ................. + [mCm(1^0)(2^m)] } - 1

first term of the expansion [mC0(1^m)(2^0)] = 1 which will get cut by -1

now remaining terms are the multiple of 2 , so we can get 2 common from all the terms

so when we divide it by 2^(n+3) we will get remainder 2^(n+2)

and not get confuse as i hav taken 2^n = m , i hav just taken it to save time....

The remainder will be [mC1(2^1)] + [mC2(2^2)]...+[[mC(n+2)](2^(n+2)]. Can you be clarify how we can equate that to mean that the remainder is (2^(n+2))?

Aditya Ganeshan - 6 years, 9 months ago

Remainder Headache!

2 solutions

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