Question

Solve this:

(2^5^40^80) divided by 7

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Answer 1

Question:

Find the remainder when  $2^{\displaystyle5^{\displaystyle40^{\displaystyle80}}}$  is divided by 7.

Solution:

Since 7 is a prime number and  gcd(2, 7) = 1,  we use Fermat's theorem.

$2^6\equiv 1\pmod7\ \ \Longrightarrow\ \ 2^{6n}\equiv 1\pmod7$

So we have to write  $5^{\displaystyle 40^{\displaystyle 80}}$  in the form  6a + b.  For this, we have to get the remainder on dividing  $5^{\displaystyle 40^{\displaystyle 80}}$  by 6.

Here it seems  gcd(5, 6) = 1,  but 6 is not a prime number. Here we use Euler's theorem.

$\phi(6)=2\ \ \ \longrightarrow\ \ \ 5^2\equiv 1\pmod6\ \ \Longrightarrow\ \ 5^{2n}\equiv 1\pmod6$

So we have to write  $40^{\displaystyle 80}$  in the form  2a + b.  And we have to find the remainder when  $40^{\displaystyle 80}$  is divided by 2.

Since 40 is divisible by 2, then so will be  $40^{\displaystyle 80},$  hence 0 is the remainder.

Let  $40^{\displaystyle 80}=2a$

$5^{2n}\equiv1\pmod6\ \ \Longrightarrow\ \ 5^{2a}\equiv 1\pmod6\ \ \Longrightarrow\ \ 5^{\displaystyle 40^{80}}\equiv 1\pmod6$

Hence the remainder when  $5^{\displaystyle 40^{\displaystyle 80}}$  is divided by 6 is 1.

Let  $5^{\displaystyle 40^{\displaystyle 80}}=6a+1$

Now,

$\large2^{6a+1}=2^{6a}\times 2\equiv 1\times 2=\mathbf{2}\pmod7\ \ \ \left[\because\ 2^6\equiv 1\pmod7\right]\\ \\ \\ \therefore\ \ \Large\boxed{\text{ 2^{\displaystyle 5^{\displaystyle 40^{\displaystyle 80}}}\equiv \bold{2}\pmod7 }}$

Hence, 2 is the answer.