Question

What will be remainder if 30^72^87 is divided by 11?

Answer 1

We can solve this question by Fermat's little theorem. Let me give a prelude to the theorem.

Fermat's little theorem
It states that if p is a prime number, then a^p-1 =1 (mod p), given that p & a are co-primes. 
Here, 11 is a prime number. Also, 30 and 11 are co-primes.

30^10 = 1 (mod 11).

It is required to simplify 30^72^87 to the form 30^10^x to apply Fermat's theorem.

It is enough to find the unit's digit of expansion 72^87 since any value of 30^10^x (mod 11) will be 1.

To find the units digit of 72^87, it's enough to find last digit of 2^87.

Taking cyclicity of 2, last digit of 2^87 is 8.

Now, we have 30^72^87 = 30^10^x . 30^2^87 = 1 . 30^8 = 30 ^8 (mod 11)

Applying basic remainder theorem, 30^8 (mod 11) = 8^8 (mod 11) = 64 ^4 (m0d 11) = (-2)^4 (mod 11) = 16 (mod 11) = 5.