Question

What is remainder when 3^40 is divided by 16?​

Answer 1

Answer:

We need to find out the remainder of 3^40 when divided by 23

Rem [3^40 / 23]

= Rem [81^10 / 23]

= Rem [12^10 / 23]

= Rem [144^5 / 23]

= Rem [6^5 / 23]

= Rem [(36*36*6) / 23]

= Rem [(-10)*(-10)*6 / 23]

= Rem [600 / 23]

= 2

Step-by-step explanation:

Answer 2

Explanation:

2 ≡ 1 mod 3.

∴ 2 ≡ 1 mod 81 (that is, $\mathsf{3^4}$)

∴ $\mathsf{2^4}$ ≡ $\mathsf{1^4\:mod\:3^4}$

∴ $\mathsf{16}$ ≡ $\mathsf{1\:mod\:3^4}$

∴ $\mathsf{16}$ ≡ $\mathsf{1\:mod\:(3^4)^{10}}$

∴ $\mathsf{16}$ ≡ $\mathsf{1\:mod\:3^{40}}$

Hence, the above statement can be expressed as

$\mathsf{3^{40}=16n+1}$, where 16 is the divisor, n is the quotient, and 1 is the remainder.

Hence, the remainder to be obtained is 1.

Answer: 1

I hope this helps. :D