Use Binomial Theorem to find the remainder of $32^{{32}^{32}}$

Question

Use Binomial Theorem to find the remainder when $32^{{32}^{32}}$ is divided by $7$.

$$32^{{32}^{32}}=2^{5.{2}^{{160}}}$$ $$5.(3-1)^{160}=5\left(C_{0}3^{160} -C_{1}3^{159} \cdots -C_{159}3 +C_{160}\right)$$

I want to prove it using Binomial Theorem. I don't know what mod thing is.

Answer 1

Notice that $\ (a+bc)^{\large k}\,\overset{\rm BT}=\, a^{\large k} + bn\ $ for some integer $\,n,\,$ by BT = Binomial Theorem, so

$\quad \color{#c00}{32^{32}}\! = (-1+3\cdot 11)^{32} \overset{\rm BT}= \color{#c00}{1+3n}$

$\quad \color{#0a0}{32^{\large 3}}\! = (2^5)^3\! = (2^3)^5\! = (1+7)^5 \overset{\rm BT}= \color{#0a0}{1+7k}$

$32^{\large\color{#c00}{32^{\Large 32}}}\! = 32^{\large\color{#c00}{ 1+3n}} = 32(\color{#0a0}{32^{\large 3}})^{\large n} = 32(\color{#0a0}{1+7k})^{\large n} \overset{\rm BT}= (4\!+\!7\cdot 4)(1+7j) = 4+7(\cdots)$

Answer 2

As $2^3=1+7$

$2^{100}=(3-1)^{100}=3a+1$ where $a$ is some fixed integer

$2^{5\cdot2^{100}}=2^{5(3a+1)}=(2^3)^{5a}\cdot2^5=(1+7)^{5a}\cdot2^5=(1+7b)32=4+7(4+32b)$