The remainder when $\left((64)^{(64)}\right)^{(64)}$ is divided by 7 is equal to

Question

Aicharya · Accepted Answer

Let $\mathrm{N}=\left((64)^{64}ight)^{64}$ $\mathbf{N}=(64)^{64^2}$ $\mathbf{N}=(1+63)^{64^2}, 	ext { let } 64^2=\mathrm{n}$ Expanding by binomial $\begin{aligned} & \mathrm{N}=(1+63)^{\mathrm{n}}=1+{ }^{\mathrm{n}} \mathrm{C}_1 63+{ }^{\mathrm{n}} \mathrm{C}_2(63)^2+\ldots . \ & =1+63 \lambda=1+7(9 \lambda) \end{aligned}$ Remainder when divided by 7 is 1

The remainder when $\left((64)^{(64)}\right)^{(64)}$ is divided by 7 is equal to

Solution