Mathematics - Binomial Theorem Question with Solution | TestHub

Given,

${64}^{{32}^{32}}={\left(63+1\right)}^{32\times 32\times 32....32\text{times}}$

$\Rightarrow {64}^{{32}^{32}}={\left(63+1\right)}^{32\lambda }$

Now we know that by binomial theorem,

${\left(1+63\right)}^{32\lambda }={}^{{}^{32\lambda }}C_0·1+{}^{{}^{32\lambda }}C_1·1·63+{}^{{}^{32\lambda }}C_2·1·{63}^2+ .......$

$\Rightarrow {\left(1+63\right)}^{32\lambda }=1+9\alpha$

So, when ${64}^{{32}^{32}}$ is divided by $9$ we get $1$ as remainder.