Question Number 29700 by 803jaideep@gmail.com last updated on 11/Feb/18
$$\mathrm{32}^{\mathrm{32}^{\mathrm{32}} } \:/\mathrm{7}...\mathrm{find}\:\mathrm{remainder} \\ $$
Answered by Tinkutara last updated on 13/Feb/18
Commented by Tinkutara last updated on 13/Feb/18
Proof by Binomial Theorem.
Commented by rahul 19 last updated on 13/Feb/18
$$\mathrm{this}\:\mathrm{seems}\:\mathrm{correct}. \\ $$
Commented by 803jaideep@gmail.com last updated on 14/Feb/18
$$\mathrm{2}^{\mathrm{5}^{\mathrm{32}} } \neq\mathrm{2}^{\mathrm{160}\:} \:\mathrm{as}\:\mathrm{5}^{\mathrm{32}} \neq\mathrm{160} \\ $$
Commented by Tinkutara last updated on 14/Feb/18
$${In}\:{second}\:{line} \\ $$$$\mathrm{32}^{\mathrm{32}} =\left(\mathrm{2}^{\mathrm{5}} \right)^{\mathrm{32}} =\mathrm{2}^{\mathrm{160}} \:{is}\:{correct}. \\ $$
Commented by 803jaideep@gmail.com last updated on 14/Feb/18
$$\mathrm{kk}...\:\mathrm{thn}\:\mathrm{i}\:\mathrm{misunderstood}\:\mathrm{it} \\ $$