If last two digits of $ 7^{7^{7}}=k$, find $k-38$?

Question

I have managed to get the answer as $07$ using modular arithmetic but answer seems wrong to me because the answer after evaluation of $k-38$ is $5$.

Please help.

Answer 1

We can calculate that $7^7=823543$.

For the last two digits of $7^n$, we have this repeated cycle for $n=1;n=2;n=3;n=4;...$: $07\rightarrow49 \rightarrow43 \rightarrow01\rightarrow07 \rightarrow49 \rightarrow43 \rightarrow01 \rightarrow07 \rightarrow49 \rightarrow43 \rightarrow01 \rightarrow...$

We can conclude that if $k$ is a non-negative integer:

• $n\equiv1\pmod {4}$ then $7^n$ have two last digits $07$.

• $n\equiv2\pmod {4}$ then $7^n$ have two last digits $49$.

• $n\equiv3\pmod {4}$ then $7^n$ have two last digits $43$.

• $n\equiv0\pmod {4}$ then $7^n$ have two last digits $01$.

$43\equiv3\pmod {4}$, so $823543\equiv3\pmod {4}$.

We can conclude thet the last two digits of the expression is $43$, hence $k-38=43-38=5.$

Also, based on the comment on your problem, we all know that $7^{7^{7}}\ne7^{49}$, but $(7^7)^7=7^{49}$.