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.
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1Can you show us your modular arithmetic? – Clayton Mar 31 '18 at 02:55
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https://math.stackexchange.com/questions/390685/the-last-2-digits-of-7777 – lab bhattacharjee Mar 31 '18 at 03:01
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I had written the expression as 7^49 – Akash Roy Mar 31 '18 at 03:42
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1Sorry I have actually calculated wrong the last 3 digit I have found is 343 therefore last two digits 43 7^2 congruent to 49 mod 100 implies 7^48 congruent to 49 mod 100 thus I multiplied 7 on both side and got 7^49 congruent to 343 mod 100 or 43 mod 100 therefore 43-38 =5. Solved my own silly doubt lol – Akash Roy Mar 31 '18 at 03:42
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I found $5$ mentally. No big deal because I knew that $7^4=2401\equiv 1 \mod {100}.$ – DanielWainfleet Mar 31 '18 at 15:34
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Are you familiar with Euker's totient function? It makes Q's like this easy and fast. – DanielWainfleet Mar 31 '18 at 15:37
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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 $(7^7)^7=7^{49}$.
user061703
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