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Find the last two digits of $7^{7^{7^{10217}}}$.

We need to find $7^{7^{7^{10217}}}$ (mod $100$) and $\phi(100) = 40$
So $7^{40} \equiv 1$(mod$100$)
I don't know how to proceed after this

Gauri Sharma
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  • @user236182: It is indeed an exact duplicate, The only difference is in the solution. The one below can be verified at a glance. – André Nicolas Jan 28 '16 at 22:29

1 Answers1

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Hint: It is true that $7^{40}\equiv 1\pmod{100}$. But already $7^4\equiv 1\pmod{100}$. So all we need to find out about the exponent is what it is modulo $4$. Almost no calculation is needed, since the bottom $7$ in the exponent is congruent to $-1$ modulo $4$.

André Nicolas
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