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I have to find the last two decimal digits of the decimal number $9^{201}$. These can be thought as the remainder leaved out by dividing by 100. I've applied Euler's theorem and since 100 is coprime with 9 and since $\phi(100) = 40$ I've got $9^{40} \equiv 1 \pmod{100}$.

Now since

$$9^{201} = 9^{(40 \cdot 5 + 1)} = (9^{40})^5 \cdot 9 \equiv 1^5 \cdot 9 \pmod{100} \equiv 1 \cdot 9 = 9 \pmod{100}$$

This means that $9^{201}$ and $9$ leave out the same remainder when divided by $100$ so I can conclude the last digit is $9$.

But I don't find a rigorous way to find out the penultimate digit. I've was mentioned different heuristic methods but I don't find them appropriate. In this case it should be $0$, but I'm not sure why. If instead of $9$ there were a two digit number like $78$ it would be clear that the two last digits are $78$.

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    $\begingroup$ You're finished. You have correctly shown that $9^{201}\equiv 9\pmod {100}$ so the last two digits are $09$. $\endgroup$ – lulu Aug 25 '20 at 17:16
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    $\begingroup$ Maybe you can find answer to your question by searching for "last two digits" here. This gives about 1700 results. $\endgroup$ – miracle173 Aug 25 '20 at 17:18
  • $\begingroup$ What should be not rigorous in this calculation ? If the base is not coprime to $100$ , the problem is slightly more difficult. $\endgroup$ – Peter Aug 26 '20 at 6:52
  • $\begingroup$ For future: use \pmod{100}, not \quad mod \, 100. $\endgroup$ – metamorphy Aug 26 '20 at 11:27
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You already have the answer. Consider a simpler example: The last two digits of 709 are "09" because $709 \equiv 9 \pmod{100}$. "mod 100" always* gives the last two digits (even if we don't always write them both).

* In base 10, before anyone gets pedantic.

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