# Euler's theorem application last two digits of a number

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 $$78$$.

• You're finished. You have correctly shown that $9^{201}\equiv 9\pmod {100}$ so the last two digits are $09$. – lulu Aug 25 '20 at 17:16
• Maybe you can find answer to your question by searching for "last two digits" here. This gives about 1700 results. – miracle173 Aug 25 '20 at 17:18
• What should be not rigorous in this calculation ? If the base is not coprime to $100$ , the problem is slightly more difficult. – Peter Aug 26 '20 at 6:52
• For future: use \pmod{100}, not \quad mod \, 100. – metamorphy Aug 26 '20 at 11:27

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).