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My question is to find the last two digits of $3^{400}$ and I found that the last digit is $1$ but I'm not sure about the other one. Any help would be appreciated, thank you for your time.

Edit:

I've found that $3^{\phi(100)} = 1 \pmod{100} ; \phi(100) = 40$ and $3^{400} = 1 \pmod{100}$ So the last digit of $3^{400}$ is $1$

Siong Thye Goh
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  • please include as much details for your attempt as possible in your post. Also, tell us what do you know, your background knowledge, to get a suitable reply. – Siong Thye Goh Jun 15 '20 at 08:07
  • I recommend to start with $3^{n}$ for smaller values of $n$. Perhaps you can find a pattern? You could try $3^{10}$ or $3^{20}$... – Matti P. Jun 15 '20 at 08:08
  • @SiongThyeGoh thank you for your input, I've added my work above. –  Jun 15 '20 at 08:14
  • @MattiP.thank you, I'll try that. –  Jun 15 '20 at 08:14
  • you are right I calculate $3^{13}\equiv 23$ and then easily you find $3^{40}\equiv01$ – Mojbn Jun 15 '20 at 14:04

2 Answers2

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Great job. Euler's theorem is the right tool since $3$ and $100$ are coprime.

By computing modulo $100$, you have actually found the last two digits. The last two digits are indeed just $1$ or if you like $01$.

The last digit is computed by $\pmod{10}$.

In general, to compute the last $n$ digits, we compute the remainder $\pmod{10^n}$.

Siong Thye Goh
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  • Thank you for your answer, how can I conclude that the last two digits are $01$. I thought that my approach is to find the last digit $1$. –  Jun 15 '20 at 08:23
  • for an integer, you can view it as $n=100q+r$ where $0 \le r < 100$. $r$, the last two digits is just the remainder when you divide $n$ by $100$. – Siong Thye Goh Jun 15 '20 at 08:25
  • if it helps, view $n=100q+10a+b$ where $a$ and $b$ are single digits. By taking modulo $100$, we have $10a+b \equiv{1} \pmod{100}$. We then conclude that $a=0$ and $b=1$. – Siong Thye Goh Jun 15 '20 at 08:27
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The multiplicative group of integers mod 100 has order of $\phi (100) = 100 \cdot \frac 12 \cdot \frac 45 = 40$

$a^{40} \equiv 1 \pmod {100}\\ (a^{40})^{10} \equiv 1 \pmod {100}$

The last two digits are $01$

Doug M
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  • Thank you, how can the last step leads to the last two digits are $01$, I am a little confuse since I am new to this subject, sorry. –  Jun 15 '20 at 08:24
  • If we consider the set of integers modulo 100, what are the last two digits of all elements of the equivalence class associated with $\underline 1$ ? – Doug M Jun 15 '20 at 08:35
  • you should probably use parentheses: $(a^{40})^{10}$ – J. W. Tanner Jun 15 '20 at 12:47