Find the last digit and last two digits of an integer. For finding the last digit of $7^{99}$ using Euler's theorem we have for integer $7$ and $10$
$7^4 \equiv 1 \pmod {10}$. Since the last digit is the remainder when divided by $10$, it gives the answer $7^{99} \equiv 1 \pmod {10}$ and the last digit is $3$.

But for finding last two digits we have to find the remainder when the number $7^{100}$ is divided by $100$. By Euler's theorem $7^{99}\equiv 1 \pmod{100}$ Now $7^{100}=7^{99}\cdot 7 \equiv 7 \pmod {10}$ So the last two digit is $07$.

Did I make any mistake? Please tell.

  • $\begingroup$ By Euler's theorem, $7^{40} \equiv 1 \pmod{100}.$ Where did $99$ come from? You used $4$, rather than $9$ in the first problem, so it's odd that you made this mistake in the second problem. $\endgroup$
    – B. Goddard
    Jun 28, 2018 at 15:39
  • $\begingroup$ @B.Goddard thanks I couldn't find it $\endgroup$
    – John757
    Jun 28, 2018 at 15:42

2 Answers 2


You are right when you state that $7^4\equiv1\pmod{10}$. Actually, $7^4\equiv1\pmod{100}$. Therefore, $7^{99}\equiv7^3\pmod{100}$. And $7^3\equiv43\pmod{100}$. Therefore, the answer is $43$.

Concerning your approach, note that what Euler's theorem tells us is that $7^{40}\equiv1\pmod{100}$.


Euler's $\phi(100)=40$, not $99$, since $100$ is not prime. You might also look at the Carmichael function $\lambda(100)=20$, which is (with a small variation) the result of applying Euler to the component prime powers of a composite number, and combining via $\text{lcm}$, giving you the largest multiplicative order possible under that modulus and the value which all orders divide. This gives $7^{20}\equiv 1$ and thus $7^{100}\equiv 1 \bmod 100$ directly (last two digits $01$).

This means that $7^{99}\equiv 7^{-1} \bmod 100$ and either we can check the powers of $7$, either directly$\bmod 100$ or through the components of $4$ and $25$ and combine, or we can find the inverse through the extended Euclidean algorithm. In this case we could probably "notice" that $7^2=49$ and use the repetition of squares $\bmod 4n$ to get $7^4(\equiv 49^2\equiv 99^2\equiv -1^2)\equiv 1\bmod 100 $ as observed by others, and thus $7^{-1}\equiv 7^3\equiv 43\bmod 100.$


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