Find the remainder when $7^{7^7}$ is divided by $10$.

Question

I am a beginner in Number Theory so just starting. I tried this problem the following way, I tried using Euler Theorem on the problem and got $7^4 \equiv 1 \text{(mod 10})$. Then I focused on the exponent $7^7$ and reduced it by mod 10 as $7^4 \times 7^3$ to $7^3 \equiv (-3)^3 \equiv (-1)(3)^3 \equiv (-1)(-3) \equiv 3$ and again repeated on the base and got $3$ as the answer, Is this correct? If I am wrong please correct me. And If you have any alternate solutions then greatly appreciated. Thanks for your time.

Answer 1

The answer is $3$. Observe the following:

• $7^{1} \equiv 7 \pmod{10}$

• $7^{2} \equiv 9 \pmod{10}$

• $7^{3} \equiv 3 \pmod{10}$

• $7^{4} \equiv 1 \pmod{10}$.

This pattern repeats. That is, powers of $7$ modulo $10$ repeat in cycles of $4$.

This means that if we can compute $x = 7^{7} \pmod{4}$, then we can easily use the above observation to compute $7^{x} \pmod{10}$. To find $x$, we once again note that $7^{2} \equiv 1 \pmod{4}$, which means that

$$7^{7} \equiv 7 \times (7^{2})^{3} \equiv 7 \times 1 \pmod{4} \equiv 3.$$

Thus, $7^{7^{7}} \equiv 7^{3} \equiv 3 \pmod{10}$.