A number theory problem by farhan miraz shihab

Good question. The last digit of the powers of every one-digit number repeats in cycles of $4$ . For powers of $7$ , for instance, the cycle is $7, 9, 3, 1$ . Knowing this, a shortcut to finding the last digit of a power is to raise the power to the $\bmod\ 4$ of the exponent. If it is $1$ , raise it to the $1^{st}$ ; $2$ , raise it to the $2^{nd}$ ; $3$ , raise it to the $3^{rd}$ . The one caveat to this shortcut is this: if the exponent $\bmod\ 4$ , however, is $0$ , you must raise the power to the $4^{th}$ , not only because raising it to $0$ would make the last digit $1$ every time, but also because if $4$ divides the exponent, then it simply means that the power's last digit is the $4^{th}$ number in its cycle of last digits.

Now why am I only taking the $\bmod\ 4$ of the exponent's last two numbers? Because every multiple of $100, 1000, 10000...\ 10^n\ \forall\ n \geq 2$ is perfectly divisible by $4$ . Think about it. Let's take the number $639$ . $639 = 600 + 39$ , and because $600$ is a multiple of $100 -$ itself a multiple of $4 -$ then $639 \bmod\ 4 \equiv 39 \bmod\ 4 \equiv 3$ . Thus, it is much faster to simply take the $\bmod\ 4$ of the last two digits of the exponent rather than the whole exponent itself $-$ a shortcut especially useful for large numbers.