What is the remainder when $4^{10}+6^{10}$ is divided by $25$?

Question

Without using calculator, how to decide? Must go with last two digits of $4^{10}+6^{10}$, can tell the last digit is $2$. How to tell the tenth digit of the sum?

Thanks!

Answer 1

• $4^5=1024 \equiv -1\pmod{25} \implies 4^{10}\equiv 1 \pmod{25}$

• $6^4 = 1296 \equiv -4\pmod{25} \implies 6^8 \equiv 16\pmod {25}$

Answer 2

Bill Dubuque gave an excellent hint in the comments. Here's the full solution.

$$ 4^{10}+6^{10}=(5-1)^{10}+(5+1)^{10}=2\left(5^{10}+\binom{10}{2}5^8+\cdots+\binom{10}{8}5^2+1\right)\equiv^*2\pmod{25} $$ (Note that $\equiv^*$ follows because all the terms except the last are multiples of $5^2=25$, so they are congruent to $0$ modulo $25$.)

Answer 3

The last digit of $4^x$ is $4,6,4,6,4,6\dots \implies 4^{10}$ has last digit $6$

The last digit of $6^x$ is always $6 \implies 6^{10}$ has last digit $6$

Therefore $4^{10} + 6^{10}$ has last digit $2$.

Any multiple of $25$ has last digit $0$ or $5\implies 25$ cannot be a divisor!

Answer 4

$$4^4=256 \equiv 6 \mod (25)$$ $$ 4^8 \equiv 36 \equiv 11 \mod (25)$$ $$ 4^{10} \equiv 16\times 11= 176 \equiv 1 \mod (25)$$

$$6^2 =36 \equiv 11 \mod (25)$$

$$6^4 \equiv 121 \equiv -4 \mod (25)$$ $$6^8 \equiv 16 \mod (25)$$ $$6^{10} \equiv 176 \equiv 1 \mod (25)$$

$$4^{10} + 6^{10} \equiv 2 \mod (25)$$