What is the remainder when $4^{10}+6^{10}$ is divided by $25$? 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!
 A: 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 $25$.)
A: *

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

*$6^4 = 1296 \equiv -4\pmod{25} \implies 6^8 \equiv 16\pmod {25}$
A: $$4^4=256 \equiv 6 \mod (25)$$
$$ 4^8 \equiv 36 \equiv 11 \mod (25)$$
$$ 4^{10} \equiv 16\times 11= 176 \equiv 1 \mod (25)$$
$$6^2 =36 \equiv 11 \mod (25)$$
$$6^4 \equiv 121 \equiv -4 \mod (25)$$
$$6^8 \equiv 16 \mod (25)$$
$$6^{10} \equiv 176 \equiv 1 \mod (25)$$ 
$$4^{10} + 6^{10} \equiv 2 \mod (25)$$
A: 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! 
