Probability that $2^a+3^b+5^c$ is divisible by 4 
If $a,b,c\in{1,2,3,4,5}$, find the probability that $2^a+3^b+5^c$ is divisible by 4.

For a number to be divisible by $4$, the last two digits have to be divisible by $4$
$5^c= \_~\_25$ if $c>1$
$3^1=3,~3^2=9,~3^3=27,~3^4=81,~ 3^5=243$
$2^1=2,~2^2=4,~2^3=8,~2^4=16,~2^5=32$
Should I add all possibilities? Is there a simpler method?

Observe that
$$2^a+3^b+5^c \equiv 2^a+(-1)^b+1 \pmod{4}$$
So for this to be $0 \pmod 4$, we have the following scenarios


*

*$a \geq 2$, $b$ is odd and $c$ is any number.

*$a=1$, $b$ is even and $c$ is any number.


The number of three tuples $(a,b,c)$ that satisfy the first case =$(4)(3)(5)=60$ and the number of three tuples $(a,b,c)$ that satisfy the second case =$(1)(2)(5)=10.$
Probability is $\frac{70}{125}$.
