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?
 A: 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 $\frac{70}{125}$.
A: Let's consider the remainders when dividing by $4$. If $a=1$ then $2^a=2$ has the remainder $2$. Otherwise, $2^a$ has the remainder zero. If $b$ is odd then $3^b$ has remainder $3$, but if $b$ is even $3^b$ has remainder $1$. Whatever $c$ is, $5^c$ will have remainder $1$. (All of these statements are provable by induction, but the patterns are obvious.)
Then there are just two basic ways for the sum to have remainder zero when divided by $4$:
Case 1: $a=1$, $b$ is even, $c$ is anything. This gives remainders $2+1+1$ or zero.
Case2: $a>1$, $b$ is odd, $c$ is anything. This gives remainders $0+3+1$ or zero.
Now count each case and add the counts.
A: $2^{\color\red{a}}+3^\color\green{b}+5^\color\magenta{c}\equiv0\pmod4\iff$


*

*$\Big(\big(\color\red{a}=1\big)\wedge\big((\color\green{b}=2)\vee(\color\green{b}=4)\big)\Big)\vee$

*$\Big(\big(\color\red{a}\neq1\big)\wedge\big((\color\green{b}\neq2)\wedge(\color\green{b}\neq4)\big)\Big)$



Therefore, the probability is $\dfrac{\color\red1\cdot\color\green2\cdot\color\magenta5+(5-\color\red1)\cdot(5-\color\green2)\cdot\color\magenta5}{5\cdot5\cdot5}=\dfrac{14}{25}$
