# 2018 UNCO Math Contest II Problems/Problem 7

## Problem

Let $x = 2^A + 10^B$ where $A$ and $B$ are randomly chosen with replacement from among the positive integers less than or equal to twelve. What is the probability that $x$ is a multiple of $12$?

## Solution

To have a number divisible by $12$, it must be divisible by $3$ and $4$.

Consider mod 3 of x: $x \: {\equiv} \: 0 \: (mod \: 3)$ $2^A + 10^B\:{\equiv}\:0\:(mod\:3)$ $2^A + 1\:{\equiv}\:0\:(mod\:3)$ $2^A\:{\equiv}\:2\:(mod\:3)$

Here, since $2^A\:{\equiv}\:1 \: or \: 2 \: (mod \: 3)$ for A is even and odd respectively $\therefore \: A$ is odd

Consider mod 4 of x: $x \: {\equiv} \: 0 \: (mod \: 4)$ $2^A + 10^B\:{\equiv}\:0\:(mod\:4)$ $2^A + 2^B\:{\equiv}\:0\:(mod\:4)$ $\because\:$ we know A is odd, for $A=1$, taking $mod\:4$ gives $2$ as result; for $A>1$, taking $mod\:4$ gives $0$ as result, so we split the case for $A=1$ and $A>1$ here.

For $A=1$, $2 + 2^B\:{\equiv}\:0\:(mod\:4)$ $2^B\:{\equiv}\:2\:(mod\:4)$ $\therefore \: B=1$

For $A>1$, $2^B\:{\equiv}\:0\:(mod\:4)$ $\therefore \: B>1$ $\therefore\:$ Concluding our above conditions, we have $(A,B) = (1,1)$ or $(A,B) \in (\{3,5,7,9,11\},\:\{2,3,4,5,6,7,8,9,10,11,12\})$

By counting the number of solutions, the required probability $=\frac{1\times1 + 5\times11}{12\times12} = \frac{56}{144} = \frac{7}{18}$