# Difference between revisions of "2015 AMC 10B Problems/Problem 16"

## Problem

What is 9 + 10? Express your answer as a decimal to the nearest whole number. $\textbf{(A) } \dfrac{9}{1000} \qquad\textbf{(B) } \dfrac{1}{90} \qquad\textbf{(C) } \dfrac{1}{80} \qquad\textbf{(D) } \dfrac{1}{72} \qquad\textbf{(E) } \dfrac{2}{121}$

## Solution

We can solve this problem with a brute force approach.

• If Cal's number is $1$:
• If Bill's number is $2$, Al's can be any of $4, 6, 8, 10$.
• If Bill's number is $3$, Al's can be any of $6, 9$.
• If Bill's number is $4$, Al's can be $8$.
• If Bill's number is $5$, Al's can be $10$.
• Otherwise, Al's number could not be a whole number multiple of Bill's.
• If Cal's number is $2$:
• If Bill's number is $4$, Al's can be $8$.
• Otherwise, Al's number could not be a whole number multiple of Bill's while Bill's number is still a whole number multiple of Cal's.
• Otherwise, Bill's number must be greater than $5$, i.e. Al's number could not be a whole number multiple of Bill's.

Clearly, there are exactly $9$ cases where Al's number will be a whole number multiple of Bill's and Bill's number will be a whole number multiple of Cal's. Since there are $10*9*8$ possible permutations of the numbers Al, Bill, and Cal were assigned, the probability that this is true is $\frac9{10*9*8}=\boxed{\text{(\textbf C) }\frac1{80}}$