# 1997 AHSME Problems/Problem 22

## Problem

Ashley, Betty, Carlos, Dick, and Elgin went shopping. Each had a whole number of dollars to spend, and together they had $56$ dollars. The absolute difference between the amounts Ashley and Betty had to spend was $19$ dollars. The absolute difference between the amounts Betty and Carlos had was $7$ dollars, between Carlos and Dick was $5$ dollars, between Dick and Elgin was $4$ dollars, and between Elgin and Ashley was $11$ dollars. How many dollars did Elgin have?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

## Solution

Working backwards, if $6 \le E \le 10$, then $6 \pm 11 \le A \le 10 \pm 11$. Since $A$ is a positive integer, $17 \le A \le 21$.

Since $17 \le A \le 21$, we know that $17 \pm 19 \le B \le 21 \pm 19$. But if $B=36$, which is the smallest possible "plus" value, then $E + A + B = 6 + 17 + 36 = 59$, which is too much money.

Hence, $17 - 19 \le B \le 21 - 19$. But since $B$ must be a positive integer, that leaves only two possibilities: $B = 1$ or $B=2$, which correspond with $E = 9$ and $E = 10$.

Concentrating only on $E=9$, we have $E=9$ leading to $A = 9 + 11 = 20$, which leads to $B = 20 - 19 = 1$, which leads to $C = 1 + 7 = 8$. Thus far we have given out $9 + 20 + 1 + 8 = 38$ dollars. This means that Dick must have $56 - 38 = 18$ dollars. However, the difference between Carlos and Dick is not $5$ dollars.

Thus, the right answer must be $\boxed{E}$. Verifying, if $E = 10$, then $A = 10 + 11 = 21$, $B = 21 - 19 = 2$, which leads to $C = 2 + 7 = 9$. Thus far, we have given out $10 + 21 + 2 + 9 = 42$ dollars, leaving $56 - 42 = 14$ dollars for Dick. Dick does indeed have $5$ dollars more than Carlos, and $4$ dollars more than Elgin.