# 1996 AJHSME Problems/Problem 14

## Problem

Six different digits from the set $$\{ 1,2,3,4,5,6,7,8,9\}$$ are placed in the squares in the figure shown so that the sum of the entries in the vertical column is 23 and the sum of the entries in the horizontal row is 12. The sum of the six digits used is $[asy] unitsize(18); draw((0,0)--(1,0)--(1,1)--(4,1)--(4,2)--(1,2)--(1,3)--(0,3)--cycle); draw((0,1)--(1,1)--(1,2)--(0,2)); draw((2,1)--(2,2)); draw((3,1)--(3,2)); label("23",(0.5,0),S); label("12",(4,1.5),E); [/asy]$ $\text{(A)}\ 27 \qquad \text{(B)}\ 29 \qquad \text{(C)}\ 31 \qquad \text{(D)}\ 33 \qquad \text{(E)}\ 35$

## Solution

Looking at the vertical column, the three numbers sum to $23$. If we make the numbers on either end $9$ and $8$ in some order, the middle number will be $6$. This is the minimum for the intersection.

Looking at the horizontal row, the four numbers sum to $12$. If we minimize the three numbers on the right to $123$, the first number has a maximum value of $6$. This is the maximum for the intersection

Thus, the minimum of the intersection is $6$, and the maximum of the intersection is $6$. This means the intersection must be $6$, and the other numbers must be $9$ and $8$ in the column, and $123$ in the row. The sum of all the numbers is $12 + 23 - 6 = 29$, and the answer is $\boxed{B}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 