# Difference between revisions of "1990 AJHSME Problems/Problem 11"

## Problem

The numbers on the faces of this cube are consecutive whole numbers. The sums of the two numbers on each of the three pairs of opposite faces are equal. The sum of the six numbers on this cube is $[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5,2)--(5,5)--(2,5)--(0,3)); draw((3,3)--(5,5)); label("15",(1.5,1.2),N); label("11",(4,2.3),N); label("14",(2.5,3.7),N); [/asy]$ $\text{(A)}\ 75 \qquad \text{(B)}\ 76 \qquad \text{(C)}\ 78 \qquad \text{(D)}\ 80 \qquad \text{(E)}\ 81$

## Solution

The only possibilities for the numbers are $11,12,13,14,15,16$ and $10,11,12,13,14,15$.

In the second case, the common sum would be $(10+11+12+13+14+15)/3=25$, so $11$ must be paired with $14$, which it isn't.

Thus, the only possibility is the first case and the sum of the six numbers is $81\rightarrow \boxed{\text{E}}$.

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