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1990 AJHSME Problems/Problem 11

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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

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....

\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)/6=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}}.

See Also

1990 AJHSME (ProblemsResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
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