1989 AJHSME Problems/Problem 20

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Problem

The figure may be folded along the lines shown to form a number cube. Three number faces come together at each corner of the cube. What is the largest sum of three numbers whose faces come together at a corner?

[asy] draw((0,0)--(0,1)--(1,1)--(1,2)--(2,2)--(2,1)--(4,1)--(4,0)--(2,0)--(2,-1)--(1,-1)--(1,0)--cycle); draw((1,0)--(1,1)--(2,1)--(2,0)--cycle); draw((3,1)--(3,0)); label("$1$",(1.5,1.25),N); label("$2$",(1.5,.25),N); label("$3$",(1.5,-.75),N); label("$4$",(2.5,.25),N); label("$5$",(3.5,.25),N); label("$6$",(.5,.25),N); [/asy]

$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

Solution

It is clear that $6$, $5$, and $4$ will not come together to get a sum of $15$.

The faces $6$, $5$, and $3$ come together at a common vertex, making the maximal sum $6+5+3=14\rightarrow \boxed{\text{D}}$.

See Also

1989 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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
All AJHSME/AMC 8 Problems and Solutions