Difference between revisions of "1993 AJHSME Problems/Problem 14"

(Problem)
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<math>\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6</math>
 
<math>\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6</math>
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==Solution==
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The square connected both to 1 and 2 cannot be the same as either of them, so must be 3.
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<cmath> \begin{tabular}{|c|c|c|}\hline 1 & 3 &\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular} </cmath>
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The last square in the top row cannot be either 1 or 3, so it must be 2.
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<cmath> \begin{tabular}{|c|c|c|}\hline 1 & 3 & 2\\ \hline & 2 &\\ \hline & & B\\ \hline\end{tabular} </cmath>
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The other two squares in the rightmost column with A and B cannot be two, so they must be 1 and 3 and therefore have a sum of <math>1+3=\boxed{\text{(C)}\ 4}</math>.
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==See Also==
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{{AJHSME box|year=1993|num-b=13|num-a=15}}

Revision as of 21:45, 22 December 2012

Problem

The nine squares in the table shown are to be filled so that every row and every column contains each of the numbers $1,2,3$. Then $A+B=$

\[\begin{tabular}{|c|c|c|}\hline 1 & &\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]

$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution

The square connected both to 1 and 2 cannot be the same as either of them, so must be 3.

\[\begin{tabular}{|c|c|c|}\hline 1 & 3 &\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]

The last square in the top row cannot be either 1 or 3, so it must be 2.

\[\begin{tabular}{|c|c|c|}\hline 1 & 3 & 2\\ \hline & 2 &\\ \hline & & B\\ \hline\end{tabular}\]

The other two squares in the rightmost column with A and B cannot be two, so they must be 1 and 3 and therefore have a sum of $1+3=\boxed{\text{(C)}\ 4}$.

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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
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