Difference between revisions of "1993 AJHSME Problems/Problem 14"
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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> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{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 . Then
Solution
The square connected both to 1 and 2 cannot be the same as either of them, so must be 3.
The last square in the top row cannot be either 1 or 3, so it must be 2.
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 .
See Also
1993 AJHSME (Problems • Answer Key • Resources) | ||
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 |