Difference between revisions of "1993 AJHSME Problems/Problem 14"
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==Problem== | ==Problem== | ||
| + | |||
The nine squares in the table shown are to be filled so that every row and every column contains each of the numbers <math>1,2,3</math>. Then <math>A+B=</math> | The nine squares in the table shown are to be filled so that every row and every column contains each of the numbers <math>1,2,3</math>. Then <math>A+B=</math> | ||
| − | + | <cmath> \begin{tabular}{|c|c|c|}\hline 1 & &\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular} </cmath> | |
| − | 1 & & \\ \hline | ||
| − | |||
| − | |||
| − | \end{tabular} | ||
<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. | ||
| + | |||
| + | <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. | ||
| + | |||
| + | <cmath> \begin{tabular}{|c|c|c|}\hline 1 & 3 & 2\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular} </cmath> | ||
| + | |||
| + | 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}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:43, 11 August 2021
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 
![\[\begin{tabular}{|c|c|c|}\hline 1 & &\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]](http://latex.artofproblemsolving.com/7/1/d/71d34d1b77e1c00605daec63e973a4c807800789.png)
Solution
The square connected both to 1 and 2 cannot be the same as either of them, so must be 3.
![\[\begin{tabular}{|c|c|c|}\hline 1 & 3 &\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]](http://latex.artofproblemsolving.com/2/7/3/2733469d5b47a5b1c8ab4f2e7d6159bab3f4dd48.png)
The last square in the top row cannot be either 1 or 3, so it must be 2.
![\[\begin{tabular}{|c|c|c|}\hline 1 & 3 & 2\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]](http://latex.artofproblemsolving.com/5/9/c/59ce0aaadc45387c1c3b79e9363cab4c05c560e7.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
