1993 AJHSME Problems/Problem 14
Revision as of 11:43, 11 August 2021 by Mathfun1000 (talk | contribs) (Fixing the second table in the solution)
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. 
