Difference between revisions of "1993 AJHSME Problems/Problem 14"
Mathfun1000 (talk | contribs) m (Fixing the second table in the solution) |
|||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 16: | Line 16: | ||
The last square in the top row cannot be either 1 or 3, so it must be 2. | 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 &\\ \hline & & B\\ \hline\end{tabular} </cmath> | + | <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>. | 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>. | ||
| Line 22: | Line 22: | ||
==See Also== | ==See Also== | ||
{{AJHSME box|year=1993|num-b=13|num-a=15}} | {{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. 
