1997 AHSME Problems/Problem 16
Problem
The three row sums and the three column sums of the array
![\[\left[\begin{matrix}4 & 9 & 2\\ 8 & 1 & 6\\ 3 & 5 & 7\end{matrix}\right]\]](http://latex.artofproblemsolving.com/a/b/e/abe9ce134dc17ff83eff2313ff77c9b7ea7e5289.png)
are the same. What is the least number of entries that must be altered to make all six sums different from one another?
Solution
If you change 
numbers, then you either change one number in each column and row (ie sudoku-style):
![\[\left[\begin{matrix}* & 9 & 2\\ 8 & * & 6\\ 3 & 5 & *\end{matrix}\right]\]](http://latex.artofproblemsolving.com/3/a/1/3a1bebd3f81fc24c316f18c18cbfdec5e4c3c9a1.png)
Or you leave at least one row and one column unchanged:
![\[\left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & 5 & 7\end{matrix}\right]\]](http://latex.artofproblemsolving.com/5/3/7/537daac744439fadd38eaab2323d7386ce55a813.png)
In the first case, you are changing just one common number in two sums, so you wind up with three pairs of sums. (In the example given, row 
is the same as column .)
In the second case, since two of the sums are unchanged, and the sums started out equal, they must remain equal. (In the second example given, row and column are untouched.)
Either way, changes is not enough. However, building on the second example, if you change either the untouched column or the untouched row, you will get a possible answer:
![\[\left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & * & 7\end{matrix}\right]\]](http://latex.artofproblemsolving.com/3/b/f/3bfc1d4190ab83c614b737513c52e702f34c301f.png)
Letting the 
be a zero does indeed give 
different sums, so the answer is 
, which is option 
.
See also
| 1997 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
