1997 AHSME Problems/Problem 16

Problem

The three row sums and the three column sums of the array

$\left[\begin{matrix}4 & 9 & 2\\ 8 & 1 & 6\\ 3 & 5 & 7\end{matrix}\right]$

are the same. What is the least number of entries that must be altered to make all six sums different from one another?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution

If you change $3$ numbers, then you either change one number in each column and row (ie sudoku-style):

$\left[\begin{matrix}* & 9 & 2\\ 8 & * & 6\\ 3 & 5 & *\end{matrix}\right]$

Or you leave at least one row and one column unchanged:

$\left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & 5 & 7\end{matrix}\right]$

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, the sum in row $x$ is the same as in column $x$ .)

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 $3$ and column $3$ are untouched.)

Either way, $3$ 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:

$\left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & * & 7\end{matrix}\right]$

Letting the $*$ be a zero does indeed give $6$ different sums, so the answer is $4$ , which is option $\boxed{D}$ .

1997 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1997 AHSME Problems/Problem 16

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