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Difference between revisions of "1997 AHSME Problems/Problem 16"
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<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math>
 
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math>
  
 +
==Solution==
 +
 +
If you change <math>3</math> numbers, then you either change one number in each column and row (ie sudoku-style):
 +
 +
<cmath> \left[\begin{matrix}* & 9 & 2\\ 8 & * & 6\\ 3 & 5 & *\end{matrix}\right] </cmath>
 +
 +
Or you leave at least one row and one column unchanged:
 +
 +
<cmath> \left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & 5 & 7\end{matrix}\right] </cmath>
 +
 +
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 <math>x</math> is the same as column <math>x</math>.)
 +
 +
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 <math>3</math> and column <math>3</math> are untouched.)
 +
 +
Either way, <math>3</math> 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:
 +
 +
<cmath> \left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & * & 7\end{matrix}\right] </cmath>
 +
 +
Letting the <math>*</math> be a zero does indeed give <math>6</math> different sums, so the answer is <math>4</math>, which is option <math>\boxed{D}</math>.
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1997|num-b=15|num-a=17}}
 
{{AHSME box|year=1997|num-b=15|num-a=17}}

Revision as of 14:21, 9 August 2011

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, row $x$ is the same as 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 $\boxed{D}$.

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
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