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Difference between revisions of "1997 AHSME Problems/Problem 16"

(Created page with "==Problem== The three row sums and the three column sums of the array <cmath> \left[\begin{matrix}4 & 9 & 2\\ 8 & 1 & 6\\ 3 & 5 & 7\end{matrix}\right] </cmath> are the same. ...")
 
 
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are the same. What is the least number of entries that must be altered to make all six sums different from one another?
 
are the same. What is the least number of entries that must be altered to make all six sums different from one another?
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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>
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==Solution==
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If you change <math>3</math> numbers, then you either change one number in each column and row (ie sudoku-style):
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<cmath> \left[\begin{matrix}* & 9 & 2\\ 8 & * & 6\\ 3 & 5 & *\end{matrix}\right] </cmath>
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Or you leave at least one row and one column unchanged:
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<cmath> \left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & 5 & 7\end{matrix}\right] </cmath>
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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 <math>x</math> is the same as in column <math>x</math>.)
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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.)
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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:
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<cmath> \left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & * & 7\end{matrix}\right] </cmath>
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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>.
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== See also ==
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{{AHSME box|year=1997|num-b=15|num-a=17}}
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{{MAA Notice}}

Latest revision as of 14:13, 5 July 2013

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 $\boxed{D}$.

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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