Difference between revisions of "1997 AHSME Problems/Problem 16"
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<cmath> \left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & 5 & 7\end{matrix}\right] </cmath> | <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 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>.) |
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.) | 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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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>. | 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 14:22, 9 August 2011
Problem
The three row sums and the three column sums of the array
![\[\left[\begin{matrix}4 & 9 & 2\\ 8 & 1 & 6\\ 3 & 5 & 7\end{matrix}\right]\]](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, the sum in row 
is the same as in 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 | ||
