Difference between revisions of "1996 AJHSME Problems/Problem 14"
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* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
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Latest revision as of 23:24, 4 July 2013
Problem
Six different digits from the set
![\[\{ 1,2,3,4,5,6,7,8,9\}\]](http://latex.artofproblemsolving.com/4/5/6/4565e9ab182fabdfa76eaec506300115b3de96bf.png)
are placed in the squares in the figure shown so that the sum of the entries in the vertical column is 23 and the sum of the entries in the horizontal row is 12.
The sum of the six digits used is
![[asy] unitsize(18); draw((0,0)--(1,0)--(1,1)--(4,1)--(4,2)--(1,2)--(1,3)--(0,3)--cycle); draw((0,1)--(1,1)--(1,2)--(0,2)); draw((2,1)--(2,2)); draw((3,1)--(3,2)); label("$23$",(0.5,0),S); label("$12$",(4,1.5),E); [/asy]](http://latex.artofproblemsolving.com/d/f/5/df58d3f5c6f96c4c683cc96e62e3ea5bceab26ed.png)
Solution
Looking at the vertical column, the three numbers sum to 
. If we make the numbers on either end 
and 
in some order, the middle number will be 
. This is the minimum for the intersection.
Looking at the horizontal row, the four numbers sum to 
. If we minimize the three numbers on the right to 
, the first number has a maximum value of . This is the maximum for the intersection
Thus, the minimum of the intersection is , and the maximum of the intersection is . This means the intersection must be , and the other numbers must be and in the column, and in the row. The sum of all the numbers is 
, and the answer is 
See Also
| 1996 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
