Difference between revisions of "1999 AMC 8 Problems/Problem 11"
(Undo revision 136967 by Mangolassi (talk)) (Tag: Undo) |
|||
| (3 intermediate revisions by 3 users not shown) | |||
| Line 22: | Line 22: | ||
four numbers are divided into two pairs with equal sums. | four numbers are divided into two pairs with equal sums. | ||
| − | == | + | ==See Also== |
{{AMC8 box|year=1999|num-b=10|num-a=12}} | {{AMC8 box|year=1999|num-b=10|num-a=12}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 17:23, 7 November 2020
Problem
Each of the five numbers 1, 4, 7, 10, and 13 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is
![[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw((1,-1)--(2,-1)--(2,2)--(1,2)--cycle); [/asy]](http://latex.artofproblemsolving.com/5/a/6/5a63267c9764e69c4a9d9a68d30bd7e85138a60a.png)
Solution
Solution 1
The largest sum occurs when 
is placed in the center. This sum is 
. Note: Two other common sums, 
and 
, are also possible.
Solution 2
Since the horizontal sum equals the vertical sum, twice this sum will be the sum
of the five numbers plus the number in the center. When the center number is
, the sum is the largest, ![\[[10 + 4 + 1 + 7 + 2(13)]=2S\\ 48=2S\\ S=\boxed{\text{(D)}\ 24}\]](http://latex.artofproblemsolving.com/3/a/f/3af0ca6b4d1929f6bb0d9b6c6052720125a54868.png)
The other
four numbers are divided into two pairs with equal sums.
See Also
| 1999 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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. 
