Difference between revisions of "1999 AMC 8 Problems/Problem 11"
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− | Problem | + | ==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 | 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> | ||
+ | |||
+ | <math>\text{(A)}\ 20 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 22 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 30</math> | ||
+ | |||
+ | ==Solution== | ||
+ | ===Solution 1=== | ||
+ | The largest sum occurs when <math>13</math> is placed in the center. This sum is <math>13 + 10 + 1 = 13 + 7 + 4 = \boxed{\text{(D)}\ 24}</math>. Note: Two other common sums, <math>18</math> and <math>21</math>, are also possible. | ||
+ | ===Solution 2=== | ||
− | <math> | + | 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 | |
− | + | <math>13</math>, the sum is the largest, <cmath>[10 + 4 + 1 + 7 + 2(13)]=2S\\ 48=2S\\ S=\boxed{\text{(D)}\ 24}</cmath> | |
+ | The other | ||
+ | four numbers are divided into two pairs with equal sums. | ||
+ | ==See Also== | ||
− | + | {{AMC8 box|year=1999|num-b=10|num-a=12}} | |
+ | {{MAA Notice}} |
Revision as of 23:34, 4 July 2013
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
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, 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.