1999 AMC 8 Problems/Problem 11
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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
(A) 20 (B) 21 (C) 22 (D) 24 (E) 30
splution
(D) 24: The largest sum occurs when 13 is placed in the center. This sum is 13 + 10 + 1 = 13 + 7 + 4 = 24. Note: Two other common sums, 18 and 21, are possible.
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 | ||
