1999 AMC 8 Problems/Problem 11
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
solution
(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.
OR
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 13, the sum is the largest, [10 + 4 + 1 + 7 + 2(13)]=2 = 48=2 = 24. 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 | ||
