1999 AMC 8 Problem 11
Attempt Problem 11 of the 1999 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
11.
Each of the five numbers and 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
Answer: D
Small hint:
The center square is counted in both sums
Big hint:
Maximize the center number, then split the doubled total equally
Solution:
Let be the number in the center. The horizontal sum plus the vertical sum counts every number once, except the center number is counted twice.
So twice the common sum is . This is largest when .
The largest common sum is therefore at most , and it is attainable: put in the center, and at the ends of one row, and and at the ends of the other. Both sums are .
Thus, D is the correct answer.
Problem 11 in other years
- 1985 AMC 8
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