1985 AIME Problems/Problem 8
Problem
The sum of the following seven numbers is exactly 19: , , , , , , . It is desired to replace each by an integer approximation , , so that the sum of the 's is also 19 and so that , the maximum of the "errors" , the maximum absolute value of the difference, is as small as possible. For this minimum , what is ?
Solution
If any of the approximations is less than 2 or more than 3, the error associated with that term will be larger than 1, so the largest error will be larger than 1. However, if all of the are 2 or 3, the largest error will be less than 1. So in the best case, we write 19 as a sum of 7 numbers, each of which is 2 or 3. Then there must be five 3s and two 2s. It is clear that in the best appoximation, the two 2s will be used to approximate the two smallest of the , so our approximations are and and the largest error is , so the answer is .
See also
1985 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |