1985 AIME Problems/Problem 8

Problem

The sum of the following seven numbers is exactly 19: $a_1 = 2.56$ , $a_2 = 2.61$ , $a_3 = 2.65$ , $a_4 = 2.71$ , $a_5 = 2.79$ , $a_6 = 2.82$ , $a_7 = 2.86$ . It is desired to replace each $a_i$ by an integer approximation $A_i$ , $1\le i \le 7$ , so that the sum of the $A_i$ 's is also 19 and so that $M$ , the maximum of the "errors" $\| A_i-a_i\|$ , the maximum absolute value of the difference, is as small as possible. For this minimum $M$ , what is $100M$ ?

Explanation of the Question

Note: please read the explanation AFTER YOU HAVE TRIED reading the problem but couldn't understand.

For the question. Let's say that you have determined 7-tuple $(A_1,A_2,A_3,A_4,A_5,A_6,A_7)$ . Then you get the absolute values of the $7$ differences. Namely, $|A_1-a_1|, |A_2-a_2|, |A_3-a_3|, |A_4-a_4|, |A_5-a_5|, |A_6-a_6|, |A_7-a_7|$ Then $M$ is the greatest of the $7$ absolute values. So basically you are asked to find the 7-tuple $(A_1,A_2,A_3,A_4,A_5,A_6,A_7)$ with the smallest $M$ .

Solution

If any of the approximations $A_i$ 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 $A_i$ 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 $a_i$ , so our approximations are $A_1 = A_2 = 2$ and $A_3 = A_4 = A_5 = A_6 = A_7 = 3$ and the largest error is $|A_2 - a_2| = 0.61$ , so the answer is $061$ .

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
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1985 AIME Problems/Problem 8

Contents

Problem

Explanation of the Question

Solution

See also