1985 AIME Problems/Problem 8

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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 $M$, and the rest would just be a piece of cake.


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 $\boxed{061}$.

See also

1985 AIME (ProblemsAnswer KeyResources)
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