AIME 1985 · 第 8 题

AIME 1985 — Problem 8

专题: Contest Math
难度: L4
来源: AIME

题目详情

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

解析

Solution 1

Let $\{x\}$ denote the fractional part of $x$ .

Because $\{2.86\} - \{2.56\} = 0.3$ , we see that the maximum spread of the numbers is $0.3$ , and the median spread of the numbers is $0.15$ .

Now, we take inspiration from rounding decimals. Note how if we have one positive integer $a$ and another positive integer $a+1$ , then $a+1-a=1$ , and the median spread is $1/2$ , or $0.5$ . Then, the following piecewise function is acquired. If a number $n$ satisfies $a < n < a+1$ , then if $\{n\} < 0.5$ , we take $\lfloor{n} \rfloor$ , and if $\{n\} \ge 0.5$ , we take $\lceil{n} \rceil$ .

Applying this to our scenario, we have $2 < a_i < 3$ and the median spread is $2.5$ . So, to determine a judgement in our scenario, we use the following piecewise function of $A_i$ .

If $\{a_i\} < 2.5 + 0.15 = 2.65$ , then take $\lfloor{a_i} \rfloor$ , and if $\{a_i\} \ge 2.65$ , take $\lceil{a_i} \rceil$ .

For all 7 numbers, we apply these and get 2, 2, 3, 3, 3, 3, 3. Adding them results in $3 \cdot 5 + 4 = 19$ , which satisfies the constraint, and the greatest value of $M$ is determined from $2.61 - \lfloor{2.61} \rfloor = 0.61$ .

This is guarenteed as notice that all $a_i > 2.5$ , and in general the rounding principle states that if a number $n$ has the property that $\{n\} < \{2.5\}$ , then take $\lfloor{n} \rfloor$ and if $\{n\} \ge \{2.5\}$ , then take $\lceil{n} \rceil$ , in which we can adjust using the same principle here.

In addition to this problem, this idea is reminiscent for ALL problems of this category, so one additional concept to add to your problem solving toolkit.

Thus, $M = 0.61$ , and $100M = 100 \times 0.61 =$ $\boxed{061}$

~Pinotation

Solution 2

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 $\boxed{061}$ .