Question

The sum of the following seven numbers is exactly $19:$ $\begin{aligned} a_1&=2.56,\\ a_2&=2.61,\\ a_3&=2.65,\\ a_4&=2.71,\\ a_5&=2.79,\\ a_6&=2.82,\\ a_7&=2.86. \end{aligned}$ It is desired to replace each $a_i$ by an integer approximation $A_i,$ $1\leq i\leq7,$ so that the sum of the $A_i$ ’s is also $19$ and so that $M,$ the maximum of the “errors” $|A_i-a_i|,$ is as small as possible. For this minimum $M,$ what is $100M?$

Answer

Answer: 61

Concepts:estimation optimization extremal argument

Difficulty rating: 2160

Small Hint:

Starting from seven $3$ ’s, the integer sum must be reduced by $2$

Big Hint:

To minimize the worst error, round the two smallest $a_i$ ’s down to $2$

Solution:

Choose $A_1=A_2=2$ and $A_3=\cdots=A_7=3.$ The sum is $19,$ and the largest error is $|A_2-a_2|=0.61.$

If $M\lt0.61,$ then $A_2,\ldots,A_7$ must all equal $3,$ and $A_1$ can only be $2$ or $3.$ Their sum would therefore be at least $20,$ a contradiction. Thus the minimum is $M=0.61,$ and $100M=61.$

1985 AIME Problem 8

Answer: 61

Difficulty rating: 2160

Small Hint:

Big Hint:

Solution:

Problem 8 in Other Years