1985 AIME Problem 8

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

Starting from seven

3

’s, the integer sum must be reduced by

2

To minimize the worst error, round the two smallest

a_i

’s down to

2

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.

Problem 8 in other years