
1. There are different positive integers written on the board. Their (arithmetic) mean is a decimal number, with the decimal part exactly 0,2016. What is the least possible value of the mean?

(Patrik Bak)


Solution. Let $s$ be the sum, $n$ the number and $p$ the integer part of the mean of the numbers on the board. Then we can write

$$
\frac{s}{n}=p+\frac{2016}{10000}=p+\frac{126}{625}
$$

which gives

$$
625(s-p n)=126 n \text {. }
$$

Numbers 126 and 625 are coprime, thus $625 \mid n$. Therefore $n \geqslant 625$.

The numbers on the board are different, that is

$p=\frac{s}{n}-\frac{126}{625} \geqslant \frac{1+2+\cdots+n}{n}-\frac{126}{625}=\frac{n(n+1) / 2}{n}-\frac{126}{625} \geqslant \frac{625+1}{2}-\frac{126}{625}>312$.

The integer $p$ is thus at least 313 and the value of the mean at least 313,2016 .

This value can be attained by numbers $1,2, \ldots, 624$ and 751 . We get

$$
\frac{1+2+\cdots+624+751}{625}=\frac{312 \cdot 625+751}{625}=313+\frac{126}{625}=313,2016 .
$$
