Average digit sum – 4

Number Theory Level 4

What is the average of the digit sums $s(n)$ of all non-negative multiples of 11 that are less than $10^6$ (90910 numbers)?

In other words, evaluate

$\frac 1{90910} \displaystyle \sum_{n=0}^{90909} s(11n)$

Other parts

Average digit sum – 5
Average digit sum – 3

Note

The digit sum $s(n)$ of a number $n$ is the sum of all digits of $n$ . For example, $s(2019)=2+0+1+9=12$ . In this problem, we will only consider base 10.

The answer is 27.

2 solutions

Chris Lewis
Mar 4, 2019

If $x$ is a multiple of $11$ , so is $999999-x$ . If both members of this pair are positive, then the sum of their digit sums will always be $54$ . It follows that the average digit sum is half of this, or $\boxed{27}$ .

Kyle T
Mar 7, 2019

<?php
$arr = array();
for($i=11;$i<=pow(10,6);$i+=11){
$arr[] = array_sum(str_split($i));
}
echo array_sum($arr)/count($arr); //27
?>

Average digit sum – 4

The answer is 27.

2 solutions

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