*This post was part of the Chalkdust 2016 Advent Calendar.*

Welcome to the twelfth day of the 2016 Chalkdust Advent Calendar. Today, we have another puzzle for you to enjoy, plus the answer to the puzzles from 06 December.

Today’s puzzle is taken from Daniel Griller‘s talk at the MathsJam conference earlier this year.

### Odd factors

Pick a number. Call it $n$. Write down all the numbers from $n+1$ to $2n$ (inclusive). Under each of these, write its largest odd factor. What is the sum of these odd factors?

Now for the solutions to the puzzles from 06 December.

### Digital sums

*Source: mscroggs.co.uk Advent calendar, day 6*

When you add up the digits of a number, the result is called the digital sum.

How many different digital sums do the numbers from 1 to 10^{91} have?

As this puzzle is part of a larger advent calendar (with prizes!), I’m not going to give you the answer here!

### Wipeout

*Source: nrich Secondary Advent calendar, day 10*

You are given the numbers 1,2,3,4,5,6 and are allowed to erase one. If you erase 5, the mean of the remaining numbers will be 3.2. Is it possible to erase a number so that the mean of the remaining number is an integer?

If you are given the numbers 1,2,3,4,…,$N$, can you erase one number so that the mean of the remaining numbers is an integer?

For the first part, erasing 6 will leave numbers that sum to 15, with a mean of 3.

For the second part, if $N$ is even, erasing $N$ from the list 1,2,3,4,…,$N$ will leave numbers that sum to $\tfrac12N(N-1)$. $N$ is even, so $\tfrac12N$ is an integer; therefore $N-1$ is a factor of the sum, so the numbers have an integer mean.

If $N$ is odd, removing the middle number from the list leaves an integer mean. I’ll let you work out why this is and will return in a few days with the answer to today’s puzzle…