Puzzles

Incredibly, the result will always be \(n^2\).

To see why, imagine writing every number, \(n+1\leq k\leq 2n\), in the form $$k=2^ab$$ where \(b\) is an odd number and also the \(k\)'s largest odd factor. The next largest number whose largest odd factor is \(b\) will be \(2^{a+1}b=2k\). But this will be larger than \(2n\), so outside the range. Therefore each number in the range has a different largest odd factor.

Each of the largest odd factors must be one of \(1, 3, 5, ..., 2n-1\), as they cannot be larger than \(2n\). But there are \(n\) odd numbers here and \(n\) numbers in the range, so each number \(1, 3, 5, ..., 2n-1\) is the highest odd factor of one of the numbers (as the highest odd factors are all different).

Therefore, the sum of the odd factors is the sum of the first \(n\) odd numbers, which is \(n^2\).

If \(n^2\) has all odd digits then the units digit of \(n\) must be odd. It can be checked that \(n\) cannot be a one digit number (except 1 or 3 as given in the question) as the tens digit will be even.

Therefore \(n\) can be written as \(10A+B\) where \(A\) is a positive integer and \(B\) is an odd positive integer.

Now consider the tens digit of this.

\(100A\) has no effect on this digit. The tens digit of \(20AB\) will be the units digit of \(2AB\) which will be even. The tens digit of \(B^2\) is even (as checked above). Therefore the tens digit of \(n^2\) is even.

Hence 1 and 9 are the only square numbers where all the digits are odd.

For which bases is this not true?

They are all equal to one third.

The sum of the first \(n\) odd numbers is \(n^2\) (this can be proved by induction). This means that:

What is \(\frac{\mathrm{sum\ of\ the\ first\ }n\mathrm{\ odd\ numbers}}{\mathrm{sum\ of\ the\ first\ }n\mathrm{\ even\ numbers}}\)?