Advent calendar 2021
2 December
The number \(7n\) has 37 factors (including 1 and the number itself). How many factors does \(8n\) have?
There was a typo in this puzzle. It originally read "38 factors" when it was meant to say "37 factors".
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If \(p_1\), \(p_2\), etc are distinct prime numbers, then the factors of the number \(N=p_1^{a_1}p_2^{a_2}...\) all have the form \(p_1^{b_1}p_2^{b_2}...\),
where \(b_1\) is 0 or 1 or ... or \(a_1\) (and similar conditions for \(a_2\), \(a_3\), etc). There are therefore \(a_1+1\) possible values of \(b_1\), \(a_2+1\) possible values
of \(b_2), and so on. Therefore, \(N\) has \((a_1+1)(a_2+1)(...)\) factors.
From this we see that the only way to have a number with 37 factors is if it is a prime number to the power of 36, and so \(n=7^{35}\).
\(8n\) is therefore equal to \(2^3\times7^{35}\). This has 4×36=144 factors.
Extension
Which other numbers can 37 be replaced with leaving a puzzle that still has a unique solution?
