Puzzles

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.

Which other numbers can 37 be replaced with leaving a puzzle that still has a unique solution?

SAM is 781.

30030 is 2×3×5×7×11×13, so the answer is the number of ways of sharing six elements into unlabelled boxes (A000110).

The answer is 203.

For any digit \(V\), \(VVV\) is a multiple of 111, and \(111=37\times3\). 37 is prime, so \(VI\) must be a multiple of 37 (as \(X\) is less than 10 so cannot be a multiple of 37). Therefore \(VI\) is either 37 or 74.

If \(VI\) was 74, then \(VVV\) is 777. But then \(VI\times X\) is even and \(VVV\) is odd. This is impossible, so \(VI\) must the 37.

Now that we know that \(VI\) is 37, we see that

and so \(X=9\) and the final solution is

730

The number can contain no 1 (as all numbers are multiples of 1). It's first and last digits cannot be 2 (as it would then be even and so a multiple of 2).

The lowest three-digit palindrome containing no 0 or 1 and not ending in 2 is 323. This is not a multiple of 2 or 3 and so is the solution.

The squarefree numbered lockers will be locked. (Squarefree number are numbers that are not divisible by any square number apart from 1.) There are 608 squarefree numbers between 1 and 1000.

The square numbered lockers will be open at the end of the process. There are 969 non-square numbers between 1 and 1000, so this many lockers will be closed.