Puzzles

The numbers 10, 20, ..., 1000 end with a 0. There are 100 of these numbers.

The numbers 100, 101, ..., 109, 200, ..., 209, 300, ..., 309, ..., 908, 909, 1000 contain a 0 in their tens column. There are \(10\times9+1=91\) of these numbers.

The number 1000 has a 0 in its hundreds column.

In total, there are 100+91+1=192 zeros.

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?

The geometric mean of a number \(N\) is always \(\sqrt{N}\). You can see why this is true by considering the pairs of factors that multiply to make the number: if \(N\) if not square and has \(k\) pairs of factors, then the product of these factors is \(N^k\), so the geometric mean is \((N^k)^{1/(2k)}=N^{1/2}\); if \(N\) is square and has \(k\) pairs of factors plus the square root, then the product of its factors is \(N^k\sqrt{N}=N^{k+1/2}\), so the geometric mean is \((N^{k+1/2})^{1/(2k+1)}=N^{1/2}\). Interestingly, this means that the geometric mean of the factors of a number is an integer only when the number is square.

Therefore the only number whose factors have a geometric mean of 22 is \(22^2\), or 484.

Santa's house could be found by following the directions 3,7,1,1,4,3,3,4,9,3,3,9.

Either both pieces must be on the diagonal, or one pieces is in the lower right half and the other is in the reflected position in the upper right half.

There are \(\left(\begin{array}{c}29\\2\end{array}\right)=406\) ways to pick two squares on the diagonal. There are 406 squares below the diagonal.

Therefore there are 406+406 = 812 ways to arrange the pieces.

We're looking for a number \(n\) such that the digits of \(n\times48\) add up to \(n\). If we try numbers in order, we find that 9 works, so today's number is 432.

The largest number you can make with the digits in the red boxes is 984.

To make sequences of 4 numbers, we can insert 4s into three different places in the length 3 sequences given to obtain:

There are some possibilities missing: those containing \(i, 4, i+1\). These can be found by taking the sequences of length 2, picking a number \(i\), adding 1 to every number larger than \(i\), then replacing \(i\) with \(i\ 4\ i+1\).

This gives a total of 3×3+2×1=11 sequences for 4 numbers.

To make sequences with 5 numbers, we can insert 5s into four different places in the length 4 sequences. This gives 4×11=44 sequences.

The missing sequences can then be found by taking the sequences of length 3, then doing the same process as above:

There are 3×3=9 of these, giving 44+9 = 53 total sequences of length 5.

To make sequences with 6 numbers, we can insert 6s into five different places in the length 5 sequences. This gives 5×53=265 sequences. We can also make sequence by picking a number to replace in the length 4 sequences. This gives 4×11=44 more sequences. Therefore there are 265+44 = 309 sequences in total.