Puzzles

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

The timetable was:

The thief was Kip Urples and the robbery took place between 1:21 and 2:52.

157

The largest rectangle will be 11 by 12 and have area 132.

If 1/10890 has a finite number of digits in base \(b\), then \(b^n/10890\) is an integer for some integer \(n\). In order for this to be possible \(b\) must be a mutiple of each prime factor of 10890: the smallest such \(b\) will be when the exponent of each prime factor is 1.

The prime factorisation of 10890 is 2×3³×5×11². Therefore the smallest base in which 1/10890 has a finite number of digits is 2×3×5×11=330.

The smallest number you can make with the digits in the red boxes is 269.

730

Let \(f(a,b)\) be the number of \(a\)-dimensional sides on a \(b\)-dimensional hypercube. A \((b+1)\)-dimensional hypercube is formed by making two copies of a \(b\)-dimensional hypercube and adding edges between the corresponding vertices on the two copies. Therefore the \((b+1)\)-dimensional hypercube will contain two copies of each \(a\)-dimensional side in the \(b\)-dimensional hypercube; it will also contain a new \(a\)-dimensional side formed by each \((a-1)\)-dimensional side of the \(b\)-dimensional hypercube. Therefore \(f(a,b+1)=2f(a,b)+f(a-1,b)\).

(This can be seen more clearly by looking at the 1-dimensional sides (edges) and 2-dimensional sides (faces) on a 3-dimensional hypercube (cube) and a 2-dimensional hypercube (a square).)

Using the above relation, we find that there are 112 6-dimensional sides on a 8-dimensional hypercube.