Puzzles

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.

153

In general, the expansion of \((x+y+z)^n\) will have \(\frac12n(n-1)\) terms.

285

If I do this for \(x\) and \(y\) from 1 to \(n\) (inclusive) then the result is the sum of the first \(n\) square numbers.

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 product of the numbers in the red boxes is 135.