Advent calendar 2018

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.

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.

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 vertices of the 26-gon lie on a circle. The triangle is therefore right-angled if (and only if) the longest side is a diameter of the circle. In other words, the triangle is right angled if (and only if) two of its vertices are opposite vertices of the 26-gon.

There are 13 different pairs of opposite points on the 26-gon. For each of these, there are 24 remaining vertices that could be the third vertex of the triangle. Therefore there are 13×24=312 different right angled triangles.

By the same method as this puzzle, the answer is 108.

If the equation has two integer solutions, then it can be written as \((x+\alpha)(x+\beta)\), where \(\alpha\) and \(\beta\) are integers. Expanding this and setting it equal to \(x^2+bx+414720=0\), we find that \(b=\alpha+\beta\) and \(414720=\alpha\beta\).

414720 has 55 different pairs of factors. Each of these pairs leads to two values of \(b\) (a positive and a negative value). Therefore there are 110 values of \(b\) that give the equation two integer solutions.

There are 6 available choices for each digit, so there are 6² and 6³ such two- and three-digit numbers. 6²+6³=252.

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.

121

In general, if you do this with the numbers from \(n+1\) to \(2n\) (inclusive), the result will be \(n^2\) (see this puzzle).

445

By the same method used to solve this puzzle, the answer if 150.

The product of the numbers in the red boxes is 108.

The largest dodecagon will be regular. It's area will be 172.

233

Let \(f(n)\) be the number of ways to write \(n\) as the sum of 1s and 2s. When you write \(n\) as the sum of 1s and 2s, the sum must end in either 1 or 2: if it ends in 1, then there are \(f(n-1)\) ways to write the rest before this 1 as the sum of 1s and 2s; if it ends in 2, there are \(f(n-2)\) ways to write the rest. This means that \(f(n)=f(n-1)+f(n-2)\), or in other words the Fibonacci numbers.