# Maths puzzles

Thursday 9th July 2020

**Puzzle 1**

Can you show that \((x^2-7x+11)^{(x^2-11x+30)}=1\) has five different whole number solutions?

Challenge: is it possible to have a similar equation with six different integer answers? Please send answers to your local Area Coordinator.

For \(a^b=1\) we have three different scenarios:

- \(a=1\) and \(b\) is any number
- \(a=-1\) and \(b\) is even
- \(b=0\) and \(a\neq0\)

**Scenario 1:**

- \(x^2-7x+11=1\)
- \(x^2-7x+10=0\)
- \((x-2)(x-5)=0\)
- \(x=2\) or \(x=5\)

**Scenario 2:**

- \(x^2-7x+11=-1\)
- \(x^2-7x+12=0\)
- \((x-3)(x-4)=0\)
- \(x=3\) or \(x=4\)

We have to check that for those values of \(x\), \(b\) is even:

- for \(x=3\) \(x^2-11x+30=6\) so even
- for \(x=4\) \(x^2-11x+30=2\) so even

**Scenario 3:**

- \(x^2-11x+30=0\)
- \((x-5)(x-6)=0\)
- \(x=5\) or \(x=6\)

We have to check that for those values of \(x\), \(a\neq0\):

- for \(x=5\) \(x^2-7x+11=1\neq0\)
- for \(x=6\) \(x^2-7x+11=5\neq0\)

Five solutions are: \(x=2, 3, 4, 5, 6\)

**Puzzle 2**

\(2\times3\times5\times7\times11\times13\) has 64 distinct factors. Can you explain why, and can you find the prime factorisation for a lower number with 64 distinct factors?

Each prime factor can be in or not in the prime factorisation of a factor, so there are \(2\times2\times2\times2\times2\times2\) possible factors. There are lower possibilities and the lowest is \(2^3\times3^3\times5\times7\)