# 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?

For $$a^b=1$$ we have three different scenarios:

1. $$a=1$$ and $$b$$ is any number
2. $$a=-1$$ and $$b$$ is even
3. $$b=0$$ 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 $$2^3\times3^3\times5\times7$$