### A Biggy

Find the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power.

### Why 24?

Take any prime number greater than 3 , square it and subtract one. Working on the building blocks will help you to explain what is special about your results.

### Fac-finding

Lyndon chose this as one of his favourite problems. It is accessible but needs some careful analysis of what is included and what is not. A systematic approach is really helpful.

# Factoring a Million

##### Age 14 to 16 Challenge Level:

No one completed the solution to this problem so we will put it into Tough Nuts for you to return to when you have time. There were several attempts and some of you were able to identify that there are 784 possible arrangements of three factors of a million.

The first step was to identify that 1000 000 = 10 6 = 2 6 x 5 6 and then to consider this as the product of three factors i.e.

 10 6 = 2 6 x 5 6 = 2 a 5 p x 2 b 5 q x 2 c 5 r (where a+b+c = 6 and p+q+r = 6).

However there are repetitions here because 2 3 5 3 x 2 2 5 2 x 2 1 5 1 is the product of the same three factors as 2 2 5 2 x 2 3 5 3 x 2 1 5 1 .

So there is still some work to do! Good luck.

Giles Cooper & Mike Hood thinks there are 139 such factors, he has produced a list of factors for example

1 x 1 x 1000000

1 x 2 x 500000

1 x 4 x 250000

...

...

80 x 100 x 125

100 x 100 x 100

Editors note:

This is correct, however I also liked the attempt by Mike Hood, who tried to use the idea of combinations of factors. This idea seemed to be a less exhaustive'' approach and begins to give a better insight into what is going on. His solution uses the fact that $1,000,000 = 2^6 x 5^6$. I think this is worth pursuing. Think about the number of ways you can combine the six 2s and the six 5s in the three factors. You then only have to consider the number of unique combinations of each of the two sets of arrangements. For example - you could choose $2^6$, $2^0$, $2^0$ and $5^6$, $5^0$, $5^0$ and there are only two unique arrangements of these two sets of factors.