Skip to main content
### Number and algebra

### Geometry and measure

### Probability and statistics

### Working mathematically

### For younger learners

### Advanced mathematics

# Factorial Fun

### Why do this problem?

The problem does not require much knowledge but it calls for careful reasoning and accurate use of standard notation. The problem provides scaffolding steps to help the problem solver think through the ideas needed to solve the problem.

### Possible approach

Ask the class to count all the factors of $n!$ for $n = 2 , 3, 4$ and $5$ and suggest that they look for the most efficient way to do this. Then suggest that this problem might help them find the best way to do it.

This is not written in to the problem itself because EVERY time we try to solve a problem, unless it is very easy, we should think about first trying simple cases.

### Key questions

How do we use the fact $24=2^3 \times 3$ to deduce that $24$ has $8$ factors? (Combinatorics question)

### Possible support

Try the problems Fac-finding, Powerful Factorial and Factoring Factorials. They are all special cases of this problem.

The problem Em'power'ed also focusses on equating the powers of prime factors.

Or search by topic

Age 16 to 18

Challenge Level

- Problem
- Getting Started
- Student Solutions
- Teachers' Resources

This is not written in to the problem itself because EVERY time we try to solve a problem, unless it is very easy, we should think about first trying simple cases.

If we find the prime factors of $n$ how do we find out how many times these prime factors are repeated in $n!$ ?

The problem Em'power'ed also focusses on equating the powers of prime factors.