# Practice Paper 1 Question 4

What is the units digit of the number \(\sum_{n=1}^{1337}(n!)^4\)

## Related topics

## Hints

- Hint 1What can you say about the units digit of \(n!\) for \(n\ge5\)?
- Hint 2What about the units digits for \(n < 5\)?
- Hint 3It's sufficient to take only the units digit when multiplying or raising the number to any power.

## Solution

All factorials greater that \(5!\) have both the factors \(5\) and \(2\), hence the units digit equal to 0. This means that we only need to worry about \(n \in \{1,2,3,4\}\). It's sufficient to take only the units digit when multiplying or raising to any power.

We have:

- \(1^4\rightarrow1\)
- \((2!)^4=2^4\rightarrow6\)
- \((3!)^4=6^4\rightarrow6\)
- \((4!)^4=24^4\rightarrow4^4\rightarrow6\)

The units digit for the sum is therefore the units digit of \(1+6+6+6\), i.e. \(9\).

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