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# Practice Paper 1 Question 4

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

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## Hints

• Hint 1
What can you say about the units digit of $$n!$$ for $$n\ge5$$?
• Hint 2
What about the units digits for $$n < 5$$?
• Hint 3
It'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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