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The Computer Laboratory

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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