What is the unit digit of $1^{1!} + 2^{2!} + 3^{3!} + ... + 9^{9!}$ ?

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Explanation

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Official Explanation

Official

Solution

To find: The unit digit of the expression

1^{1!} + 2^{2!} + 3^{3!} + … + 9^{9!}

Approach and Working:

Each digit has its own unit digit cycle (the maximum cycle is 4 for any number). Following are the unit digits with their corresponding cycle:

- For 1: 1 (cycle 1)

- For 2: 2 4 8 6 (cycle 4)

- For 3: 3 9 7 1 (cycle 4)

- For 4: 4 6 (cycle 2)

- For 5: 5 (cycle 1)

- For 6: 6 (cycle 1)

- For 7: 7 9 3 1 (cycle 4)…

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What this question tests

What it tests

Your understanding of divisibility rules, least common multiples, and greatest common factors.

Common trap

Forgetting that 1 is not prime and that every integer divides 0.

What is the unit digit of 11!+22!+33!+...+99!1^{1!} + 2^{2!} + 3^{3!} + ... + 9^{9!}11!+22!+33!+...+99! ?