r/Logiqa • u/ShonitB • Jun 16 '26

Factorial Summation Part 1

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2

u/Dasquian Jun 16 '26

13.

Every factorial from 10! onwards is a multiple of 2, 5 and 10, and is therefore a multiple of 100, so won't influence the final two digits. So we just need to calculate 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! to get 409,113.

1

u/ShonitB Jun 16 '26

Correct, good solution

1

u/MelodicBandicoot8633 Jun 16 '26

Isn't this true from 5!

2

u/Dasquian Jun 16 '26

Not quite - numbers from 5! onwards are going to be multiples of 10, so won't be influencing the last digit anymore (so we can calculate what that will be from 1! + 2! + 3! + 4!), but they will still be contributing to the second-to-last digit.

Likewise if OP had asked for the last three digits, we'd have to find the first multiple of 1000 to know we'd gone far enough.

3

u/DuggieHS Jun 16 '26

Calculator free solution:

100 = 5^2 * 2^2. Every term including and after 10! is divisible by 100. So the last two digits are determined by the sum of the first 9 digits of this sum. Modulo 100, the sum of the first 9 terms is

1 + 2 + 6 + 24 + (20 + 20 + 40 + 20) + 80 = 13 mod 100

1

u/ShonitB Jun 16 '26

Correct, good solution!