## Task 4 - 330614

If $n$ is a natural number, the product of all natural numbers from 1 to $n$ is denoted by $n$! (read: "n factorial"). For example, $4! = 1 \cdot 2 \cdot 3 \cdot 4 = 24$.

What are the last three digits of the number that would result from calculating

$$
1! + 2! + 3! + 4! + \ldots + 99! + 100!
$$
By starting with $1!=1$ and successively multiplying the last obtained result by $2,3,4, \ldots$ etc., and each time only considering the last three digits, one finds:

$$
1!=1, \quad 2!=2, \quad 3!=6, \quad 4!=24, \quad 5!=120, \quad 6!=720
$$

the last three digits of 7! are ...040 the last three digits of 8! are ...320 the last three digits of 9! are ...880 the last three digits of 10! are ...800 the last three digits of 11! are ...800 the last three digits of 12! are ...600 the last three digits of 13! are ...800 the last three digits of 14! are ...200

From 15! onwards, the last three digits are always ...000, and thus do not change the result in the required addition. Therefore, by adding the results listed here, we already get:

The last three digits of $1!+2!+\ldots+100!$ are ...313

Solutions of the I. Round 1993 taken from [5]

### 3.34.2 II. Round 1993, Class 6