10. Let's denote with 40! the number obtained by multiplying all integers from 1 to 40, that is $40! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 38 \cdot 39 \cdot 40$. Among the integers greater than 40 that are divisors of 40!, find the five smallest and indicate their sum.
(A) 225
(B) 215
(C) 219
(D) 217
(E) 223
Problem 10. The answer is (A).

The 5 smallest divisors of 40! greater than 40 are: $42, 44, 45, 46, 48$, all of which factorize with factors that we find among the numbers to be multiplied to obtain 40! (for example $44=4 \cdot 11$, $46=2 \cdot 23$, etc.). Conversely, prime numbers greater than 40 do not divide 40!, since a prime number divides a product of integers if and only if it divides one of the factors, and clearly none of the integers that are multiplied to obtain 40! is divisible by a prime number greater than 40. Therefore, the required value is found by adding the 5 numbers listed above, which is 225. (Problem proposed by P. Francini)