19. After drawing a triangular scheme like the one next to it, Roberto writes the integers from 1 to 8 in the circles of the top row. Then, in each of the other circles, he writes the product of the numbers contained in the two circles above it that are connected to it with an arrow (thus he gets 2 $6, 12, \ldots$ and so on). How many zeros will the number he has to write in the bottom circle end with?
(A) 35
(B) 36
(C) 34
(D) 32
(E) 33

![](https://cdn.mathpix.com/cropped/2024_04_17_e21ecd5de9c42efa6d1eg-2.jpg?height=437&width=425&top_left_y=781&top_left_x=2316)
(19) The correct answer is $(A)$.

The number of trailing zeros of a number is the minimum between the exponent of 2 and that of 5 in its prime factorization. Let's start by calculating the exponent of 5 present in the prime factorization of each number written, remembering that in a product the exponents add up. We know that in the first row the exponents of 5 are $0,0,0,0,1,0,0,0$, and therefore in the second $0,0,0,1,1,0,0$; in the third $0,0,1,2,1,0$; in the fourth $0,1,3,3,1$; in the fifth $1,4,6,4$; in the sixth $5,10,10$; in the seventh 15,20; finally in the eighth 35. Proceeding in the same way with the first 2, it is easily seen that the power of 2 that divides the last number is well above $2^{35}$. The number written in the lowest circle therefore ends with exactly 35 zeros.

Problem proposed by Giovanni Barbarino.