16. After drawing a triangular scheme like the one shown here, Alberto writes the integers from 1 to 7 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 obtaining 2, $6,12, \ldots$ and so on). How many zeros will the number that he will write in the bottom circle end with?
(A) 15
(B) 1
(C) 16
(D) 13
(E) 14

![](https://cdn.mathpix.com/cropped/2024_04_17_fd2d0ad36e2f3d54c0ceg-2.jpg?height=387&width=406&top_left_y=1399&top_left_x=968)

## Unione Matematica Italiana  PROGETTO OLIMPIADI DI MATEMATICA

Ministry of Education,

UNIVERSITY AND RESEARCH

![](https://cdn.mathpix.com/cropped/2024_04_17_fd2d0ad36e2f3d54c0ceg-3.jpg?height=48&width=59&top_left_y=362&top_left_x=157)

## The Games of Archimedes - Biennio Competition  November 23, 2017

- The test consists of 16 problems. Each question is followed by 5 answers indicated by the letters (A), (B), (C), (D), (E). Only one of these answers is correct, the other 4 are wrong.
- Each correct answer is worth 5 points, each wrong answer is worth 0 points. For each unanswered or illegible answer, 1 point will be assigned.
- For each of the problems, you must write the letter corresponding to the answer you believe is correct in the grid provided below. No erasures or corrections on the grid are allowed. The use of any type of calculator or communication device is not permitted.

You have 110 minutes to complete the test. Good luck and have fun!

## NAME

 LAST NAME CLASS date of birth: email (optional):![](https://cdn.mathpix.com/cropped/2024_04_17_fd2d0ad36e2f3d54c0ceg-3.jpg?height=120&width=1234&top_left_y=1218&top_left_x=140)
(16) 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 $0,0,0,0,1,0,0$, and thus in the second $0,0,0,1,1,0$; in the third $0,0,1,2,1$; in the fourth $0,1,3,3$; in the fifth $1,4,6$; in the sixth 5,10; and finally in the seventh 15. 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 $2^{15}$. The number written in the lowest circle therefore ends with exactly 15 zeros.

Problem proposed by Giovanni Barbarino.

N.B. As usual, the proposed problems have been, in addition to being checked and selected, in many cases extensively modified or reformulated by the competition organizers. Therefore, in indicating the names of the proposers, it is understood that the responsibility for the selection of the questions, as well as for any errors, imprecisions, or questionable formulations, lies exclusively with the organizers themselves and in no way with the proposers, whom we wish to thank for their significant contribution, the richness of the ideas offered, the originality, and the quality of the proposals presented.

The organizers of the Archimedes Games, Alessandro D'Andrea, Paolo Francini, Andrea Sambusetti