Maths Olympiad Prep

Library / CEMC Fermat /442 of 566

CEMC Fermat, 2000

Number theory Difficulty 3.0 AMC 10/12 Find the answer Canada

If the expression 156×285×55715^6 \times 28^5 \times 55^7 was evaluated, it would end with a string of consecutive zeros. How many zeros are in this string?

(A) 1010 (B) 1818 (C) 2626 (D) 1313 (E) 55

We don't reproduce this publisher's solutions. Their own solution is here — work the problem first.
Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project. Solutions are the publisher's, linked not copied.