Find the number of trailing zeros in the expa… | GMAT Sprint

Free Practice Question

PS · Problem Solving

705-805

Divisibility/Multiples/Factors

Find the number of trailing zeros in the expansion of $(20!21!22! ......... *33!)^{3!}$ .

Answer Choices

Correct answer marked below

468

Correct

469

470

467

471

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Explanation

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Official Explanation

Official

Number of trailing zeros in 20!, 21!, 22!, 23!, and 24! will be 4 (20/5=4. For 21!, 22!, 23! and 24!, instead of 20 you'll have 21, 22, ... but the result will be the same) → total of

4 \times 5 = 20

trailing zeros for these 5 terms.

(Note: This won't always be correct: for example 20 and 50 have one trailing zero each but

20 \times 50 = 1,000

has three trailing zeros not two. That's because extra 2 in 20 and extra 5 in 50 "produced" one more trailing zero. In our case though, we won't have any extra 5-s in any factorial, as all are already used for existing trailing zeros.)

Number of trailing zeros in 25!, 26!, 27!, 28!, and 29! will be

5 + 1 = 6

(

25/5 + 25/5^2 = 6

) → total of

6 \times 5 = 30

trailing zeros for these 5 terms.…

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What this question tests

What it tests

Your understanding of divisibility rules, least common multiples, and greatest common factors.

Common trap

Forgetting that 1 is not prime and that every integer divides 0.

Find the number of trailing zeros in the expansion of (20!∗21!∗22!.........∗33!)3!(20!*21!*22! ......... *33!)^{3!}(20!∗21!∗22!.........∗33!)3!.