If 60! is written out as an integer, with how many

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Mon Nov 12, 2018 9:16 pm

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If 60! is written out as an integer, with how many consecutive 0's will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56

OA C

Source: Manhattan Prep

Source: Beat The GMAT — Problem Solving | Mon Nov 12, 2018 9:16 pm

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Nov 13, 2018 3:26 am

BTGmoderatorDC wrote:If 60! is written out as an integer, with how many consecutive 0's will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56

60! = 60*59*58*....*3*2*1.

Since 10=2*5, EVERY COMBINATION OF 2*5 contained within the prime-factorization of 60! will yield a 0 at the end of the integer representation of 60!.
The prime-factorization of 60! is composed of FAR MORE 2'S than 5's.
Thus, the number of 0's depends on the NUMBER OF 5's contained within 60!.

To count the number of 5's, simply divide increasing POWERS OF 5 into 60.

Every multiple of 5 within 60! provides at least one 5:
60/5 = 12 --> twelves 5's.
Every multiple of 5Â² provides a SECOND 5:
60/5Â² = 2 --> two more 5's.
Thus, the total number of 5's contained within 60! = 12+2 = 14.

The correct answer is C.

Another example:

If 200! is written out as an integer, with how many consecutive 0's will that integer end?

Every multiple of 5 within 200! provides at least one 5:
200/5 = 40 --> forty 5's.
Every multiple of 5Â² provides a SECOND 5:
200/5Â² = 8 --> eight more 5's.
Every multiple of 5Â³ provides a THIRD 5:
200/5Â³ = 1 --> one more 5.
Thus, the total number of 5's contained within 200! = 40+8+1 = 49.

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Tue Nov 13, 2018 9:52 am

The reason why "the counting of 5Â´s recipe" (mentioned and shown in the post above) works is explained here:

https://www.beatthegmat.com/if-n-is-the ... tml#819465

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8132; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Sun Jan 27, 2019 6:37 pm

BTGmoderatorDC wrote:If 60! is written out as an integer, with how many consecutive 0's will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56

OA C

Source: Manhattan Prep

To determine the number of trailing zeros in a number, we need to determine the number of 5-and-2 pairs within the prime factorization of 60!. Note that each 5-and-2 pair (which is equivalent to 10) creates a trailing zero.

Since we know there are fewer 5s in 60! than 2s, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.

To determine the number of 5s within 60!, we can use the following shortcut in which we divide 60 by 5, then divide the quotient of 60/5 by 5 and continue this process until we no longer get a nonzero quotient.

60/5 = 12

12/5 = 2 (we can ignore the remainder)

Since 2/5 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 5 within 60!.

Thus, there are 12 + 2 = 14 factors of 5 within 60! And this means also that there are 14 5-and-2 pairs within 60!, and thus there are 14 trailing zeros.

Answer: C

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