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

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

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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

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

https://www.beatthegmat.com/if-n-is-the-greatest-positive-integer-for-which-5-n-is-a-fac-t304112.html#819465

Regards,
Fabio.

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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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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 5² provides a SECOND 5:
60/5² = 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 5² provides a SECOND 5:
200/5² = 8 --> eight more 5's.
Every multiple of 5³ provides a THIRD 5:
200/5³ = 1 --> one more 5.
Thus, the total number of 5's contained within 200! = 40+8+1 = 49.

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