BTGmoderatorDC wrote: ↑

Tue May 05, 2020 1:51 am

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

A trailing 0 is formed when 2 is multiplied to 5. Thus, we must count the number of 2s and that of 5s in 60!. Lesser of the count of 2s and 5s will determine the no. of trailing 0s.

Since 5 > 2, there would be fewer 5s than 2s in 60!. So, we should count no. of 5s in 60!.

We know that 60! = 1.2.3.4.5...60.

To get the no. of 5s, let's read the table of 5 till 60.

It's 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60.

Each of the 12 numbers 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 have one 5 except 25 and 50; they have two 5s each.

So, there are 12 + 1 + 1 = 14 fives.

So, there would be 14 consecutive 0s in 60!.

The correct answer: C

Hope this helps!

-Jay
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BTGmoderatorDC wrote: ↑

Tue May 05, 2020 1:51 am

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 that number.

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!

Since there are 14 factors of 5 within 60!, there are 14 5-and-2 pairs and thus 14 trailing zeros.

Answer: C