How many positive integer n are there, such that n! ends with exactly

Question

How many positive integer n are there, such that n! ends with exactly 30 zeroes?

A. 0
B. 1
C. 2
D. 3
E. 4

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GMATinsight · Answer

Every zero = Every power of 10

Every 10 = Pair of 2*5

Power of 2 in any factorial < Power of 5 in any factorial (Because bigger prime has less frequent multiples than any smaller prime)

i.e. Power of 10 in any factorial = Power of 5 in that factorial

Power of 5 in x! =   +   +   + ... and so on

i.e. 120!, power of 5 = 24 + 4 + 0 = 28 NOT ACCEPTABLE
i.e. 121!, power of 5 = 24 + 4 + 0 = 28 NOT ACCEPTABLE
i.e. 122!, power of 5 = 24 + 4 + 0 = 28 NOT ACCEPTABLE
i.e. 123!, power of 5 = 24 + 4 + 0 = 28 NOT ACCEPTABLE
i.e. 124!, power of 5 = 24 + 4 + 0 = 28 NOT ACCEPTABLE 
But
 i.e. 125!, power of 5 = 25 + 5 + 1 = 31 NOT ACCEPTABLE

i.e. There is no value of n which gives 30 powers of 5

Hence, n = {NULL} = 0 values

Answer: Option A

newyork2012 · Answer

1.
n/5+n/25=30
5n+n=750
6n=750
n=125=5^3

But when n=125 , we will have to add one more 0 to the end ... we will have 31 0s

2.
n/5+n/25+n/125=30
31n=30*125
no solution

To sum up, no solution

Posted from my mobile device

ScottTargetTestPrep · Answer

Typically, we are given the value of n and asked to determine the number of trailing zeros of n!. For example, how many trailing zeros does 10! have? However, this problem is asking for the reverse.

Recall that the number of trailing zeros of n! depends on the number of 5-and-2 pairs that are contained in n!. However, since there are fewer 5s than 2s, it really depends on the number of 5s that are contained in n!. So let’s determine the smallest value of n such that n! ends with exactly 30 zeroes. Such a value of n must be a multiple of 5. That is, if there exists a value of n such that n! (where n is a multiple of 5) has exactly 30 zeros, then (n + 5)! must have at least 31 zeros (notice n + 5 is the next multiple of 5, which makes (n + 5)! having at least one more factor of 5 than n!). So let’s list the number of zeros of n! (where n is a multiple of 5) starting with 5!:

n! Number of trailing zeros
 5! 1
 10! 2
 15! 3
 20! 4
 25! 6 (Notice 25 has 2 factors of 5, so the number of trailing zeros jumps by 2 from 20!.) 
 30! 7 
 35! 8 
 40! 9 
 45! 10 
 50! 12 (Notice 50 has 2 factors of 5, so the number of trailing zeros jumps by 2 from 45!. 
 Also, we can do things a little faster, increase the value of n by 25, since we can 
 see that the number of trailing zeros will be increased by 6.)
 75! 18 
 100! 24 (If we continue with the shortcut, the next one will be 125!. However, since 125 has 
 3 factors of 5, so let’s go back to the increments of 5.)
 105! 25
 110! 26
 115! 27
 120! 28
 125! 31 (Notice 125 has 3 factors of 5, so the number of trailing zeros jumps by 3 from 124!.)

As we can see from the list above, there are no values of n such that n! has exactly 30 trailing zeros. (Notice that n! has 28 trailing zeros for n = 120, 121, 122, 123 and 124, but it has 31 trailing zeros when n = 125.)

Answer: A

Kinshook · Answer

Asked: How many positive integer n are there, such that n! ends with exactly 30 zeroes?

Power of 5 in 125! = 25 + 5 + 1 = 31
n<125
Power of 5 in 120! = 24 + 4 = 28
n>120
Power of 5 in 121! = 24 + 4 = 28
Power of 5 in 122! = 24 + 4 = 28
Power of 5 in 123! = 24 + 4 = 28
Power of 5 in 124! = 24 + 4 = 28

Since 125 has 5^3, number of zeros jump from 28 to 31.

There is no n feasible such that n! ends with exactly 30 zeros

IMO A

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How many positive integer n are there, such that n! ends with exactly

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