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# What is the value of N?

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Re: What is the value of N? [#permalink]
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Bunuel wrote:
What is the value of N?

(1) N! ends with 28 zeroes

(2) (N+2)! ends with 31 zeroes and (N-1)! ends with 28 zeroes

Kudos for a correct solution.

stastement 1: If N=124 --> total number of zeroes=124/5+124/25=28. If N=123 --> total number of zeroes=123/5+123/25=28.

statement 2: works with both 123 and 124.

1+2) works with both 123 and 124.

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Re: What is the value of N? [#permalink]
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Hi all,
Again a value DS question.
1). Statement 1 is insufficient.
To find the number of zeros all we need is find number of five’s.
For N values 120,121,122,123,124… the number of zeros(Number of five's) is 28
So insufficient
2). Statement 2 is insufficient.
N values again can be 123,124,125…
Together also insufficient,
Because N can be 123 or 124.
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Re: What is the value of N? [#permalink]
CrackVerbalGMAT wrote:
Hi all,
Again a value DS question.
1). Statement 1 is insufficient.
To find the number of zeros all we need is find number of five’s.
For N values 120,121,122,123,124… the number of zeros(Number of five's) is 28
So insufficient
2). Statement 2 is insufficient.
N values again can be 123,124,125…
Together also insufficient,
Because N can be 123 or 124.

hi N cannot be 125....
because it is given N! has 28 zeroes in end..
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Re: What is the value of N? [#permalink]
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I dont understand why has everyone picked 124, 125 as the starting point for N

Not clear - Can somebody pls explain
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Re: What is the value of N? [#permalink]
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Bunuel wrote:
Bunuel wrote:
What is the value of N?

(1) N! ends with 28 zeroes

(2) (N+2)! ends with 31 zeroes and (N-1)! ends with 28 zeroes

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION

In any N!, the number of trailing zeroes = number of factors of 2 and 5. For example, for 6! (which is 720), the ending 0 is created by taking the numbers 2 and 5 in (6)(5)(4)(3)(2)(1) and multiplying them together. It's also important to realize that since every second number is divisible by 2 but every fifth number is a multiple of 5, the number of occurrences of 5 is less than 2, hence the limiting factor is 5.

Sine we know we're working with giant numbers, we can start getting an idea of what we're looking at by seeing how many zeroes 100! ends with

Every multiple of 5 between that 1 and 100 provides one factor of 5, and at 25, 50, 75, and 100, you get an extra 5 (for example, 75 = 3 * 5 * 5, so 75 provides two factors of 5). So the number of factors in 100! is 24 as a starting point for us.

Now since N! ends with 28 zeroes, we need to four more 5’s in 100!, so N should be 120. But one thing is to be noticed here is 120!, 121!, 122!, 123! and 124! also have 28 zeroes as from 120! to 124! there is no more extra 5 factor.

Hence Statement 1 is not sufficient.

As per Statement 2 the (N+2)! has 31 zeroes.

So let’s check which numbers can have 30 zeroes. We know from the above that 120!-124! should have 28 zeroes.

But when it comes to 125!, we have 5^3 factor so 125!-129! have 31 zeroes.

So N+2 can be any number between 125 to 129, but the statement also says N-1 has 28 zeroes that means N-1 can be any number from 120! to 124!.

So N can be 125 or 124 or 123 so again Statement 2 is not sufficient.

Let’s combine Statements 1 and 2: in this case, N could be 124 or it could be 123, so E is the correct answer.

For similiar questions check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.

Hope this helps.
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Re: What is the value of N? [#permalink]
buddyisraelgmat wrote:
I dont understand why has everyone picked 124, 125 as the starting point for N

Not clear - Can somebody pls explain

hi

firstly the value of zeroes depends on power of 5 as power of 2 will always be greater than power of 5...
as we see (N+2)! increases the zeroes by 3 digits...it means between N and N+2, there is 5^3 multiple..125,250... here 125 fits in ..
because no of zeroes=125/5+125/5^2+125/5^3=25+5+1=31...
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Re: What is the value of N? [#permalink]
Bunuel wrote:
Bunuel wrote:
What is the value of N?

(1) N! ends with 28 zeroes

(2) (N+2)! ends with 31 zeroes and (N-1)! ends with 28 zeroes

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION

In any N!, the number of trailing zeroes = number of factors of 2 and 5. For example, for 6! (which is 720), the ending 0 is created by taking the numbers 2 and 5 in (6)(5)(4)(3)(2)(1) and multiplying them together. It's also important to realize that since every second number is divisible by 2 but every fifth number is a multiple of 5, the number of occurrences of 5 is less than 2, hence the limiting factor is 5.

Sine we know we're working with giant numbers, we can start getting an idea of what we're looking at by seeing how many zeroes 100! ends with

Every multiple of 5 between that 1 and 100 provides one factor of 5, and at 25, 50, 75, and 100, you get an extra 5 (for example, 75 = 3 * 5 * 5, so 75 provides two factors of 5). So the number of factors in 100! is 24 as a starting point for us.

Now since N! ends with 28 zeroes, we need to four more 5’s in 100!, so N should be 120. But one thing is to be noticed here is 120!, 121!, 122!, 123! and 124! also have 28 zeroes as from 120! to 124! there is no more extra 5 factor.

Hence Statement 1 is not sufficient.

As per Statement 2 the (N+2)! has 31 zeroes.

So let’s check which numbers can have 30 zeroes. We know from the above that 120!-124! should have 28 zeroes.

But when it comes to 125!, we have 5^3 factor so 125!-129! have 31 zeroes.

So N+2 can be any number between 125 to 129, but the statement also says N-1 has 28 zeroes that means N-1 can be any number from 120! to 124!.

So N can be 125 or 124 or 123 so again Statement 2 is not sufficient.

Let’s combine Statements 1 and 2: in this case, N could be 124 or it could be 123, so E is the correct answer.

I got a little doubt

Zeros in 120! to 124! is 29

Correct me i assumed it like: dividing 120! with 5,25&75 i will get 24+4+1 number of 5's i.e 29

while taking 100!= 24 and then counting 5's in 105,110,115,120 will take to 28 . whats missing here?
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Re: What is the value of N? [#permalink]
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ikishan wrote:

I got a little doubt

Zeros in 120! to 124! is 29

Correct me i assumed it like: dividing 120! with 5,25&75 i will get 24+4+1 number of 5's i.e 29

while taking 100!= 24 and then counting 5's in 105,110,115,120 will take to 28 . whats missing here?

hi,
to find power of 5 or zeroes in any factorial, we have to divide the number with increasing power of 5.... 5^1,5^2,5^3 and so on....
you have correctly divided by 5,25 but 75 is wrong as next number would be 5^3=125..
and you cannot divide 120 by 125...
so answer is 24+4=28 and not 29...
hope it is clear...
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Re: What is the value of N? [#permalink]
chetan2u wrote:
ikishan wrote:

I got a little doubt

Zeros in 120! to 124! is 29

Correct me i assumed it like: dividing 120! with 5,25&75 i will get 24+4+1 number of 5's i.e 29

while taking 100!= 24 and then counting 5's in 105,110,115,120 will take to 28 . whats missing here?

hi,
to find power of 5 or zeroes in any factorial, we have to divide the number with increasing power of 5.... 5^1,5^2,5^3 and so on....
you have correctly divided by 5,25 but 75 is wrong as next number would be 5^3=125..
and you cannot divide 120 by 125...
so answer is 24+4=28 and not 29...

hope it is clear...

Thanks man, that filled clarity.
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Re: What is the value of N? [#permalink]
The question tests the concept of “skipping zeroes” in factorials as it relates to the number of trailing zeroes a Factorial has.

1st concept is that the number of trailing zeroes that a Factorial will have depends on the Count of Pairs of Prime Factors (2 * 5) that are in the Factorial’s prime factorization.

When the Factorials are listed consecutively, the no of trailing zeroes will stay the same for 5 consecutive values of N where N! Is a Factorial

then the no. of trailing zeroes will increase by 1 trailing zero for the next Multiple of 5 you hit moving up in ascending order.

However, if the next number is a multiple of (5)^2 = 25, the trailing zeroes will “skip 1” and the next group of Factorials will have + 2 more trailing zeroes.

If the next multiple is (5)^3, then the trailing zeroes will “skip 2” and the next group of Factorials will have +3 trailing zeroes.

For example:

15! thru 19! ——- 3 trailing zeroes

20 is the next multiple of 5, and it is only a multiple of (5)^1 so the trailing zeroes will increase by + 1

20! thru 24! ——-4 trailing zeroes

However, now the next multiple of 5 will be a multiple of (5)^2 so the trailing zeroes will “skip 1” in the consecutive ordering

25! - 29! ———6 trailing zeroes

If you understand this concept and the fact that the GMAT always provides questions that are answerable, then you can eliminate S1 right off the bat. There will be 4 values of N for which N! has 4 trailing zeroes.

S1 not sufficient.

However, doing some work in statement 1 makes statement 2 a lot easier.

If we use a round about estimate to say that every 25 consecutive Values of N “groupings” there will be “around” 6 trailing zeroes, we can test:

130!

130/5 = 26

26/5 = 5

5/5 = 1

130! thru 134! will have —- 26 + 5 + 1 = 31 trailing zeroes

From here we can go down a couple of sets and try 120!

120/5 = 24

24/5 = 4

120! thru 124! will have 28 trailing zeroes

S1 not sufficient: N can be 120, 121, 122, 123, or 124

S2: (n + 2)! Has 31 trailing zeroes
(n - 1) has 28 trailing zeroes.

If (n + 2)! = 125! — 125! Has 31 trailing zeroes——then n = 123

(N - 1)! = 122! ———which has 28 trailing zeroes.

N = 123 is a possibility.

However: 126! Will also have 31 trailing zeroes and 123! Will have 28 trailing zeroes

N = 124 is a possibility.

S2 not sufficient.

Using both statement together, it is still possible for N to be 123 or 124

E neither nor

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