If n is an odd natural number and n! ends with 32 zeros, the

Thanks for this reply.It seems I was missing 1 of the fives in 125!

Posted from my mobile device

Bunuel

i am not agree with option A as it is given as correct answer.

i am getting 135!, 137! and 139! for 32 trailing zeros. so i think it should be 3.

Please provide your comments on this.

Thanks

131! has 32 trailing zeros: 131/5 + 131/25 + 131/125 = 26 + 5 + 1 = 32.
133! has 32 trailing zeros: 133/5 + 133/25 + 133/125 = 26 + 5 + 1 = 32.

But 135! has 33 trailing zeros: 135/5 + 135/25 + 135/125 = 27 + 5 + 1 = 33.

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

I got the right answer but it took me close to 4 minutes to get to it.

by trial and error i plugged in first 100 and realized i was too low, then went to 125 and realized i was still too low, Finally went to 131 and hit jackpot.
I got the answer by using the below formula
(n/5) +(n/5^2) + (n/5^3) = 32

the final numbers are 131 and 133 (since they are odd Numbers only)

Is there a faster solution to this?

No, I think you've done this in the fastest possible way. It's clearly not a real GMAT question, so don't worry about how long it took - if you can solve it at all, you're doing very well.

There are a few real GMAT questions which test the prime factorizations of factorials, as this question does, but in every one of them, it is practical to work out the answer in two minutes without knowing any special formula. The real GMAT questions I've seen on this topic (there are three or four in the GMATFocus product, and one in the OG) are never more complicated than this one:

What is the largest integer k for which 5^k is a divisor of 50!

To answer that question without a formula, you only need to look at the ten multiples of five inside 50!, which can be done by hand within two minutes. I've never seen a real GMAT question in this format where you'd need to look at more than ten different multiples if you were solving by hand, so the question posted above is far more complicated than any real question I've encountered.

Right now, the GMAT timer on 2 minutes 22 second to solve this problem correctly. I doubt too, that it is possible to solve this problem at this time.

Hi,

just read this Q and few responses above..

I would have a point here..

We do not require to calculate anything here..
32 zeroes too is just a number which will not effect the answer..

A zero will be added when a 2 and a 5 is added...
every alternate term is even so we are worried only about a 5..

And 5 comes after every four number and two each will be odd and two even in these..
so there will be two ODD natural numbers before a 5 is added in the product and in turn an extra ZERO..

ans is 2..

The '32' does matter here - the answer is not always '2'. You might try a few simpler questions to see why:

* For how many odd numbers n does n! have exactly 4 trailing zeros? (the answer is two, 21 and 23)

* For how many odd numbers n does n! have exactly 6 trailing zeros? (the answer is three, 25, 27, 29)

* For how many odd numbers n does n! have exactly 5 trailing zeros? (the answer is none)

hi,

I get your point..
I took somehow after crossing over to another 0, how many n odd natural numbers will have same trailing zeroes..
1) when a five is added due to addition of a multiple of 10.. x1 and x3 are the two..
2) when a five is added due to addition of a multiple of odd number.. x7 and x9 are the two..
3) in the case of

this happens because of addition of a number which is multiple of 5^2 or 5^3 and so on..
again as I took the case that we are xg over from x number of 0s to x+1 0s..

But I think read extra in the Q..
Thanks

We have geometric series:

\(\frac{n}{5} + \frac{n}{5^2} + \frac{n}{5^3} + … ≈ 32\)

\((\frac{n}{5})/(1-\frac{1}{5}) ≈ 32\)

\(\frac{n}{4} ≈ 32\)

\(n ≈ 128\)

Nearest odd is 129, but 129! has only 31 trailing zero. In order to add additional zero we need to have additional multiple of 5 in factorial.
Hence, for 32 trailing zeros we have:

130!, 131!, 132!, 133! and 134! Starting from 135 factorial we’ll get additional multiple of 5 and additional trailing zero.

Picking up odd numbers we have – 131! and 133!

Answer A.

We need to remember:
Multiples of 5 add 1 traling zero,
- \(5^2\) - 2 traling zeros
- \(5^3\) - 3 traling zeros
- etc.

Hi,

Dont need to go with formula.

We konw.. 32 zerores means.. 32 5's and 32 2's.. In any factorial of number, 2s is always be higher than 5's.

Now here Question is howmany odd numbers are possible...

means for any number of zeroes.. there will always 2 odd number..

starting for 1 to 5...1 and 3 are odd number.. whenever it hits 5 , we are adding one more zero to it.

Hope this logic helps.

Hi pratik1709,

I think above logic might not work always.

if the question had asked number of possible values of odd number n such that 33 trailing zeroes.
then there are three such possible odd numbers, 135!, 137!, 139!

it depends on which parity of n adds that extra zero(es).

say odd number n adds extra zero(es), then there are 3 possible odd number, eg n - 135(odd) adds extra zero to previous 134!, then 135!, 137! and 139! are possible odd numbers of n such that n! has 33 trailing zeroes
say even number n adds extra zero(es), then there are 2 possible odd number.e.g n = 130(even) adds extra zero to previous 129!, then 131! and 133! are only possible numbers of n such that n! has 32 trailing zeroes

Hello,

Is it true that the answer above has to be 2 or none? For any number of zeros, there will be 2 possible values of n, UNLESS that number is 5 taken to a power:

20! to 24! has four trailing zeros - 20/5 = 4 (so 21 and 23 would be possible answers for four zeros)
25! has six trailing zeros - 25/5 + 25/25 = 6

So no values of n! have five trailing zeros.

n=odd, and n!=32 zeros = sum of the quotients of \(n/5+n/5^2+n/5^m=32…(where:5^m≤n)\);
for n!=125! sum of the quotients is \(125/5+125/25+125/125=25+5+1=31\)
for n!=130! sum of the quotients is \(130/5+130/25+130/125=26+5+1=32\)
for n!=135! sum of the quotients is \(135/5+135/25+135/125=27+5+1=33\)
possible values of n are the odds from 130≤n<135 n={131,133}=2

Answer (A)

Hi! how do we reach upto those numbers? isn't there any method to solve this?

Links to the theory and similar questions are already given above. Here they ate again:

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