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factorial and divisibility

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by mehrasa » Sun Sep 11, 2011 12:22 pm
what is the least N such that N! is divisible by 1000?
8
10
15
20
25

I think the answer will be 10
1000= 2^3*5^3 and 10 has both 2 and 5 as its prime factors
what do u think?!
Last edited by mehrasa on Sun Sep 11, 2011 7:53 pm, edited 1 time in total.
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by pemdas » Sun Sep 11, 2011 1:54 pm
perform prime factorization of 1000 and include all primes into N!
1000 (2) 500
500 (2) 250
250 (2) 125
125 (5) 25
25 (5) 5
5 (5) => 2^3 *5^3

5! will include 2, 4(2^2) and 5 (5^1) we need two more fives (5^2), 15 will include 10 and 15
Hence 15! c
mehrasa wrote:what is the least N such that N! is divisible by 1000?
8
10
15
20
25

I think the answer will be 10
1000= 2^3*5^2 and 10 has both 2 and 5 s its prime factors
what do u think?!
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by mehrasa » Sun Sep 11, 2011 7:58 pm
As I understood you mean that by 10! we can not have three 5 while 15! will give us three 5 and of course three 2....
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by Geva@EconomistGMAT » Sun Sep 11, 2011 11:31 pm
mehrasa wrote:As I understood you mean that by 10! we can not have three 5 while 15! will give us three 5 and of course three 2....
pemdas's explanation is correct. In order to make 1000=10^3, you need three "building blocks" of 5, and three "building blocks" of 2. 10! is not enough, as it only has two powers of 5: one in 5, one in 10. You need the next multiple of 5 to add the necessary third power of 5 needed to make 1000.

Note that the 2 are irrelevant: there are many powers of 2 in any factorial, many more than powers of 5. 8 alone inludes 3 powers of 2, each of which can be paired with a 5 to make another power of 10. Focus on the limiting factor - the powers of 5.
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by saketk » Mon Sep 12, 2011 8:32 am
mehrasa wrote:what is the least N such that N! is divisible by 1000?
8
10
15
20
25

I think the answer will be 10
1000= 2^3*5^3 and 10 has both 2 and 5 as its prime factors
what do u think?!
The key here is to look at the power of 5 which is 3.

if you choose 10! then you will maximum of 2 5's.

the least number which satisfies the condition is 15! = 15! has 3 5's, making it the perfect fit here.
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by DominateTheGMAT » Sat Jan 25, 2014 8:44 am
All of the explanations above are spot-on, but sometimes I find that a more thorough video explanation helps to make things even clearer. I took the liberty of recording a quick video that explains not only how to solve this question, but in it I also review some of the underlying concepts in this problem that will enable you to adapt and answer other GMAT "number theory" questions involving factors, multiples, and divisibility.

Here it is: https://youtu.be/q9ON4jdT-sY

Hope it helps!
Brett Ethridge
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