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Math Expert V
Joined: 02 Sep 2009
Posts: 61385
If k and n are both integer and k > n > 0 is k!/n! divisible by 30 ?  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 50% (02:05) correct 50% (01:47) wrong based on 44 sessions

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If k and n are both integer and $$k > n > 0$$ is $$\frac{k!}{n!}$$ divisible by 30 ?

(1) $$k + n = 30$$

(2) $$k - n = 6$$

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Math Expert V
Joined: 02 Aug 2009
Posts: 8250
Re: If k and n are both integer and k > n > 0 is k!/n! divisible by 30 ?  [#permalink]

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If k and n are both integer and $$k > n > 0$$ is $$\frac{k!}{n!}$$ divisible by 30 ?
$$\frac{k!}{n!}$$ will surely be divisible by 30 if there are at least a difference of 5 between them as product of 5 consecutive numbers will be multiple of 2, 3 and 5 and hence 30...

(1) $$k + n = 30$$
$$k\neq{n}$$, so the closest they will be when k=16 and n=14, so $$\frac{k!}{n!}$$=$$\frac{16!}{14!}=16*15=8*30$$, hence divisible by 30.
Any further gap between k and n will surely have 15*16 in it..
Suff

(2) $$k - n = 6$$
this means consecutive six integers and their product will be at least 1*2*3*4*5*6, so divisible by 30.

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If k and n are both integer and k > n > 0 is k!/n! divisible by 30 ?  [#permalink]

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Bunuel wrote:
If k and n are both integer and $$k > n > 0$$ is $$\frac{k!}{n!}$$ divisible by 30 ?

(1) $$k + n = 30$$

(2) $$k - n = 6$$

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Analyzing the question:
30 = 2 * 3 * 5. We need a factor of 2, 3, and 5 from (k!/n!). If k and n have a difference of at least 5, we automatically get these factors from the numbers between k and n. For example, if k = 13 and n = 8, then we have k! / n! = 13 * 12 * 11 * 10 * 9 which is 5 numbers multiplied since k - n = 5. We can find a multiple of 5 for sure since there are 5 consecutive numbers multiplied, a multiple of 2 and 3 as well.

Statement 1:
If k and n are too far apart (k!/n!) will have a factor of 30 as demonstrated above. So let us focus on the closer k and n pairs, k = 16 and n = 14 -> (k!/n!) = 16 * 15. It has a factor of 30. If we take a bigger k we can see it will expand the factors for (k!/n!) so this is sufficient. (e.g. k = 17 n = 13, k!/n! = 17*16*15*14 which is more factors)

Statement2:
As demonstrated above, this is sufficient.

Ans: D
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