If k and n are both integer and k n 0 is k!/n! divisible by 30 ?

Question

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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chetan2u · Answer

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
eq{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.

D

TestPrepUnlimited · Answer

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

ajinkyajadhav · Answer

1. sufficient since,

if k>n then it has to be maximum k=16 n=14 in which case you have 16*15 which is divisble by 30
or minimum k=29 ,n =1 in case you have 29! which is divisible by 30.
hence sufficient

2. sufficient since,

if k-n=6 , k=n+6
now k!/n! = (n+6)(n+5).......(n+1) 
here since n>=1 we can say the above statment has 2 3's and a 5 hence divisible by 30.

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

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