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13 Nov 2019, 02:36
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
55% (02:15) correct
45%
(01:54)
wrong
based on 190
sessions
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Not Attempted Yet
If k and n are both integer and \(k > n > 0\) is \(\frac{k!}{n!}\) divisible by 30 ?
(1) \(k + n = 30\)
(2) \(k - n = 6\)
Are You Up For the Challenge: 700 Level Questions
(1) \(k + n = 30\)
(2) \(k - n = 6\)
Are You Up For the Challenge: 700 Level Questions
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
13 Nov 2019, 06:56
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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.
D
\(\frac{k!}{n!}\) 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.
D
13 Nov 2019, 08:59
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Bunuel
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
