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

1) K+n =30

2) k-n = 6

what are the factors of \(30 - 2*3*5\) so k should have these EXTRA factors when compared to n..

lets see the statements:-

1) \(k+n=30\) we know k >n, so \(k>\frac{30}{2}\) and \(n<\frac{30}{2}\) so if we try and make n as MAX as possible n will be 14 and thus k will be 30-14=16.. \(\frac{16!}{14!}=15*16\), which is Div by 30 sufficient

2) \(k-n=6\) this means \(\frac{k!}{n!}\) is PRODUCT of 6 consecutive integers.. ANY 6 consecutive integers will surely have MULTIPLES of 2,3,4,5,6 this PRODUCT will be div by 2*3*4*5*6=720, which is div by 30 sufficient

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

Show Tags

20 Sep 2017, 04:53

1

This post was BOOKMARKED

chetan2u wrote:

gary391 wrote:

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

1) K+n =30

2) k-n = 6

what are the factors of \(30 - 2*3*5\) so k should have these EXTRA factors when compared to n..

lets see the statements:-

1) \(k+n=30\) we know k >n, so \(k>\frac{30}{2}\) and \(n<\frac{30}{2}\) so if we try and make n as MAX as possible n will be 14 and thus k will be 30-14=16.. \(\frac{16!}{14!}=15*16\), which is Div by 30 sufficient

2) \(k-n=6\) this means \(\frac{k!}{n!}\) is PRODUCT of 6 consecutive integers.. ANY 6 consecutive integers will surely have MULTIPLES of 2,3,4,5,6 this PRODUCT will be div by 2*3*4*5*6=720, which is div by 30 sufficient

D

hi

great ...

According to statement 2: k - n = 6...

For clarification, do you mean something as such as under: k!/n!

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

1) K+n =30

2) k-n = 6

what are the factors of \(30 - 2*3*5\) so k should have these EXTRA factors when compared to n..

lets see the statements:-

1) \(k+n=30\) we know k >n, so \(k>\frac{30}{2}\) and \(n<\frac{30}{2}\) so if we try and make n as MAX as possible n will be 14 and thus k will be 30-14=16.. \(\frac{16!}{14!}=15*16\), which is Div by 30 sufficient

2) \(k-n=6\) this means \(\frac{k!}{n!}\) is PRODUCT of 6 consecutive integers.. ANY 6 consecutive integers will surely have MULTIPLES of 2,3,4,5,6 this PRODUCT will be div by 2*3*4*5*6=720, which is div by 30 sufficient

D

hi

great ...

According to statement 2: k - n = 6...

For clarification, do you mean something as such as under: k!/n!

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

Show Tags

20 Sep 2017, 06:39

gmatcracker2017 wrote:

chetan2u wrote:

gary391 wrote:

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

1) K+n =30

2) k-n = 6

what are the factors of \(30 - 2*3*5\) so k should have these EXTRA factors when compared to n..

lets see the statements:-

1) \(k+n=30\) we know k >n, so \(k>\frac{30}{2}\) and \(n<\frac{30}{2}\) so if we try and make n as MAX as possible n will be 14 and thus k will be 30-14=16.. \(\frac{16!}{14!}=15*16\), which is Div by 30 sufficient

2) \(k-n=6\) this means \(\frac{k!}{n!}\) is PRODUCT of 6 consecutive integers.. ANY 6 consecutive integers will surely have MULTIPLES of 2,3,4,5,6 this PRODUCT will be div by 2*3*4*5*6=720, which is div by 30 sufficient

D

hi

great ...

According to statement 2: k - n = 6...

For clarification, do you mean something as such as under: k!/n!

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

Show Tags

20 Sep 2017, 08:28

chetan2u wrote:

gary391 wrote:

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

1) K+n =30

2) k-n = 6

what are the factors of \(30 - 2*3*5\) so k should have these EXTRA factors when compared to n..

lets see the statements:-

1) \(k+n=30\) we know k >n, so \(k>\frac{30}{2}\) and \(n<\frac{30}{2}\) so if we try and make n as MAX as possible n will be 14 and thus k will be 30-14=16.. \(\frac{16!}{14!}=15*16\), which is Div by 30 sufficient

2) \(k-n=6\) this means \(\frac{k!}{n!}\) is PRODUCT of 6 consecutive integers.. ANY 6 consecutive integers will surely have MULTIPLES of 2,3,4,5,6 this PRODUCT will be div by 2*3*4*5*6=720, which is div by 30 sufficient

D

Hello chetan2u, Would you please explain statement1? I didn't get this one. Thank you
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