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

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Intern
Joined: 28 Dec 2010
Posts: 23

Kudos [?]: 46 [1], given: 337

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

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22 Aug 2017, 16:29
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Question Stats:

43% (01:46) correct 57% (01:30) wrong based on 46 sessions

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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
[Reveal] Spoiler: OA

Kudos [?]: 46 [1], given: 337

Math Expert
Joined: 02 Aug 2009
Posts: 5204

Kudos [?]: 5820 [1], given: 117

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

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22 Aug 2017, 19:05
1
KUDOS
Expert's post
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
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5820 [1], given: 117

Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 192

Kudos [?]: 17 [0], given: 435

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

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20 Sep 2017, 04:53
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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!

= k * (k - 1 ) *(k - 2 )*(k - 3 )*(k - 4 )*(k - 5)...

Kudos [?]: 17 [0], given: 435

Math Expert
Joined: 02 Aug 2009
Posts: 5204

Kudos [?]: 5820 [1], given: 117

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

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20 Sep 2017, 05:28
1
KUDOS
Expert's post
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!

= k * (k - 1 ) *(k - 2 )*(k - 3 )*(k - 4 )*(k - 5)...

Yes...
k!/n!....
If k-n=6......n=k-6..
So k!/(k-6)!= k(k-1)(k-2)(k-3)(k-4)(k-5)(k-6)!/(k-6)! = k(k-1)....(k-5)
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5820 [1], given: 117

Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 192

Kudos [?]: 17 [0], given: 435

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

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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!

= k * (k - 1 ) *(k - 2 )*(k - 3 )*(k - 4 )*(k - 5)...

chetan2u
hi

thanks man ..
you are great ...

Kudos [?]: 17 [0], given: 435

Manager
Joined: 09 Mar 2017
Posts: 121

Kudos [?]: 43 [0], given: 275

Location: India
Concentration: Marketing, Organizational Behavior
WE: Information Technology (Computer Software)
Re: If k and n are both integer and k>n>0 is k!/n! divisible by 30 ? [#permalink]

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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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Hit Kudus if this has helped you get closer to your goal, and also to assist others save time. Tq

Kudos [?]: 43 [0], given: 275

Re: If k and n are both integer and k>n>0 is k!/n! divisible by 30 ?   [#permalink] 20 Sep 2017, 08:28
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