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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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22 Aug 2017, 15:29
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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) kn = 6
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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, 18:05
gary391 wrote: If k and n are both integer and k>n>0 is k!/n! divisible by 30 ?
1) K+n =30
2) kn = 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 3014=16..\(\frac{16!}{14!}=15*16\), which is Div by 30 sufficient 2) \(kn=6\) 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 30sufficient D
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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, 03:53
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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) kn = 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 3014=16..\(\frac{16!}{14!}=15*16\), which is Div by 30 sufficient 2) \(kn=6\) 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 30sufficient 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)... please say to me ... thanks in advance ..man



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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, 04:28
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) kn = 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 3014=16..\(\frac{16!}{14!}=15*16\), which is Div by 30 sufficient 2) \(kn=6\) 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 30sufficient 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)... please say to me ... thanks in advance ..man Yes... k!/n!.... If kn=6......n=k6.. So k!/(k6)!= k(k1)(k2)(k3)(k4)(k5)(k6)!/(k6)! = k(k1)....(k5)
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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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20 Sep 2017, 05:39
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) kn = 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 3014=16..\(\frac{16!}{14!}=15*16\), which is Div by 30 sufficient 2) \(kn=6\) 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 30sufficient 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)... please say to me ... thanks in advance ..man chetan2u hi thanks man .. you are great ...



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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, 07:28
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) kn = 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 3014=16..\(\frac{16!}{14!}=15*16\), which is Div by 30 sufficient 2) \(kn=6\) 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 30sufficient D Hello chetan2u, Would you please explain statement1? I didn't get this one. Thank you
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Re: If k and n are both integer and k>n>0 is k!/n! divisible by 30 ?
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