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If n >2 and n^45n^2+4 = 20h, Is h an integer ? [#permalink]
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Updated on: 22 Aug 2017, 07:42
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If n >2 and n^45n^2+4 = 20h, Is h an integer? 1) n/4 is an integer 2) n/5 is an integer
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Originally posted by gary391 on 21 Aug 2017, 07:57.
Last edited by chetan2u on 22 Aug 2017, 07:42, edited 1 time in total.
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Re: If n >2 and n^45n^2+4 = 20h, Is h an integer ? [#permalink]
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22 Aug 2017, 07:41
gary391 wrote: If n >2 and n^45n^2+4 = 20h, Is h an integer?
1) n/4 is an integer
2) n/5 is an integer HI... please follow the rules for posting a Q.. otherwise for the solution.. \(n^45n^2+4 = 20h.............. h=\frac{n^45n^2+4}{20}\) for h to be an integer \(n^45n^2+4\) has to be a multiple of 20..lets see the statements I.. \(\frac{n}{4}\) is an integer.. let n/4 = s where s is the integer.. n = 4s so \(h=\frac{n^45n^2+4}{20}.......\frac{(4s)^45*(4s)^2+4}{20}.....\) so if s is 1, ans is 180/20=9 yes h is an integer.. but when s is multiple of 5, clearly numerator will be multiple of 5+4, which will not be multiple of 5 and thus 20.. so h is NOT an integerinsufficient II \(\frac{n}{5}\) is an integer.. let n/5 = s where s is the integer.. n = 5s so \(h=\frac{n^45n^2+4}{20}.......\frac{(5s)^45*(5s)^2+4}{20}.....\) so whatever be the value of s, the NUMERATOR will be MULTIPLE of 5+4, which will not be a multiple of 5 and thus multiple of 20 ans is always.. NOsufficient B
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Re: If n >2 and n^45n^2+4 = 20h, Is h an integer ? [#permalink]
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22 Aug 2017, 08:25
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Hello chetan2u, Please clarify my below doubt: After solving the equation i got it in normalised form as: (n+1)(n1)(n+2)(n2)=20h Now for statement 1, n=4k. Thus if we put in the values of n=4 and 20, we get: 4 > 5*3*6*2; This will be a multiple of 20 so h has to be integer 20 > 21*19*22*18; This will not be a multiple of 20 since it doesn't have a 5 so h can either be multiple of 5 or a fraction containing 5/multiple of 5 in numerator with a number factor of any of the remaining primes on LHS. Thus insufficient. Now for statement 1, n=5k. Thus if we put in any values of 5; LHS won't be a multiple of 5 thus h can be either be multiple of 5 or a fraction containing 5/multiple of 5 in numerator with a number factor of any of the remaining primes on LHS. Thus insufficient. What am I missing here? I got to the crux of the question and almost solved it but I am unable to understand where am I going wrong? Regards
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Re: If n >2 and n^45n^2+4 = 20h, Is h an integer ? [#permalink]
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22 Aug 2017, 08:35
gmatexam439 wrote: Hello chetan2u, Please clarify my below doubt: After solving the equation i got it in normalised form as: (n+1)(n1)(n+2)(n2)=20h Now for statement 1, n=4k. Thus if we put in the values of n=4 and 20, we get: 4 > 5*3*6*2; This will be a multiple of 20 so h has to be integer 20 > 21*19*22*18; This will not be a multiple of 20 since it doesn't have a 5 so h can either be multiple of 5 or a fraction containing 5/multiple of 5 in numerator with a number factor of any of the remaining primes on LHS. Thus insufficient. Now for statement 1, n=5k. Thus if we put in any values of 5; LHS won't be a multiple of 5 thus h can be either be multiple of 5 or a fraction containing 5/multiple of 5 in numerator with a number factor of any of the remaining primes on LHS. Thus insufficient. What am I missing here? I got to the crux of the question and almost solved it but I am unable to understand where am I going wrong? Regards hi.. you have solved it correctly but going wrong on final interpretation in red coloured portion.. LHS is NOT a multiple of 5 say 5s+4 and RHS is 20h.. so if h is a multiple of 5... 5s+4=20*5... NOT possible.. h has to remove the 5 in 20 and that h can do by being a fraction with 5 in denominator..
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Re: If n >2 and n^45n^2+4 = 20h, Is h an integer ? [#permalink]
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22 Aug 2017, 08:55
chetan2u wrote: hi..
you have solved it correctly but going wrong on final interpretation in red coloured portion.. LHS is NOT a multiple of 5 say 5s+4 and RHS is 20h.. so if h is a multiple of 5... 5s+4=20*5... NOT possible.. h has to remove the 5 in 20 and that h can do by being a fraction with 5 in denominator.. Yes, sorry. I reviewed my equation and as you pointed out 5 should be in the denominator always. Thus it will never be an integer. Damn !! Thank you so much for the quick reply. Regards
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Re: If n >2 and n^45n^2+4 = 20h, Is h an integer ? [#permalink]
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23 Aug 2017, 13:02
chetan2u wrote: gary391 wrote: If n >2 and n^45n^2+4 = 20h, Is h an integer? \(\frac{n}{4}\) is an integer.. let n/4 = s where s is the integer.. n = 4s so \(h=\frac{n^45n^2+4}{20}.......\frac{(4s)^45*(4s)^2+4}{20}.....\) so if s is 1, ans is 180/20=9 yes h is an integer.. but when s is multiple of 5, clearly numerator will be multiple of 5+4, which will not be multiple of 5 and thus 20.. so h is NOT an integer insufficient
Chetan2u, can you, please explain this moment? It seems, I miss some properties of numbers. but when s is multiple of 5, clearly numerator will be multiple of 5+4  from where does it come from? I solved in next manner: (n2)(n1)(n+1)(n+2) = 20h . n=4 > divisable. n=20 > not divisable > insufficient The disadvantage of this method  I can miss option n = 20. And in your solution it seems impossible. . n is divisable by 5 > n2, n1, n+1, n+2  none of them is divisable by 5 > so the answer is: never divisable > not integer. Sufficient.



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Re: If n >2 and n^45n^2+4 = 20h, Is h an integer ? [#permalink]
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23 Aug 2017, 19:07
DharLog wrote: chetan2u wrote: gary391 wrote: If n >2 and n^45n^2+4 = 20h, Is h an integer? \(\frac{n}{4}\) is an integer.. let n/4 = s where s is the integer.. n = 4s so \(h=\frac{n^45n^2+4}{20}.......\frac{(4s)^45*(4s)^2+4}{20}.....\) so if s is 1, ans is 180/20=9 yes h is an integer.. but when s is multiple of 5, clearly numerator will be multiple of 5+4, which will not be multiple of 5 and thus 20.. so h is NOT an integer insufficient
Chetan2u, can you, please explain this moment? It seems, I miss some properties of numbers. but when s is multiple of 5, clearly numerator will be multiple of 5+4  from where does it come from? I solved in next manner: (n2)(n1)(n+1)(n+2) = 20h . n=4 > divisable. n=20 > not divisable > insufficient The disadvantage of this method  I can miss option n = 20. And in your solution it seems impossible. . n is divisable by 5 > n2, n1, n+1, n+2  none of them is divisable by 5 > so the answer is: never divisable > not integer. Sufficient. hi.. equation is \((4s)^45s^2+4\).. now if s is a multiple of 5, (4s)^4 will become (4*5)^4=20^4, which will be div by 5 similarly 5s^2 will also be multiple of 5. left is 4, so the entire equation becomes MULTIPLE of 5 +4, which will not be div by 5
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Re: If n >2 and n^45n^2+4 = 20h, Is h an integer ? [#permalink]
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24 Aug 2017, 03:15
hi.. equation is \((4s)^45s^2+4\).. now if s is a multiple of 5, (4s)^4 will become (4*5)^4=20^4, which will be div by 5 similarly 5s^2 will also be multiple of 5. left is 4, so the entire equation becomes MULTIPLE of 5 +4, which will not be div by 5 =============================
Fuf, it seems to me, I wanted to sleep too much =) (MULTIPLE of 5) + 4  how could I understand it in another way?
Thank you very much!



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If n >2 and n^45n^2+4 = 20h, Is h an integer ? [#permalink]
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31 Oct 2017, 12:27
(n^4  5n^2 + 4) => (n^4  4n^2  n^2 + 4) => (n^2  4)(n^2  1) => (n2)(n1)(n+1)(n+2)
question is (n2)(n1)(n+1)(n+2)/20? => (n2)(n1)(n+1)(n+2)/ (5 * 4)? from above we can see for any value of n, two values of (n2),(n1),(n+1),(n+2) will be even, so it is definitely divisible by 4,
so can be simplified as (n2)(n1)(n+1)(n+2) divisible by 5?
Statement 1: n is multiple of 4, if n = 4, (n2)(n1)(n+1)(n+2) => 2 * 3 * 5 * 6 / 5? YES if n = 20 (LCM of 5 and 4) => (18 * 19 * 21 * 22)/5? NO Not sufficient
Statement 2: n/5 integer => n is multiple of 5 => so definitely any of (n2),(n1),(n+1),(n+2) wont be divisible by 5 => answer to question is NO Sufficient
Answer (B)




If n >2 and n^45n^2+4 = 20h, Is h an integer ?
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