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# If n >2 and n^4-5n^2+4 = 20h, Is h an integer ?

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If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

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21 Aug 2017, 06:57
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If n >2 and n^4-5n^2+4 = 20h, Is h an integer?

1) n/4 is an integer

2) n/5 is an integer
[Reveal] Spoiler: OA

Last edited by chetan2u on 22 Aug 2017, 06:42, edited 1 time in total.
Math Expert
Joined: 02 Aug 2009
Posts: 5537
Re: If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

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22 Aug 2017, 06:41
3
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Expert's post
gary391 wrote:
If n >2 and n^4-5n^2+4 = 20h, Is h an integer?

1) n/4 is an integer

2) n/5 is an integer

HI...

otherwise for the solution..
$$n^4-5n^2+4 = 20h.............. h=\frac{n^4-5n^2+4}{20}$$
for h to be an integer $$n^4-5n^2+4$$ 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 $$h=\frac{n^4-5n^2+4}{20}.......\frac{(4s)^4-5*(4s)^2+4}{20}.....$$
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

II
$$\frac{n}{5}$$ is an integer..
let n/5 = s where s is the integer.. n = 5s
so $$h=\frac{n^4-5n^2+4}{20}.......\frac{(5s)^4-5*(5s)^2+4}{20}.....$$
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.. NO

sufficient

B
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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

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Re: If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

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22 Aug 2017, 07:25
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Hello chetan2u,

After solving the equation i got it in normalised form as:
(n+1)(n-1)(n+2)(n-2)=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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Posts: 5537
Re: If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

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22 Aug 2017, 07:35
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Expert's post
gmatexam439 wrote:
Hello chetan2u,

After solving the equation i got it in normalised form as:
(n+1)(n-1)(n+2)(n-2)=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..

_________________

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

BANGALORE/-

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Joined: 28 Mar 2017
Posts: 825
Re: If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

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22 Aug 2017, 07:55
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^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

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23 Aug 2017, 12:02
chetan2u wrote:
gary391 wrote:
If n >2 and n^4-5n^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 $$h=\frac{n^4-5n^2+4}{20}.......\frac{(4s)^4-5*(4s)^2+4}{20}.....$$
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:
(n-2)(n-1)(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 -> n-2, n-1, n+1, n+2 - none of them is divisable by 5 ---> so the answer is: never divisable ---> not integer. Sufficient.
Math Expert
Joined: 02 Aug 2009
Posts: 5537
Re: If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

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23 Aug 2017, 18:07
1
KUDOS
Expert's post
DharLog wrote:
chetan2u wrote:
gary391 wrote:
If n >2 and n^4-5n^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 $$h=\frac{n^4-5n^2+4}{20}.......\frac{(4s)^4-5*(4s)^2+4}{20}.....$$
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:
(n-2)(n-1)(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 -> n-2, n-1, n+1, n+2 - none of them is divisable by 5 ---> so the answer is: never divisable ---> not integer. Sufficient.

hi..
equation is $$(4s)^4-5s^2+4$$..
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
_________________

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

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Joined: 26 Jun 2017
Posts: 115
Location: Russian Federation
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Re: If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

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24 Aug 2017, 02:15
hi..
equation is $$(4s)^4-5s^2+4$$..
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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Joined: 02 Apr 2014
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If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

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31 Oct 2017, 11:27
(n^4 - 5n^2 + 4) => (n^4 - 4n^2 - n^2 + 4) => (n^2 - 4)(n^2 - 1) => (n-2)(n-1)(n+1)(n+2)

question is (n-2)(n-1)(n+1)(n+2)/20? => (n-2)(n-1)(n+1)(n+2)/ (5 * 4)?
from above we can see for any value of n, two values of (n-2),(n-1),(n+1),(n+2) will be even, so it is definitely divisible by 4,

so can be simplified as (n-2)(n-1)(n+1)(n+2) divisible by 5?

Statement 1:
n is multiple of 4,
if n = 4, (n-2)(n-1)(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 (n-2),(n-1),(n+1),(n+2) wont be divisible by 5 => answer to question is NO
Sufficient

If n >2 and n^4-5n^2+4 = 20h, Is h an integer ?   [#permalink] 31 Oct 2017, 11:27
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