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If n > 2 and n^4 - 5n^2 + 4 = 20h, is h an integer? 13 Aug 2020, 00:02
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28% (02:41) correct 72% (02:38) wrong based on 158 sessions

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If \(n > 2\) and \(n^4 - 5n^2 + 4 = 20h\), is h an integer?

(1) \(\frac{n}{4}\) is an integer

(2) \(\frac{n}{5}\) is an integer


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If n > 2 and n^4 - 5n^2 + 4 = 20h, is h an integer? Updated on: 02 Mar 2025, 10:59
Originally posted by GMATinsight on 13 Aug 2020, 00:24.
Last edited by Bunuel on 02 Mar 2025, 10:59, edited 2 times in total.
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Bunuel
If \(n > 2\) and \(n^4 - 5n^2 + 4 = 20h\), is h an integer?


(1) \(\frac{n}{4}\) is an integer

(2) \(\frac{n}{5}\) is an integer
Show more

Answer: Option B

GMATinsight's Solution
\(n^4 - 5n^2 + 4 = 20h\)
i.e. \((n^2-4) * (n^2-1) = 20h\)

i.e. \((n-2)(n+2)*(n-1)(n+1) = 20h\)

i.e. \((n-2)(n-1)*(n+1)(n+2) = 20h\)

STatment 1) \(\frac{n}{4}\) is an integer

@n=4, \((4-2)(4-1)*(4+1)(4+2) = 20h\) i.e. h is an Integer
@n=20, \((20-2)(20-1)*(20+1)(20+2) = 20h\) i.e. h is NOT an Integer

NOT SUFFICIENT

Statement 2: \(\frac{n}{5}\) is an integer

i.e. n is a multiple of 5

For any value of n that is multiple of 5, (n-2)(n-1)*(n+1)(n+2) will NOT be a multiple of 5 hence h is always a NON-INTEGER

SUFFICIENT

ANswer: Option B
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If n > 2 and n^4 - 5n^2 + 4 = 20h, is h an integer? 13 Aug 2020, 01:14
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Bunuel
If \(n > 2\) and \(n^4 - 5n^2 + 4 = 20h\), is h an integer?

(1) \(\frac{n}{4}\) is an integer

(2) \(\frac{n}{5}\) is an integer
Show more

IMO B is the correct answer.

We have n^4 - 5n^2 + 4 = (n^2 - 1) * (n^2 - 4) = (n - 1) * (n + 1) * (n - 2) * (n + 2)
Then h = [(n - 1) * (n + 1) * (n - 2) * (n + 2)] / 20

(1) If n/4 is an integer. Let's say n/4 = k then n = 4k
h = [(4k - 1) * (4k + 1) * (4k - 2) * (4k + 2)] / 20
h = [(4k - 1) * (4k + 1) * (2k - 1) * (2k + 1)] / 5

Since n > 2 then k >= 1. We will try each example for k from 1 to 5
If k = 1 then h = 3 * 5 * 1 * 3 / 5 (Integer)
If k = 2 then h = 7 * 9 * 3 * 5 / 5 (Integer)
...
If k = 5 then h = 19 * 21 * 9 * 11 / 5 (Not integer)
--> (1) is INSUFFICIENT

(2) If n/5 is an integer. Let's say n/5 = k then n = 5k
h = [(5k - 1) * (5k + 1) * (5k - 2) * (5k + 2)] / 20
For sure (5k - 1) * (5k + 1) * (5k - 2) * (5k + 2) cannot divided by 5 then h is for sure not an integer
--> (2) is SUFFICIENT
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If n > 2 and n^4 - 5n^2 + 4 = 20h, is h an integer? Updated on: 14 Aug 2020, 06:59
Originally posted by GMATWhizTeam on 13 Aug 2020, 04:31.
Last edited by GMATWhizTeam on 14 Aug 2020, 06:59, edited 1 time in total.
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Bunuel
If \(n > 2\) and \(n^4 - 5n^2 + 4 = 20h\), is h an integer?


(1) \(\frac{n}{4}\) is an integer

(2) \(\frac{n}{5}\) is an integer

Show more

Solution


Step 1: Analyse Question Stem


    • n > 2
    • \(n^4 – 5n^2 + 4 = 20h\)
    • We need to find if h is an integer.
      o Now, n will be an integer only if \( n^4 – 5n^2 + 4 = 20h\) is a multiple of 20.
      o i.e. h is an integer if \((\frac{(n^4 -5n^2 + 4)}{20})_R= 0\)

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE


Statement 1: \(\frac{n}{4}\) is an integer.
    • According to this statement: n = 4k, where k is an integer.
    • So, \((\frac{(n^4 -5n^2 + 4)}{20})_R= (\frac{4^4*k^4 -5*4^2*k^2 + 4}{20})_R = (\frac{256*k^4 -80*k^2 + 4}{20})_R \)
    • Depending on the values of k there can be many cases, for example, consider the following two cases:
      o Case 1: if k = 5,
         \( (\frac{256*k^4 -80*k^2 + 4}{20})_R =(\frac{256*5^4}{20})_R -(\frac{80*5^2}{20})_R + (\frac{4}{20})_R =0 - 0 + 4 = 4 \), which is not equal to 0.
      o Case 2: if k = 1,
      o \( (\frac{256*k^4 -80*k^2 + 4}{20})_R =(\frac{256 -80 + 4}{20})_R = 0 \)
    • We are getting two contradictory results.
Hence, statement 1 is NOT sufficient and we can eliminate answer Options A and D.

Statement 2: \(\frac{n}{5}\) is an integer.
    • According to this statement: n = 5m, where m is an integer.
    • So, \((\frac{(n^4 -5n^2 + 4)}{20})_R= (\frac{5^4*m^4 -5*5^2*m^2 + 4}{20})_R = (\frac{625*m^4 -125*m^2 + 4}{20})_R \)
    • There can be two cases:
      o Case 1: if m is even, say m = 2,
         In that case \( (\frac{625*k^4 -125*k^2 + 4}{20})_R =(\frac{625*2^4}{20})_R -(\frac{125*2^2}{20})_R +(\frac{ 4}{20})_R = 0 -0 + 4 = 4 \), which is not equal to 0.
      o Case 2: if m is odd, say m = 1,
         \( (\frac{625*m^4 -125*m^2 + 4}{20})_R =(\frac{625 -125 + 4}{20})_R = 4 \), which is not equal to 0.
      o Thus, h is not an integer.
Hence, statement 2 is sufficient.
Thus, the correct answer is Option B.
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If n > 2 and n^4 - 5n^2 + 4 = 20h, is h an integer? 13 Aug 2020, 10:00
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If \(n > 2\) and \(n^4 - 5n^2 + 4 = 20h\), is h an integer?


(1) \(\frac{n}{4}\) is an integer

(2) \(\frac{n}{5}\) is an integer


Solution


Step 1: Analyse Question Stem


    • n > 2
    • \(n^4 – 5n^2 + 4 = 20h\)
    • We need to find if h is an integer.
      o Now, n will be an integer only if \( n^4 – 5n^2 + 4 = 20h\) is a multiple of 20.
      o i.e. h is an integer if \((\frac{(n^4 -5n^2 + 4)}{20})_R= 0\)

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE


Statement 1: \(\frac{n}{4}\) is an integer.
    • According to this statement: n = 4k, where k is an integer.
    • So, \((\frac{(n^4 -5n^2 + 4)}{20})_R= (\frac{4^4*k^4 -5*4^2*k^2 + 4}{20})_R = (\frac{256*k^4 -80*k^2 + 4}{20})_R \)
    • Depending on the values of k there can be many cases, for example, consider the following two cases:
      o Case 1: if k = 0,
         \( (\frac{256*k^4 -80*k^2 + 4}{20})_R =(\frac{0 -0 + 4}{20})_R = 4 \), which is not equal to 0.
      o Case 2: if k = 1,
      o \( (\frac{256*k^4 -80*k^2 + 4}{20})_R =(\frac{256 -80 + 4}{20})_R = 0 \)
    • We are getting two contradictory results.
Hence, statement 1 is NOT sufficient and we can eliminate answer Options A and D.

Statement 2: \(\frac{n}{5}\) is an integer.
    • According to this statement: n = 5m, where m is an integer.
    • So, \((\frac{(n^4 -5n^2 + 4)}{20})_R= (\frac{5^4*m^4 -5*5^2*m^2 + 4}{20})_R = (\frac{625*m^4 -125*m^2 + 4}{20})_R \)
    • There can be two cases:
      o Case 1: if m is even, say m = 0,
         In that case \( (\frac{625*k^4 -125*k^2 + 4}{20})_R =(\frac{0 -0 + 4}{20})_R = 4 \), which is not equal to 0.
      o Case 2: if m is odd, say m = 1,
         \( (\frac{625*m^4 -125*m^2 + 4}{20})_R =(\frac{625 -125 + 4}{20})_R = 4 \), which is not equal to 0.
      o Thus, h is not an integer.
Hence, statement 2 is sufficient.
Thus, the correct answer is Option B.
Show more


The question states that n>2, if we take k=0, that makes n=0 since n=4k.
Isn't that a wrong value to test our answer on?
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If n > 2 and n^4 - 5n^2 + 4 = 20h, is h an integer? 14 Aug 2020, 07:21
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Bunuel
If \(n > 2\) and \(n^4 - 5n^2 + 4 = 20h\), is h an integer?


(1) \(\frac{n}{4}\) is an integer

(2) \(\frac{n}{5}\) is an integer


Solution


Step 1: Analyse Question Stem


    • n > 2
    • \(n^4 – 5n^2 + 4 = 20h\)
    • We need to find if h is an integer.
      o Now, n will be an integer only if \( n^4 – 5n^2 + 4 = 20h\) is a multiple of 20.
      o i.e. h is an integer if \((\frac{(n^4 -5n^2 + 4)}{20})_R= 0\)

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE


Statement 1: \(\frac{n}{4}\) is an integer.
    • According to this statement: n = 4k, where k is an integer.
    • So, \((\frac{(n^4 -5n^2 + 4)}{20})_R= (\frac{4^4*k^4 -5*4^2*k^2 + 4}{20})_R = (\frac{256*k^4 -80*k^2 + 4}{20})_R \)
    • Depending on the values of k there can be many cases, for example, consider the following two cases:
      o Case 1: if k = 0,
         \( (\frac{256*k^4 -80*k^2 + 4}{20})_R =(\frac{0 -0 + 4}{20})_R = 4 \), which is not equal to 0.
      o Case 2: if k = 1,
      o \( (\frac{256*k^4 -80*k^2 + 4}{20})_R =(\frac{256 -80 + 4}{20})_R = 0 \)
    • We are getting two contradictory results.
Hence, statement 1 is NOT sufficient and we can eliminate answer Options A and D.

Statement 2: \(\frac{n}{5}\) is an integer.
    • According to this statement: n = 5m, where m is an integer.
    • So, \((\frac{(n^4 -5n^2 + 4)}{20})_R= (\frac{5^4*m^4 -5*5^2*m^2 + 4}{20})_R = (\frac{625*m^4 -125*m^2 + 4}{20})_R \)
    • There can be two cases:
      o Case 1: if m is even, say m = 0,
         In that case \( (\frac{625*k^4 -125*k^2 + 4}{20})_R =(\frac{0 -0 + 4}{20})_R = 4 \), which is not equal to 0.
      o Case 2: if m is odd, say m = 1,
         \( (\frac{625*m^4 -125*m^2 + 4}{20})_R =(\frac{625 -125 + 4}{20})_R = 4 \), which is not equal to 0.
      o Thus, h is not an integer.
Hence, statement 2 is sufficient.
Thus, the correct answer is Option B.


The question states that n>2, if we take k=0, that makes n=0 since n=4k.
Isn't that a wrong value to test our answer on?
Show more

Heymoses46

Yes, It was a miss from our side. We have updated the solution.

Regards, :)
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If n > 2 and n^4 - 5n^2 + 4 = 20h, is h an integer? 22 Mar 2022, 05:59
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