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For how many integers n is n^4 + 6n < 6n^3 + n^2 ? 21 Feb 2020, 05:35
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For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

A. 2
B. 3
C. 4
D. 5
E. 6

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For how many integers n is n^4 + 6n < 6n^3 + n^2 ? 21 Feb 2020, 05:45
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For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

A. 2
B. 3
C. 4
D. 5
E. 6
Show more

\(n^4 + 6n < 6n^3 + n^2\)
i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)

Check the critical value plotting and solutions obtained in image attached.

Answer: Option C
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Screenshot 2020-02-21 at 6.14.56 PM.png
Screenshot 2020-02-21 at 6.14.56 PM.png [ 237.03 KiB | Viewed 9366 times ]

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For how many integers n is n^4 + 6n < 6n^3 + n^2 ? 18 May 2020, 23:47
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Bunuel
For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

A. 2
B. 3
C. 4
D. 5
E. 6

\(n^4 + 6n < 6n^3 + n^2\)
i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)

Check the critical value plotting and solutions obtained in image attached.

Answer: Option C
Show more


i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \) ; could you please explain how you arrived at this from the previous step
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For how many integers n is n^4 + 6n < 6n^3 + n^2 ? 19 May 2020, 01:46
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Bunuel
For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

A. 2
B. 3
C. 4
D. 5
E. 6

\(n^4 + 6n < 6n^3 + n^2\)
i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)

Check the critical value plotting and solutions obtained in image attached.

Answer: Option C


i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \) ; could you please explain how you arrived at this from the previous step
Show more

sume3tss

CHeck the new steps in Blue

i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \(n^2*(n-1)*(n+1) - 6n*(n-1)*(n+1) < 0\)

i.e. \((n-1)*(n+1)*(n^2 - 6n) < 0\)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)
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For how many integers n is n^4 + 6n < 6n^3 + n^2 ? 19 May 2020, 04:17
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Bunuel
For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

A. 2
B. 3
C. 4
D. 5
E. 6

\(n^4 + 6n < 6n^3 + n^2\)
i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)

Check the critical value plotting and solutions obtained in image attached.

Answer: Option C


i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \) ; could you please explain how you arrived at this from the previous step

sume3tss

CHeck the new steps in Blue

i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \(n^2*(n-1)*(n+1) - 6n*(n-1)*(n+1) < 0\)

i.e. \((n-1)*(n+1)*(n^2 - 6n) < 0\)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)
Show more


I understood the above part, but I am not able to understand how it was plotted onto the graph.
Can you please help.
Thanks
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For how many integers n is n^4 + 6n < 6n^3 + n^2 ? 19 May 2020, 05:12
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Bunuel
For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

A. 2
B. 3
C. 4
D. 5
E. 6

\(n^4 + 6n < 6n^3 + n^2\)
i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)

Check the critical value plotting and solutions obtained in image attached.

Answer: Option C


i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \) ; could you please explain how you arrived at this from the previous step

sume3tss

CHeck the new steps in Blue

i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \(n^2*(n-1)*(n+1) - 6n*(n-1)*(n+1) < 0\)

i.e. \((n-1)*(n+1)*(n^2 - 6n) < 0\)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)


I understood the above part, but I am not able to understand how it was plotted onto the graph.
Can you please help.
Thanks
Show more

DhruvS

We plot the critical values on Number line

Critical values are the values of n for which the function result in a zero outcome

Function \(n*(n-6)*(n-1)*(n+1)\) will becomes 0 if
n = 0, or
(n-6) = 0 i.e. n = 6 or
(n-1) = 0 i.e. n = 1 or
(n-+1) = 0 i.e. n =-1 or

So we have 4 critical values i.e. {-1, 0, 1, 6} so we mark those values on Number Line and make loop between adjacent Critical values

Then we try with any value in a loop to see whether function will assume positive value or negative values and that's how we find favorable ranges.
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For how many integers n is n^4 + 6n < 6n^3 + n^2 ? 20 May 2020, 03:09
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Bunuel
For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

A. 2
B. 3
C. 4
D. 5
E. 6

\(n^4 + 6n < 6n^3 + n^2\)
i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)

Check the critical value plotting and solutions obtained in image attached.

Answer: Option C


i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \) ; could you please explain how you arrived at this from the previous step

sume3tss

CHeck the new steps in Blue

i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \(n^2*(n-1)*(n+1) - 6n*(n-1)*(n+1) < 0\)

i.e. \((n-1)*(n+1)*(n^2 - 6n) < 0\)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)


I understood the above part, but I am not able to understand how it was plotted onto the graph.
Can you please help.
Thanks

DhruvS

We plot the critical values on Number line

Critical values are the values of n for which the function result in a zero outcome

Function \(n*(n-6)*(n-1)*(n+1)\) will becomes 0 if
n = 0, or
(n-6) = 0 i.e. n = 6 or
(n-1) = 0 i.e. n = 1 or
(n-+1) = 0 i.e. n =-1 or

So we have 4 critical values i.e. {-1, 0, 1, 6} so we mark those values on Number Line and make loop between adjacent Critical values

Then we try with any value in a loop to see whether function will assume positive value or negative values and that's how we find favorable ranges.
Show more

GMATinsight Thank you for your reply. Please confirm my understanding below:
We plotted the values on the graph and now since n*(n-6)*(n-1)*(n+1)<0, i.e. the expression is less <0. We need to consider only the adjacent critical values and make two ranges==> -1<n<0 and 1<n<6. As there are no integers between -1<n<0, hence the only possible values are 2,3,4,5, according to 1<n<6.

Also can we not consider -1<n<6 to include all values ?
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Schools: IIM (A) ISB '24
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For how many integers n is n^4 + 6n < 6n^3 + n^2 ? 21 May 2020, 06:44
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GMATinsight
Bunuel
For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

A. 2
B. 3
C. 4
D. 5
E. 6

\(n^4 + 6n < 6n^3 + n^2\)
i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)

Check the critical value plotting and solutions obtained in image attached.

Answer: Option C


i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \) ; could you please explain how you arrived at this from the previous step

sume3tss

CHeck the new steps in Blue

i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \(n^2*(n-1)*(n+1) - 6n*(n-1)*(n+1) < 0\)

i.e. \((n-1)*(n+1)*(n^2 - 6n) < 0\)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)


I understood the above part, but I am not able to understand how it was plotted onto the graph.
Can you please help.
Thanks

DhruvS

We plot the critical values on Number line

Critical values are the values of n for which the function result in a zero outcome

Function \(n*(n-6)*(n-1)*(n+1)\) will becomes 0 if
n = 0, or
(n-6) = 0 i.e. n = 6 or
(n-1) = 0 i.e. n = 1 or
(n-+1) = 0 i.e. n =-1 or

So we have 4 critical values i.e. {-1, 0, 1, 6} so we mark those values on Number Line and make loop between adjacent Critical values

Then we try with any value in a loop to see whether function will assume positive value or negative values and that's how we find favorable ranges.

GMATinsight Thank you for your reply. Please confirm my understanding below:
We plotted the values on the graph and now since n*(n-6)*(n-1)*(n+1)<0, i.e. the expression is less <0. We need to consider only the adjacent critical values and make two ranges==> -1<n<0 and 1<n<6. As there are no integers between -1<n<0, hence the only possible values are 2,3,4,5, according to 1<n<6.

Also can we not consider -1<n<6 to include all values ?
Show more


DhruvS

NO, we can NOT incluce values in range -1<n<6 simply because the function is not negative between 0 and 1
'
also because function is not negative at values n = 0 and n = 1

I hope it clear you doubt. :)
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For how many integers n is n^4 + 6n < 6n^3 + n^2 ? 21 May 2020, 09:56
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DhruvS
GMATinsight
Bunuel
For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

A. 2
B. 3
C. 4
D. 5
E. 6

\(n^4 + 6n < 6n^3 + n^2\)
i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)

Check the critical value plotting and solutions obtained in image attached.

Answer: Option C


i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \) ; could you please explain how you arrived at this from the previous step

sume3tss

CHeck the new steps in Blue

i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \(n^2*(n-1)*(n+1) - 6n*(n-1)*(n+1) < 0\)

i.e. \((n-1)*(n+1)*(n^2 - 6n) < 0\)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)


I understood the above part, but I am not able to understand how it was plotted onto the graph.
Can you please help.
Thanks

DhruvS

We plot the critical values on Number line

Critical values are the values of n for which the function result in a zero outcome

Function \(n*(n-6)*(n-1)*(n+1)\) will becomes 0 if
n = 0, or
(n-6) = 0 i.e. n = 6 or
(n-1) = 0 i.e. n = 1 or
(n-+1) = 0 i.e. n =-1 or

So we have 4 critical values i.e. {-1, 0, 1, 6} so we mark those values on Number Line and make loop between adjacent Critical values

Then we try with any value in a loop to see whether function will assume positive value or negative values and that's how we find favorable ranges.

GMATinsight Thank you for your reply. Please confirm my understanding below:
We plotted the values on the graph and now since n*(n-6)*(n-1)*(n+1)<0, i.e. the expression is less <0. We need to consider only the adjacent critical values and make two ranges==> -1<n<0 and 1<n<6. As there are no integers between -1<n<0, hence the only possible values are 2,3,4,5, according to 1<n<6.

Also can we not consider -1<n<6 to include all values ?


DhruvS

NO, we can NOT incluce values in range -1<n<6 simply because the function is not negative between 0 and 1
'
also because function is not negative at values n = 0 and n = 1

I hope it clear you doubt. :)
Show more


GMATinsight Understood. Thank You so much. :)
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For how many integers n is n^4 + 6n < 6n^3 + n^2 ? 21 May 2020, 10:12
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Bunuel
For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

A. 2
B. 3
C. 4
D. 5
E. 6

\(n^4 + 6n < 6n^3 + n^2\)
i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)

Check the critical value plotting and solutions obtained in image attached.

Answer: Option C


i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \) ; could you please explain how you arrived at this from the previous step

sume3tss

CHeck the new steps in Blue

i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \(n^2*(n-1)*(n+1) - 6n*(n-1)*(n+1) < 0\)

i.e. \((n-1)*(n+1)*(n^2 - 6n) < 0\)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)


I understood the above part, but I am not able to understand how it was plotted onto the graph.
Can you please help.
Thanks

DhruvS

We plot the critical values on Number line

Critical values are the values of n for which the function result in a zero outcome

Function \(n*(n-6)*(n-1)*(n+1)\) will becomes 0 if
n = 0, or
(n-6) = 0 i.e. n = 6 or
(n-1) = 0 i.e. n = 1 or
(n-+1) = 0 i.e. n =-1 or

So we have 4 critical values i.e. {-1, 0, 1, 6} so we mark those values on Number Line and make loop between adjacent Critical values

Then we try with any value in a loop to see whether function will assume positive value or negative values and that's how we find favorable ranges.

GMATinsight Thank you for your reply. Please confirm my understanding below:
We plotted the values on the graph and now since n*(n-6)*(n-1)*(n+1)<0, i.e. the expression is less <0. We need to consider only the adjacent critical values and make two ranges==> -1<n<0 and 1<n<6. As there are no integers between -1<n<0, hence the only possible values are 2,3,4,5, according to 1<n<6.

Also can we not consider -1<n<6 to include all values ?


DhruvS

NO, we can NOT incluce values in range -1<n<6 simply because the function is not negative between 0 and 1
'
also because function is not negative at values n = 0 and n = 1

I hope it clear you doubt. :)


GMATinsight Understood. Thank You so much. :)
Show more

DhruvS

The tradition of thanking someone on GMAT CLUB is by pressing +1KUDOS button on the post that helped you... :) :cool:
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For how many integers n is n^4 + 6n < 6n^3 + n^2 ? 21 May 2020, 10:24
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GMATinsight
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DhruvS
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Bunuel
For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

A. 2
B. 3
C. 4
D. 5
E. 6

\(n^4 + 6n < 6n^3 + n^2\)
i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)

Check the critical value plotting and solutions obtained in image attached.

Answer: Option C


i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \) ; could you please explain how you arrived at this from the previous step

sume3tss

CHeck the new steps in Blue

i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \(n^2*(n-1)*(n+1) - 6n*(n-1)*(n+1) < 0\)

i.e. \((n-1)*(n+1)*(n^2 - 6n) < 0\)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)


I understood the above part, but I am not able to understand how it was plotted onto the graph.
Can you please help.
Thanks

DhruvS

We plot the critical values on Number line

Critical values are the values of n for which the function result in a zero outcome

Function \(n*(n-6)*(n-1)*(n+1)\) will becomes 0 if
n = 0, or
(n-6) = 0 i.e. n = 6 or
(n-1) = 0 i.e. n = 1 or
(n-+1) = 0 i.e. n =-1 or

So we have 4 critical values i.e. {-1, 0, 1, 6} so we mark those values on Number Line and make loop between adjacent Critical values

Then we try with any value in a loop to see whether function will assume positive value or negative values and that's how we find favorable ranges.

GMATinsight Thank you for your reply. Please confirm my understanding below:
We plotted the values on the graph and now since n*(n-6)*(n-1)*(n+1)<0, i.e. the expression is less <0. We need to consider only the adjacent critical values and make two ranges==> -1<n<0 and 1<n<6. As there are no integers between -1<n<0, hence the only possible values are 2,3,4,5, according to 1<n<6.

Also can we not consider -1<n<6 to include all values ?


DhruvS

NO, we can NOT incluce values in range -1<n<6 simply because the function is not negative between 0 and 1
'
also because function is not negative at values n = 0 and n = 1

I hope it clear you doubt. :)


GMATinsight Understood. Thank You so much. :)

DhruvS

The tradition of thanking someone on GMAT CLUB is by pressing +1KUDOS button on the post that helped you... :) :cool:
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Thanks for reminding. :lol: :tongue_opt3
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For how many integers n is n^4 + 6n < 6n^3 + n^2 ? 24 May 2020, 06:58
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Bunuel
For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

A. 2
B. 3
C. 4
D. 5
E. 6

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Solution:

n^4 + 6n < 6n^3 + n^2

n^4 - 6n^3 - n^2 + 6n < 0

n^3(n - 6) - n(n - 6) < 0

(n^3 - n)(n - 6) < 0

n(n^2 - 1)(n - 6) < 0

n(n - 1)(n + 1)(n - 6) < 0

We see that on the left hand side of the last inequality, there are 4 factors.

If n is a negative integer less than -1, then all the 4 factors are negative. So the product of the 4 factors will be positive and will not be less than 0.

If n is -1, 0, 1 or 6, then one of the factors will be 0. So the product will be 0 and will not be less than 0.

If n is 2, 3, 4, or 5, the first 3 factors are positive while the last factor is negative. So the product will be negative and hence less than 0.

If n is a positive integer greater than 6, then all the 4 factors are positive. So the product will be positive and will not be less than 0.

Alternate Solution:

Let’s rewrite the given inequality as n^4 - n^2 < 6n^3 - 6n. Then:

n^2(n^2 - 1) < 6n(n^2 - 1)

n^2(n^2 - 1) - 6n(n^2 - 1) < 0

(n^2 - 1)(n^2 - 6n) < 0

In order to have the product of two expressions negative, one of the expressions must be negative and the other expression must be positive.

If n^2 - 1 is negative, then only n = 0 satisfies n^2 - 1 < 0. Notice that n = 0 does not satisfy n^4 + 6n < 6n^3 + n^2; thus it follows that n^2 - 1 must be positive.

Since n^2 - 1 > 0, |n| must be 2 or greater. Further, since n^2 - 1 is positive, n^2 - 6n must be negative; i.e. n^2 < 6n. Notice that this inequality is not satisfied for any negative integer and only satisfied for n = 1, 2, 3, 4 or 5. Combined with the earlier condition that |n| must be 2 or greater, we see that there are four integers satisfying the given inequality (namely 2, 3, 4 and 5).

Answer: C
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For how many integers n is n^4 + 6n < 6n^3 + n^2 ? 24 May 2020, 07:29
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Bunuel
For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

A. 2
B. 3
C. 4
D. 5
E. 6

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Asked: For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

\(n^4 + 6n - 6n^3 - n^2 <0\)
\(n(n^3 - 6n^2 - n + 6) < 0\)
\(n(n-6)(n^2-1)<0\)
n(n-1)(n+1)(n-6) < 0

By WAVY LINE method: -
1<n<6
-1<n<0

n = {2,3,4,5}

IMO C
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For how many integers n is n^4 + 6n < 6n^3 + n^2 ? 24 Jul 2023, 06:36
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Bunuel
For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

A. 2
B. 3
C. 4
D. 5
E. 6

\(n^4 + 6n < 6n^3 + n^2\)
i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)

Check the critical value plotting and solutions obtained in image attached.

Answer: Option C
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This could be a very stupid thing to ask but bear with me please. is it okay to move 6n to the left side of inequality when we dont know whether n is positive or negative?
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For how many integers n is n^4 + 6n < 6n^3 + n^2 ? 06 Nov 2025, 13:58
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