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macjas
If n is a positive integer and if (n^3 - n)/(n+1) = 240, then what is the value of n?

A. 12
B. 16
C. 17
D. 20
E. 48

\(\frac{(n^3 - n)}{(n+1)}\) = 240

\(\frac{(n^3 - n)}{(n+1)}\) = \((16 )(15)\)

\(\frac{n(n^2 - 1)}{(n+1)}\) = \((16 )(15)\)

\(\frac{n(n - 1)(n+1)}{(n+1)}\)= \((16 )(15)\)

\(n(n - 1)\) = \((16 )(15)\)

So, \(n\) = \(16\)

Hence answer is definitely (B)
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If n is a positive integer and if (n^3 - n)/(n+1) = 240, then what is the value of n?

A. 12
B. 16
C. 17
D. 20
E. 48

We can simplify the given equation:

(n^3 - n)/(n + 1) = 240

n(n^2 - 1)/(n + 1) = 240

n(n + 1)(n - 1)/(n + 1) = 240

n(n - 1) = 240

n^2 - n - 240 = 0

(n - 16)(n + 15) = 0

n = 16 or n = -15

Answer: B
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macjas
If n is a positive integer and if (n^3 - n)/(n+1) = 240, then what is the value of n?

A. 12
B. 16
C. 17
D. 20
E. 48

\(\frac{n^3-n}{n+1}=240\) --> \(\frac{n(n-1)(n+1)}{n+1}=240\) --> \(n(n-1)=240\) --> 240 is the product of two positive consecutive integers \(n-1\) and \(n\) --> \(n=16\) (16*15=240).

Answer: B.

What is the significance of "If n is a positive integer" in this question? Would we not cancel (n+1) if n is not a positive integer?
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Bunuel
macjas
If n is a positive integer and if (n^3 - n)/(n+1) = 240, then what is the value of n?

A. 12
B. 16
C. 17
D. 20
E. 48

\(\frac{n^3-n}{n+1}=240\) --> \(\frac{n(n-1)(n+1)}{n+1}=240\) --> \(n(n-1)=240\) --> 240 is the product of two positive consecutive integers \(n-1\) and \(n\) --> \(n=16\) (16*15=240).

Answer: B.

What is the significance of "If n is a positive integer" in this question? Would we not cancel (n+1) if n is not a positive integer?

We would but after getting \(n(n-1)=240\) we wouldn't be able to use the logic in highlighted part because if it were given that n is an integer, then \(n-1\) and \(n\) might not be consecutive integers and we would be left with quadratics to solve. We'd get the same answer though.
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What is the significance of "If n is a positive integer" in this question? Would we not cancel (n+1) if n is not a positive integer?[/quote]

We would but after getting \(n(n-1)=240\) we wouldn't be able to use the logic in highlighted part because if it were given that n is an integer, then \(n-1\) and \(n\) might not be consecutive integers and we would be left with quadratics to solve. We'd get the same answer though.[/quote]

Hi Bunuel

Great work, you mentioned that we might not be able to get consecutive integers, but -16 and -15 are also consecutive integers and as you have mentioned we would arrive at the same solution. I would lean towards the isolation of a single solution as the significance of the positive integer constraint. As you are well aware of, this would probably be more significant in a DS yes or no or even a DS value question.

Thanks again for the excellent contributions.
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Great work, you mentioned that we might not be able to get consecutive integers, but -16 and -15 are also consecutive integers and as you have mentioned we would arrive at the same solution. I would lean towards the isolation of a single solution as the significance of the positive integer constraint. As you are well aware of, this would probably be more significant in a DS yes or no or even a DS value question.

Thanks again for the excellent contributions.

Notice that we are not only told that n is an integer but also that n is positive: " n is a positive integer".
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We would but after getting \(n(n-1)=240\) we wouldn't be able to use the logic in highlighted part because if it were given that n is an integer, then \(n-1\) and \(n\) might not be consecutive integers and we would be left with quadratics to solve. We'd get the same answer though.

I like the idea of plugging and not solving the quadratic equation...this saves a hell of time
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[/quote]

What is the significance of "If n is a positive integer" in this question? Would we not cancel (n+1) if n is not a positive integer?[/quote]


We would but after getting \(n(n-1)=240\) we wouldn't be able to use the logic in highlighted part because if it were given that n is an integer, then \(n-1\) and \(n\) might not be consecutive integers and we would be left with quadratics to solve. We'd get the same answer though.[/quote]


Thank you. This helps :) :thumbup:
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