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For how many integer values of n is (n^2 + 6n + 24)/(n + 3) an integer 02 Nov 2023, 03:11
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28% (02:28) correct 72% (02:38) wrong based on 119 sessions

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For how many integer values of n is \(\frac{n^2 + 6n + 24}{n + 3}\) an integer?

A. 1
B. 2
C. 4
D. 6
E. 8

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For how many integer values of n is (n^2 + 6n + 24)/(n + 3) an integer 23 Jan 2024, 02:56
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For how many integer values of \(n\) is \(\frac{n^2 + 6n + 24}{n + 3}\) an integer?

A. 1
B. 2
C. 4
D. 6
E. 8


Start with completing the square in the numerator:

    \(\frac{n^2 + 6n + 24}{n + 3} = \frac{n^2 + 6n + 9 + 15}{n + 3} = \frac{(n + 3)^2 + 15}{n + 3}\).

Split the fraction:

    \(\frac{(n + 3)^2 + 15}{n + 3} = \frac{(n + 3)^2}{n + 3} + \frac{15}{n + 3} = n + 3 + \frac{15}{n + 3}\).

Since \(n + 3\) is an integer, the whole expression will be an integer whenever \(\frac{15}{n + 3}\) is an integer.

For \(\frac{15}{n + 3}\), the denominator, \(n+3\), must be a divisor of 15. Since \(15 = 3*5\), it has \((1 + 1)(1 + 1) = 4\) positive divisors and 4 negative ones: 1, 3, 5, 15, -1, -3, -5, and -15. Therefore, \(n\) can take a total of 8 values: -2, 0, 2, 12, -4, -6, -8, and -18, respectively.


Answer: E
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For how many integer values of n is (n^2 + 6n + 24)/(n + 3) an integer 02 Nov 2023, 04:10
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Bunuel
For how many integer values of n is \(\frac{n^2 + 6n + 24}{n + 3}\) an integer?

A. 1
B. 2
C. 3
D. 4
E. 6
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\(\frac{n^2 + 6n + 24}{n + 3}\) = \(\frac{n^2 + 6n + 9+15}{n + 3}\)

= \(\frac{(n+3)^2 + 15}{n + 3}\)

= \(\frac{(n+3)^2}{n + 3} + \frac{15}{n + 3}\)

= \((n+3) + \frac{15}{n + 3}\)

For \(\frac{n^2 + 6n + 24}{n + 3}\) to be an integer, \(\frac{15}{n + 3}\) must be an integer.

Hence, n+3 must be a factor of 15.

Number of factors ⇒

15 = 5 * 3

Number of positive factors = 2 * 2 = 4 factors

Total factors = 4 * 2 = 8 factors

Note: We multiplied by 2 to also account for negative factors.

Hence, there are \(8\) values of \(n\) for which \(\frac{n^2 + 6n + 24}{n + 3}\) is an integer

Option E
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For how many integer values of n is (n^2 + 6n + 24)/(n + 3) an integer 02 Nov 2023, 07:28
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in my opinion, you are correct until this: "Hence, n+3 must be a factor of 15."

now, factors of 15 = 1, 3, 5, 15 : n+ 3 cannot be 1 so n+3 can only be 3, 5, 15 which means n can be 0, 2, 12.

n can take 3 possible values. answer is C

Please let me know if I am missing anything.
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For how many integer values of n is (n^2 + 6n + 24)/(n + 3) an integer 02 Nov 2023, 07:36
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in my opinion, you are correct until this: "Hence, n+3 must be a factor of 15."

now, factors of 15 = 1, 3, 5, 15 : n+ 3 cannot be 1 so n+3 can only be 3, 5, 15 which means n can be 0, 2, 12.

n can take 3 possible values. answer is C

Please let me know if I am missing anything.
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"n+ 3 cannot be 1" → Why do you think so?

\(n + 3 = 1\)

\(n = -2\)

Hence, the numerator \(n^2 + 6n + 24 = 16\)

\(\frac{n^2 + 6n + 24 }{ n+3} = 16\) ⇒ the answer is an integer.

Also, why can't the factor be -1, -3, -5, -15?

Ex. if n + 3 = -1, so n can be 2 & so on ..
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For how many integer values of n is (n^2 + 6n + 24)/(n + 3) an integer 11 Apr 2024, 13:50
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Bunuel
For how many integer values of n is \(\frac{n^2 + 6n + 24}{n + 3}\) an integer?

A. 1
B. 2
C. 4
D. 6
E. 8

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­\(\frac{n^2 + 6n + 24}{n + 3}\) = \(\frac{n^2 + 2*n*3 + 3^2 + 15}{n + 3}\) = \(\frac{(n+3)^2 + 15}{n + 3}\)
That means, 15/(n+3) has to be an integer.
15 can be divided by 1,3,5,15 (if we consider positive values) and -1,-3,-5,-15 (if we consider negative values). The total value n can have is 8. Option (E) is correct.
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For how many integer values of n is (n^2 + 6n + 24)/(n + 3) an integer 16 Apr 2025, 00:38
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