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Difficulty: HardDivisibility, Factors, and Multiples

For how many positive integers nn is the expression n2+7n+120n+3\frac{n^2 + 7n + 120}{n + 3} equal to an integer?

  1. A
    6
  2. 9Answer
  3. C
    10
  4. D
    11
  5. E
    12

Answer

9 positive integers
By dividing the numerator by the denominator, the expression simplifies to n+4+108n+3n + 4 + \frac{108}{n + 3}. For this to yield an integer, (n+3)(n + 3) must be a positive divisor of 108. The number 108 has prime factorization 22⋅332^2 \cdot 3^3, giving it (2+1)(3+1)=12(2+1)(3+1) = 12 total positive divisors. Since nn must be a positive integer (n≥1n \ge 1), n+3n + 3 must be at least 4. Eliminating the divisors 1, 2, and 3 leaves 9 valid values for n+3n + 3, which correspond to 9 unique positive integers nn.

Step-by-Step Solution

1
Perform polynomial long division or algebraic manipulation on the numerator
\frac{n^2 + 7n + 120}{n + 3} = \frac{(n^2 + 3n) + (4n + 12) + 108}{n + 3} = \frac{n(n+3) + 4(n+3) + 108}{n + 3} = n + 4 + \frac{108}{n + 3}
Separating the expression into an integer term (n+4)(n + 4) and a proper fractional term 108n+3\frac{108}{n + 3} isolates the divisibility condition.
2
Determine the condition for the fractional term to be an integer
The expression is an integer if and only if (n+3)(n + 3) is a positive integer divisor of 108.
Since nn is a positive integer, (n+4)(n + 4) is always an integer, so 108n+3\frac{108}{n + 3} must also be an integer.
3
Calculate the total number of positive divisors of 108
Prime factorization: 108=22⋅33108 = 2^2 \cdot 3^3. Total divisors = (2+1)(3+1)=3⋅4=12(2+1)(3+1) = 3 \cdot 4 = 12.
The formula for the number of positive divisors of p1ap2bp_1^{a} p_2^{b} is (a+1)(b+1)(a+1)(b+1).
4
Apply the constraint that nn is a positive integer (n≥1n \ge 1)
Since n≥1n \ge 1, we have n+3≥4n + 3 \ge 4. The divisors of 108 are {1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108}. The divisors less than 4 are {1, 2, 3} (3 divisors).
If n+3n + 3 were 1, 2, or 3, nn would be −2-2, −1-1, or 00, none of which are positive integers.
5
Subtract invalid divisors from total divisors
Valid divisors = 12−3=912 - 3 = 9.
Each divisor d≥4d \ge 4 yields exactly one unique positive integer n=d−3n = d - 3.

Key Concept

Algebraic Divisibility and Prime Factorization Divisor Counting
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Topics

Divisibility, Factors, and Multiples

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