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Difficulty: MediumDivisibility, Factors, and Multiples

A positive integer NN has exactly 1212 positive integer divisors. If NN is a multiple of 1818, and exactly 23\frac{2}{3} of the positive integer divisors of NN are even, what is the value of NN?

  1. A
    36
  2. B
    72
  3. 108Answer
  4. D
    144
  5. E
    216

Answer

108
The integer 108 prime factorizes as 2² × 3³. It has (2+1)(3+1) = 12 total positive divisors. Its odd divisors are the 4 divisors of 3³ (namely 1, 3, 9, 27), leaving 12 - 4 = 8 even divisors. Thus, exactly 8/12 = 2/3 of its divisors are even, and 108 is divisible by 18.

Step-by-Step Solution

1
Determine the number of even and odd divisors of N.
Number of even divisors = (2/3) × 12 = 8; Number of odd divisors = 12 - 8 = 4.
The problem states that 2/3 of the 12 total divisors are even.
2
Express N as 2^a × M, where M is odd.
Total divisors = (a + 1) × (divisors of M) = 12. Since divisors of M are the odd divisors of N, (a + 1) × 4 = 12, so a + 1 = 3 and a = 2.
Odd divisors of N come entirely from the odd part M.
3
Find M such that M is odd, divisible by 9 (since N is a multiple of 18 = 2 × 3²), and has exactly 4 divisors.
M must be of the form 3^k. Since divisors count is k + 1 = 4, k = 3, so M = 3³ = 27.
If M had two prime factors p × q, each exponent would be 1, which cannot be divisible by 3².
4
Calculate N = 2^a × M.
N = 2² × 27 = 4 × 27 = 108.
Combining the even and odd prime factor components.

Key Concept

Divisibility and Counting Divisors via Prime Factorization
Estimated Time:1m 30s
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Topics

Divisibility, Factors, and Multiples

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