Question

Difficulty: MediumDivisibility, Factors, and Multiples

If kk is a positive integer that is divisible by both 8 and 9, and kk has exactly 12 positive integer divisors, what is the value of kk?

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

Answer

72
The value 72 has the prime factorization 23322^3 \cdot 3^2. It is divisible by 8=238 = 2^3 and 9=329 = 3^2, and its number of positive integer divisors is (3+1)(2+1)=12(3+1)(2+1) = 12, satisfying all conditions of the problem.

Step-by-Step Solution

1
Express the divisibility conditions in terms of prime factors.
Since kk is divisible by 8=238 = 2^3 and 9=329 = 3^2, the prime factorization of kk must be of the form k=2a3bp1c1k = 2^a \cdot 3^b \cdot p_1^{c_1} \cdots, where a3a \geq 3 and b2b \geq 2.
Any multiple of 8 must contain at least three factors of 2, and any multiple of 9 must contain at least two factors of 3.
2
Write the formula for the total number of positive integer divisors.
The number of positive integer divisors is given by (a+1)(b+1)(c1+1)=12(a+1)(b+1)(c_1+1)\cdots = 12.
The total number of divisors of a number expressed in prime factorization is found by adding 1 to each exponent and multiplying the results.
3
Determine the values of the exponents aa and bb.
Since a3a \geq 3, we have a+14a+1 \geq 4. Since b2b \geq 2, we have b+13b+1 \geq 3. Therefore, (a+1)(b+1)4×3=12(a+1)(b+1) \geq 4 \times 3 = 12.
Because the product (a+1)(b+1)(a+1)(b+1) is already at least 12, there can be no additional prime factors, so a+1=4    a=3a+1 = 4 \implies a = 3 and b+1=3    b=2b+1 = 3 \implies b = 2.
4
Calculate the value of kk.
k=2332=89=72k = 2^3 \cdot 3^2 = 8 \cdot 9 = 72.
Evaluating 23322^3 \cdot 3^2 gives the unique positive integer satisfying all given conditions.

Key Concept

Determining integer values using prime factorization constraints and total divisor count rules.
Estimated Time:2m 0s
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