A positive integer n is divisible by 12 and has exactly 15 positive integer divisors. What… — GMAT

A positive integer $n$ is divisible by 12 and has exactly 15 positive integer divisors. What is the least possible value of $n$ ?

Answer: 144

144

To find the least positive integer

n

that is a multiple of 12 and has 15 divisors, we examine the prime factorization of 12 (

2^2 \cdot 3^1

) and the divisor count formula

(e_1+1)(e_2+1)\dots = 15

. The number 15 factors as

5 \times 3

, implying

n

has the form

p^4 \cdot q^2

. Minimizing

n

requires placing the larger exponent on the smaller prime factor, so

p=2

and

q=3

, giving

n = 2^4 \cdot 3^2 = 144

Determine prime factor requirements for divisibility by 12.

Since

12 = 2^2 \cdot 3^1

n

must have prime factors 2 and 3 with exponents

a \ge 2

and

b \ge 1

Divisibility requires that all prime factors of the divisor appear in the dividend with equal or higher exponents.

Apply the divisor count formula to determine the exponent structure.

The total number of divisors is 15, which factors as

(4 + 1)(2 + 1) = 15

. Thus,

n = p^4 \cdot q^2

The number of positive divisors is given by

(e_1 + 1)(e_2 + 1) \dots = 15

Assign prime factors to minimize

n

Assigning the larger exponent 4 to the smaller prime 2 gives

2^4 \cdot 3^2 = 144

To minimize a product of prime powers, larger exponents should be paired with smaller prime bases.

Divisor count formula and prime factorization properties

Question