If n is a positive integer with exactly two distinct prime factors such that n is a multip… — GMAT

If $n$ is a positive integer with exactly two distinct prime factors such that $n$ is a multiple of $12$ and $n^2$ has exactly $27$ positive integer divisors, what is the value of $n$ ?

A
36
48Answer
C
72
D
144
E
162

Answer

Since

n

is a multiple of

12 = 2^2 \cdot 3

and has only two prime factors,

n = 2^a \cdot 3^b

with

a \ge 2

and

b \ge 1

. The square

n^2 = 2^{2a} \cdot 3^{2b}

has

(2a+1)(2b+1) = 27

divisors. The only factor pair of 27 satisfying

2a+1 \ge 5

9 \times 3

, yielding

2a+1=9 \implies a=4

and

2b+1=3 \implies b=1

. Thus,

n = 2^4 \cdot 3^1 = 48

Step-by-Step Solution

Express

n

in terms of its prime factorization.

Since

n

is a multiple of

12 = 2^2 \cdot 3

and has exactly two distinct prime factors, its prime factorization must be

n = 2^a \cdot 3^b

, where

a \ge 2

and

b \ge 1

Divisibility by 12 requires at least two factors of 2 and at least one factor of 3.

Formulate the expression for the number of positive divisors of

n^2

n^2 = (2^a \cdot 3^b)^2 = 2^{2a} \cdot 3^{2b}

. The number of positive divisors is

(2a + 1)(2b + 1) = 27

The total number of positive integer divisors of a prime power product

p^x q^y

(x+1)(y+1)

Solve for integer values of

a

and

b

using factor pairs of 27.

The factor pairs of 27 into two integers greater than 1 are

9 \times 3

. Since

a \ge 2

2a + 1 \ge 5

. Thus,

2a + 1 = 9

and

2b + 1 = 3

, giving

a = 4

and

b = 1

Setting

2a + 1 = 3

yields

a = 1

, which violates the requirement that

a \ge 2

for divisibility by 12.

Calculate

n

n = 2^4 \cdot 3^1 = 16 \cdot 3 = 48

Substituting

a=4

and

b=1

into the prime factorization of

n

gives 48.

Key Concept

Divisibility rules and prime factorization count of positive divisors

Question

Answer

Step-by-Step Solution

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