Let n be a positive integer whose prime factorization consists only of the prime factors 2… — GRE General Test

Let $n$ be a positive integer whose prime factorization consists only of the prime factors $2$ and $3$ . If $n$ has exactly $12$ positive divisors and $\gcd(n, 36) = 12$ , what is the value of $n$ ?

Answer: 96

Answer

The value of

n

96

Representing

n = 2^a \times 3^b

, the number of positive divisors is

(a+1)(b+1) = 12

. The greatest common divisor

\gcd(n, 36) = \gcd(2^a \times 3^b, 2^2 \times 3^2) = 2^{\min(a,2)} \times 3^{\min(b,2)} = 12 = 2^2 \times 3^1

. This requires

\min(a,2) = 2 \implies a \ge 2

and

\min(b,2) = 1 \implies b = 1

. Substituting

b = 1

into

(a+1)(1+1) = 12

gives

2(a+1) = 12

, so

a = 5

. Therefore,

n = 2^5 \times 3^1 = 32 \times 3 = 96

Step-by-Step Solution

Set up the prime factorization of

n

and the divisor count equation.

n = 2^a \times 3^b

and

(a + 1)(b + 1) = 12

Since the prime factors of

n

are only

2

and

3

n

must take the form

2^a \times 3^b

, where the number of positive divisors is

(a+1)(b+1)

Analyze the exponent requirements using the greatest common divisor.

a \ge 2

and

b = 1

\gcd(2^a \times 3^b, 2^2 \times 3^2) = 2^{\min(a,2)} \times 3^{\min(b,2)} = 2^2 \times 3^1

. Matching powers gives

\min(a,2) = 2 \implies a \ge 2

, and

\min(b,2) = 1 \implies b = 1

Solve for exponent

a

and calculate

n

a = 5

, giving

n = 2^5 \times 3^1 = 96

Substituting

b = 1

into

(a+1)(1+1) = 12

gives

2(a+1) = 12 \implies a = 5

, which satisfies

a \ge 2

Key Concept

Prime exponent rules for GCD and divisor counting

Estimated Time:2m 0s

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