What is the value of the positive integer n if n is a multiple of 18 and n has exactly 9 p… — GMAT

What is the value of the positive integer $n$ if $n$ is a multiple of 18 and $n$ has exactly 9 positive integer divisors?

Answer: 36

The prime factorization of 18 is

2^1 \cdot 3^2

. Any multiple

n

of 18 must take the form

n = 2^a \cdot 3^b \dots

where

a \ge 1

and

b \ge 2

. The total number of positive divisors of

n

is given by

(a+1)(b+1)\dots = 9

. Given that

a+1 \ge 2

and

b+1 \ge 3

, the only product of integers equal to 9 is

3 \times 3

. This requires

a+1 = 3 \implies a = 2

and

b+1 = 3 \implies b = 2

, with no additional prime factors present. Therefore,

n = 2^2 \cdot 3^2 = 36

Determine the prime factorization constraints imposed by 18.

18 = 2^1 \cdot 3^2

, so

n = 2^a \cdot 3^b \dots

with

a \ge 1

and

b \ge 2

Any multiple of 18 must contain at least one factor of 2 and two factors of 3.

Apply the divisor count formula to set up an equation.

(a+1)(b+1) = 9

The number of positive divisors of

2^a \cdot 3^b

is given by

(a+1)(b+1)

Solve for the exponents

a

and

b

a = 2

and

b = 2

Because 9 can only be factored as

3 \times 3

for integer components where

a+1 \ge 2

and

b+1 \ge 3

, both

a+1

and

b+1

must equal 3.

Compute the value of

n

n = 2^2 \cdot 3^2 = 36

Multiply the prime power factors together to find the value of

n

The total number of positive divisors of a positive integer

n = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}

is given by

(e_1 + 1)(e_2 + 1) \dots (e_k + 1)

Question