Let n be a positive integer that is a factor of 180. If n is divisible by 6 but is not div… — GRE General Test

Let $n$ be a positive integer that is a factor of $180$ . If $n$ is divisible by $6$ but is not divisible by $4$ , what is the maximum possible number of positive divisors of $n$ ?

A
6
B
8
12Answer
D
18
E
24

Answer

The prime factorization of

180

2^2 \times 3^2 \times 5^1

. Any factor

n

180

is of the form

2^a \times 3^b \times 5^c

. For

n

to be divisible by

6

, we must have

a \ge 1

and

b \ge 1

. For

n

not to be divisible by

4

, we must have

a < 2

, which forces

a = 1

. To maximize the total number of divisors

(a+1)(b+1)(c+1)

, we select the largest possible values for

b

and

c

, which are

b = 2

and

c = 1

. This yields

n = 2^1 \times 3^2 \times 5^1 = 90

, giving

(1+1)(2+1)(1+1) = 12

positive divisors.

Step-by-Step Solution

Find the prime factorization of 180.

180 = 2^2 \times 3^2 \times 5^1

Any positive factor

n

180

must have the form

n = 2^a \times 3^b \times 5^c

, where

0 \le a \le 2

0 \le b \le 2

, and

0 \le c \le 1

Apply the divisibility conditions to determine the possible values of the exponent

a

a = 1

Since

n

is divisible by

6 = 2 \times 3

a \ge 1

and

b \ge 1

. Since

n

is not divisible by

4 = 2^2

a < 2

. Thus,

a

must equal

1

Maximize the divisor count formula

(a+1)(b+1)(c+1)

using the available ranges for

b

and

c

b = 2

and

c = 1

To maximize the number of positive divisors, choose the maximum allowable values for

b

(

b=2

) and

c

(

c=1

Calculate the maximum number of positive divisors.

(1+1)(2+1)(1+1) = 2 \times 3 \times 2 = 12

For

n = 2^1 \times 3^2 \times 5^1 = 90

, the total number of positive divisors is

12

Key Concept

Divisor count formula and prime factor constraints

Estimated Time:1m 30s

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