A positive integer N has exactly 12 positive factors. The sum of all distinct prime factor… — State PSC Exam

A positive integer $N$ has exactly $12$ positive factors. The sum of all distinct prime factors of $N$ is $10$ . If $N$ is divisible by $12$ but not divisible by $9$ , what is the total number of positive factors of $5N$ ?

A
15
B
16
18Answer
D
24

Answer

The total number of positive factors of

5N

18

The distinct prime factors of

N

that sum to

10

and satisfy divisibility by

12

are

2, 3,

and

5

. Because

N

has

12

factors, its exponents must be

2, 1, 1

. Since

N

is divisible by

12

and not divisible by

9

N = 2^2 \times 3^1 \times 5^1 = 60

. Consequently,

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

, which possesses

(2+1)(1+1)(2+1) = 18

factors.

Step-by-Step Solution

Determine the set of distinct prime factors of

N

The distinct prime factors of

N

are

2

3

, and

5

The sum of distinct prime factors is

10

. Possible sets of distinct primes summing to

10

are

\{3, 7\}

and

\{2, 3, 5\}

. Because

N

is divisible by

12

, its prime factorization must include

2

and

3

, forcing the set of prime factors to be

\{2, 3, 5\}

Determine the exponent structure of

N

The prime factorization of

N

takes the form

p_1^2 \times p_2^1 \times p_3^1

for distinct primes

p_1, p_2, p_3

The number of positive factors is given by

(a_1+1)(a_2+1)(a_3+1) = 12

. The only integer factorization of

12

into three factors greater than

1

3 \times 2 \times 2

, requiring exponents

2, 1, 1

in some arrangement.

Assign exponents to prime factors

2

3

, and

5

N = 2^2 \times 3^1 \times 5^1 = 60

Since

N

is divisible by

12 = 2^2 \times 3^1

, the exponent of

2

must be at least

2

and the exponent of

3

at least

1

. Since

N

is not divisible by

9 = 3^2

, the exponent of

3

cannot be

2

or greater, so it must be exactly

1

. Thus, the prime

2

takes exponent

2

, leaving prime

5

with exponent

1

Calculate the prime factorization and factor count of

5N

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

, which has

18

positive factors.

Multiplying

N

5

increments the exponent of

5

from

1

2

. The factor count formula yields

(2+1) \times (1+1) \times (2+1) = 3 \times 2 \times 3 = 18

Key Concept

Factors, Multiples, and Prime Factorization

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