A positive integer N has a prime factorization of the form p^a \cdot q^b, where p and q ar… — State PSC Exam

A positive integer $N$ has a prime factorization of the form $p^a \cdot q^b$ , where $p$ and $q$ are distinct prime numbers, and $a, b \ge 1$ . If the total number of positive factors of $N^2$ is $45$ and the total number of positive factors of $N^3$ is $91$ , what is the total number of positive factors of $N$ ?

A
$12$
$15$ Answer
C
$18$
D
$20$

Answer

The total number of positive factors of

N

15

For a composite integer

N = p^a q^b

, the number of positive factors is

(a+1)(b+1)

. Given that

N^2 = p^{2a} q^{2b}

has

45

positive factors, we have

(2a+1)(2b+1) = 45

. Factoring

45

into odd pairs yields two possibilities for

(a, b)

(1, 7)

(2, 4)

. Using the second condition that

N^3 = p^{3a} q^{3b}

has

91

factors, we check

(3a+1)(3b+1) = 91

. For

(1, 7)

(4)(22) = 88 \neq 91

. For

(2, 4)

(7)(13) = 91

, which is correct. Therefore,

a=2

and

b=4

, so the number of positive factors of

N

(2+1)(4+1) = 15

Step-by-Step Solution

Set up the factor count equation for

N^2

Since

N = p^a q^b

N^2 = p^{2a} q^{2b}

. The number of positive factors of

N^2

(2a+1)(2b+1) = 45

For a prime factorization

p^x q^y

, the total number of positive factors is

(x+1)(y+1)

Find possible integer solutions for

(a, b)

The odd factor pairs of

45

are

(3, 15)

and

(5, 9)

. This gives two possible sets of exponents: Case 1:

2a+1=3 \implies a=1

and

2b+1=15 \implies b=7

. Case 2:

2a+1=5 \implies a=2

and

2b+1=9 \implies b=4

Since

a, b \ge 1

, both factors

2a+1

and

2b+1

must be odd integers greater than

1

Evaluate candidate solutions using the second condition for

N^3

For

N^3 = p^{3a} q^{3b}

, the number of factors is

(3a+1)(3b+1) = 91

. Testing Case 1 (

a=1, b=7

(3(1)+1)(3(7)+1) = 4 \times 22 = 88 \neq 91

. Testing Case 2 (

a=2, b=4

(3(2)+1)(3(4)+1) = 7 \times 13 = 91

, which matches.

The true exponent values must simultaneously satisfy both given conditions.

Calculate the total number of positive factors of

N

With

a=2

and

b=4

N = p^2 q^4

. The total number of positive factors of

N

(2+1)(4+1) = 3 \times 5 = 15

Apply the total factor formula

(a+1)(b+1)

N

Key Concept

Calculating factor counts of a prime-factored number from equations involving factor counts of its powers

Estimated Time:2m 0s

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