A positive integer N has the prime factorization N = 2^a \times 3^b \times 5^c, where a, b… — State PSC Exam

A positive integer $N$ has the prime factorization $N = 2^a \times 3^b \times 5^c$ , where $a$ , $b$ , and $c$ are positive integers. It is known that $N$ has exactly $48$ positive factors, $N/2$ has $36$ positive factors, and $N/5$ has $40$ positive factors. What is the total number of positive factors of $N/3$ ?

24Answer
B
32
C
16
D
0

Answer

The total number of positive factors of

N/3

is 24.

By writing the factor formula

(a+1)(b+1)(c+1) = 48

and comparing it with

d(N/2) = a(b+1)(c+1) = 36

and

d(N/5) = (a+1)(b+1)c = 40

, we uniquely determine the exponents

a = 3

b = 1

, and

c = 5

. Dividing

N

3

changes the prime exponent of

3

0

. Applying the factor count formula to

2^3 \times 3^0 \times 5^5

yields

(3+1)(0+1)(5+1) = 24

Step-by-Step Solution

Express the total number of factors of

N

using its prime factorization

d(N) = (a+1)(b+1)(c+1) = 48

The number of positive factors of a number

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)

Use the factor counts of

N/2

and

N/5

to determine the exponents

a

and

c

a = 3

and

c = 5

Since

d(N/2) = a(b+1)(c+1) = 36

, taking the ratio

d(N/2)/d(N)

yields

a/(a+1) = 36/48 = 3/4

, which gives

a = 3

. Similarly,

d(N/5) = (a+1)(b+1)c = 40

, so

d(N/5)/d(N) = c/(c+1) = 40/48 = 5/6

, giving

c = 5

Solve for the remaining exponent

b

b = 1

Substituting

a = 3

and

c = 5

into

d(N) = 48

gives

(3+1)(b+1)(5+1) = 48 \implies 24(b+1) = 48 \implies b+1 = 2 \implies b = 1

Calculate the number of positive factors of

N/3

d(N/3) = (3+1)(0+1)(5+1) = 4 \times 1 \times 6 = 24

Dividing

N

3

reduces the exponent of

3

from

b = 1

b-1 = 0

, so

N/3 = 2^3 \times 3^0 \times 5^5

Key Concept

Prime Factorization and Number of Factors

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic