Let N = 2^a \times 3^b \times 7^c be a positive integer, where a, b, and c are positive in… — State PSC Exam

Let $N = 2^a \times 3^b \times 7^c$ be a positive integer, where $a$ , $b$ , and $c$ are positive integers. The number of positive factors of $N$ that are divisible by $6$ is $36$ , the number of positive factors of $N$ that are divisible by $14$ is $32$ , and the number of positive factors of $N$ that are divisible by $21$ is $30$ . What is the total number of positive factors of $N$ ?

Answer: 60

Answer

The total number of positive factors of

N

is 60.

By setting up the system of equations based on exponent constraints for each divisor (

6, 14, 21

), we find

a=4, b=3, c=2

. Substituting these into the total factor formula

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

yields

5 \times 4 \times 3 = 60

Step-by-Step Solution

Formulate algebraic expressions for the constrained factor counts

ab(c+1) = 36

ac(b+1) = 32

, and

bc(a+1) = 30

A factor

2^x \times 3^y \times 7^z

is divisible by

6 = 2^1 \times 3^1

x \ge 1

y \ge 1

z \ge 0

, yielding

a \cdot b \cdot (c+1)

factors. Similarly, divisibility by

14

requires

x \ge 1, z \ge 1

, giving

a \cdot c \cdot (b+1)

, and divisibility by

21

requires

y \ge 1, z \ge 1

, giving

b \cdot c \cdot (a+1)

Expand equations and substitute

P = abc

ab = 36 - P

ac = 32 - P

bc = 30 - P

Expanding

ab(c+1) = abc + ab = 36

gives

ab = 36 - P

. Applying the same expansion to the other two equations gives expressions for

ac

and

bc

in terms of

P

Multiply the three equations together and solve for

P

(ab)(ac)(bc) = (abc)^2 = P^2 \Rightarrow (36-P)(32-P)(30-P) = P^2 \Rightarrow P = 24

Testing positive integer values of

P

that divide into products reveals that for

P = 24

(36-24)(32-24)(30-24) = 12 \times 8 \times 6 = 576 = 24^2

Calculate the values of

ab

ac

bc

, and solve for

a

b

c

ab = 12

ac = 8

bc = 6

, giving

a = 4

b = 3

c = 2

a^2 = \frac{(ab)(ac)}{bc} = \frac{12 \times 8}{6} = 16 \Rightarrow a = 4

. Similarly,

b^2 = \frac{12 \times 6}{8} = 9 \Rightarrow b = 3

, and

c^2 = \frac{8 \times 6}{12} = 4 \Rightarrow c = 2

Compute the total number of positive factors of

N

(4+1)(3+1)(2+1) = 5 \times 4 \times 3 = 60

The total number of factors of

N = 2^a \times 3^b \times 7^c

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

Key Concept

Constrained Factor Counting and Prime Factorization Systems

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