Consider the number N = 2^4 \times 3^3 \times 5^2. How many positive factors of N are also… — State PSC Exam

Consider the number $N = 2^4 \times 3^3 \times 5^2$ . How many positive factors of $N$ are also multiples of $120$ ?

For a factor of

N

to be a multiple of

120

, it must contain at least the prime factorization of

120

, which is

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

. We can factor this out from

N

N = 120 \times (2^1 \times 3^2 \times 5^1)

. The number of such factors is simply the total number of divisors of the remaining portion

(2^1 \times 3^2 \times 5^1)

, which is calculated by adding

1

to each exponent and multiplying them:

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

Determine the prime factorization of 120.

120 = 2^3 \times 3^1 \times 5^1

To find factors of N that are multiples of 120, we must know the minimum prime powers required.

Determine the prime factorization of the quotient N / 120.

\frac{2^4 \times 3^3 \times 5^2}{2^3 \times 3^1 \times 5^1} = 2^{(4-3)} \times 3^{(3-1)} \times 5^{(2-1)} = 2^1 \times 3^2 \times 5^1

Any factor of N that is a multiple of 120 can be expressed as

120 \times k

, where

k

must be a factor of the remaining prime powers of N.

Calculate the total number of positive factors for the quotient

k

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

The number of choices for

k

corresponds exactly to the number of factors of N that are multiples of 120.

Identifying restricted factors and multiples using prime factorization.

Question