Consider the positive integer N = 720, which has a prime factorization of 2^4 \times 3^2 \… — State PSC Exam

Consider the positive integer $N = 720$ , which has a prime factorization of $2^4 \times 3^2 \times 5^1$ . Which of the following statements regarding the positive factors of $N$ are correct?

The number of positive factors of $N$ that are perfect squares is $6$ .Answer
The sum of all positive even factors of $N$ is $2340$ .Answer
C
The number of positive factors of $N$ that are multiples of $15$ is $6$ .
The product of all positive factors of $N$ is equal to $720^{15}$ .Answer

Answer

The correct statements are those asserting that

N

has 6 perfect square factors, that the sum of all positive even factors of

N

is 2340, and that the product of all positive factors of

N

720^{15}

The statements confirming 6 perfect square factors, a sum of 2340 for even factors, and a factor product of

720^{15}

are all mathematically correct applications of prime factorization principles.

Step-by-Step Solution

Determine perfect square factors of

N = 2^4 \times 3^2 \times 5^1

For a factor

2^a \times 3^b \times 5^c

to be a square, exponents

a, b, c

must be even. Possible values:

a \in \{0, 2, 4\}

(3 options),

b \in \{0, 2\}

(2 options),

c \in \{0\}

(1 option). Number of square factors

= 3 \times 2 \times 1 = 6

Perfect squares require all prime factors to have even exponents.

Calculate the sum of all positive even factors

Sum of all factors

= (2^0+2^1+2^2+2^3+2^4)(3^0+3^1+3^2)(5^0+5^1) = 31 \times 13 \times 6 = 2418

. Sum of odd factors (only

2^0

)

= 1 \times 13 \times 6 = 78

. Sum of even factors

= 2418 - 78 = 2340

Even factors are obtained by subtracting odd factor sum from total factor sum.

Count factors divisible by 15

Since

15 = 3^1 \times 5^1

, factors of 720 divisible by 15 require

a \in \{0, 1, 2, 3, 4\}

(5 choices),

b \in \{1, 2\}

(2 choices), and

c \in \{1\}

(1 choice). Total factors

= 5 \times 2 \times 1 = 10

Divisibility by 15 requires at least power 1 for both prime factors 3 and 5.

Compute the product of all positive factors

Total factors

T = (4+1)(2+1)(1+1) = 30

. Product of factors

= N^{T/2} = 720^{30/2} = 720^{15}

Factors pair up such that

f_i \times f_{T+1-i} = N

, giving

T/2

pairs.

Key Concept

Properties of positive factors derived from prime factorization: square factor counts, even factor sums, constrained divisibility factor counts, and total factor product formula.

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