Question

The product of number of odd and even factors of 36000 will be

• A. 100
• B. 864
• C. 720
• D. 72

Answer 1

The correct option is D 72

Any number x can be expressed in its prime factorisation form as $x={\left(p\right)}^{a_1}{\left(q\right)}^{a_2}{\left(r\right)}^{a_3},$

where, p is even and q, r are odd numbers then, the total number of factors of the given number x are

$(a_1+1)(a_2+1)(a_3+1).$

Also, the number of odd factors of the given number, x will be $(a_2+1)(a_3+1).$

Here, $x=36000=2^5\times 3^2\times 5^3$

where, p = 2, q = 3, r = 5; $a_1=5,a_2=2,a_3=3.$
∴ Total number of factors of 36000 = (5 + 1)(2 + 1)(3 + 1)

= 6 × 3 × 4

= 72

Number of odd factors of 36000 = (2 + 1)(3 + 1)

= 3 × 4

= 12

∴ Number of even factors of 36000 = Total factor of 36000 – Number of odd factors

= 72 – 12

= 60

⇒ Product of number of odd and even factors of 36000 = 12 × 60 = 720

Hence, the correct answer is option (d).