Find the number of factors of the number represented by = $2^{3}$ x $3^{2}$ x $5^{1}$ x $7^{1}$

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Solution

The correct option is C 48
If a number N is represented as the product of primes i.e.
N = $2^{a}$ x $3^{b}$ x $5^{c}$ x $7^{d}$
the number of factors are :
(a+1) (b+1) (c+1) (d+1)
From the given question
a = 3
b = 2
c = 1
d = 1
So number of factors are
= (3+1) (2+1) (1+1) (1+1)
= 4 x 3 x 2 x 2
= 48

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