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Explanatory Answer

Method of solving this CAT Question from Number Theory - Factors: How many factors of number N are also multiples of number K? Have you ever thought of that? Well, now you should.

The prime factorization of 2⁸ * 3⁶ * 5⁴ * 10⁵ is 2¹³ * 3⁶ * 5⁹.
For any of these factors questions, start with the prime factorization. Remember that the formulae for number of factors, sum of factors, are all linked to prime factorization.
120 can be prime-factorized as 2³ * 3 * 5.
All factors of 2¹³ * 3⁶ * 5⁹ that can be written as multiples of 120 will be of the form 23 * 3 * 5 * K.
2¹³ * 3⁶ * 5⁹ = 2³ * 3 * 5 * K
=> K = 2¹⁰ * 3⁵ * 5⁸.
The number of factors of N that are multiples of 120 is identical to the number of factors of K.
Number of factors of K = (10 + 1) (5 + 1) * (8 + 1) = 11 * 6 * 9 = 594

Alternative approach

Any factor of 2¹³ * 3⁶ * 5⁹. will be of the form 2^p * 3^q * 5^r. When we are trying to find the number of factors without any constraints, we see that.
p can take values 0, 1, 2, 3......13 – 14 values.
q can take values 0, 1, 2......6 – 7 values.
r can take values 0, 1, 2, 3......9 – 10 values.
So, the total number of factors will be 14 * 7 * 10.
This is just a rehash of our formula (a + 1) (b + 1) (c + 1).
In this scenario we are looking for factors of 213 * 36 * 59 that are multiples of 120. These will also have to be of the form 2p * 3q * 5r. But as 120 is 23 * 3 * 5, the set of values p, q, r can take are limited.
p can take values 3, .......13 – 11 values
q can take values 1, 2, .......6 – 6 values
r can take values 1, 2, 3, .......9 – 9 values
So, the total number of factors that are multiples of 120 will be 11 * 6 * 9 = 594.

The question is "How many factors of the number 2⁸ * 3⁶ * 5⁴ * 10⁵ are multiples of 120?"

Hence the answer is 594

Choice C is the correct answer.