The question is about factors of number N that are also multiples of number K. A number is given as multiplication of factors. We need to find out the number of factors of that number which are factors of another number. Dealing with factors of a number is a vital component in CAT Number Systems: Factors. A range of CAT questions can be asked based on this simple concept of Factors. CAT exam has always tested the idea of Factors from Number systems. The idea of Factors questions forms an integral part of the CAT syllabus.

Question 5: How many factors of the number 2^{8} * 3^{6} * 5^{4} * 10^{5} are multiples of 120?

- 540
- 660
- 594
- 792

594

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The prime factorization of 2^{8} * 3^{6} * 5^{4} * 10^{5} is 2^{13} * 3^{6} * 5^{9}.

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 2^{3} * 3 * 5.

All factors of 2^{13} * 3^{6} * 5^{9} that can be written as multiples of 120 will be of the form 23 * 3 * 5 * K.

2^{13} * 3^{6} * 5^{9} = 2^{3} * 3 * 5 * K

=> K = 2^{10} * 3^{5} * 5^{8}.

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

Any factor of 2^{13} * 3^{6} * 5^{9}. 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 2 ^{8} * 3^{6} * 5^{4} * 10^{5} are multiples of 120?"**

Choice C is the correct answer.

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