How many factors of number N are also multiples of number K? Have you ever thought of that? Well, now you should.

Number Systems - Factors

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

540

660

594

792

Correct Answer

Choice C. 594.

Detailed Solution

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 2^{3} * 3 * 5 * K.

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 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 2^{13} * 3^{6} * 5^{9} that are multiples of 120. These will also have to be of the form 2^{p} * 3^{q} * 5^{r}. But as 120 is 2^{3} * 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.

Correct Answer: 594

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