### Number theory questions - Factors II

Two more questions in
Number Theory. These are also focused on finding the number of factors of a
given number.

Question 1

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

Question 2

Number N = 2^6 * 5^5 *
7^6 * 10^7; how many factors of N are even numbers?

Correct Answers

Question
1: 11 * 6 * 9 = 594

Question
2: 1183

Explanations

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

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
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

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.

The prime-factorization of 2

^{6}* 5^{5}* 7^{6}* 10^{7 }is 2^{13}* 5^{12}* 7^{6}
The total number of factors of N = 14*13*7

We need to find the total number of even factors. For this, let
us find the total number of odd factors and then subtract this from the total
number of factors. Any odd factor will have to be a combination of powers of only
5 and 7.

Total number of odd factors of 2

^{13}* 5^{12}* 7^{6 }= (12+1) * (6 + 1) = 13 * 7
Total number of factors = (13+1) * (12+1) * (6 + 1)

Total number of even factors = 14 * 13 * 7 - 13 * 7

Number of even factors = 13 * 13 * 7 = 1183

Alternative approach

Any factor of 2

^{13}* 5^{12}* 7^{6 }will be of the form 2^{p}* 5^{q}* 7^{r}.
Any even factor of 2

^{13}* 5^{12}* 7^{6 }will also be of the same form, except that p cannot be zero in this case
p can take values 1, 2, 3, ….13 – 13 values

q can take values 0,1, 2, …12 – 13 values

r can take values 0, 1, 2, 3,….5 – 7 values

So, the total number of even factors be 13 * 13 * 7 = 1183

Labels: CAT number theory, CAT questions, CAT Solutions, Factors

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