Question

How many factors of 803661374160 are even numbers?

Answer 1

803661374160 = 73^2 *37^2* 17^1 * 5^1 * 3^4 * 2^4

Any factor of this number should be of the form 73^a * 37^b * 17^c * 5^d * 3^e * 2^f

For the factors of this number to be an even number f should be non 0

a can take values 0, 1, 2,

b can take values 0, 1, 2

c can take values 0, 1

d can take values 0, 1

e can take value 0, 1, 2, 3, 4

f can take value 1, 2, 3, 4.


I hope this will help you please mark my answer as brainliest answer.

Answer 2

When I prime factorized the number 803661374160, I got the following:

803661374160 = 2⁴ × 3⁴ × 5 × 17 × 37² × 73²

We've to find the no. of even factors. So don't care 2⁰. Take the 4 powers of 2: 2¹, 2², 2³, 2⁴.

Here it's 3⁴. So there are 5 powers: 3⁰, 3¹, 3², 3³, 3⁴.

Here it's only 5. So there are 2 powers: 5⁰, 5¹. So do 17, means, there are only 2 powers of 17: 17⁰, 17¹.

Here it's 37². So there are 3 powers: 37⁰, 37¹, 37².

So do 73². It has also 3 powers: 73⁰, 73¹, 73².

So the total number of even factors is to multiple these nos. of powers.

I.e., 4 × 5 × 2 × 2 × 3 × 3 = 720.

I don't think this will be right, but I answered on a logic!

Hope this will be right and may be helpful.

Thank you. Have a nice day. :-))