Question

sum of all even factors of 64800​

Answer 1

Answer:

hey....mark as brainlist and plz follow I'll follow u back.....

Step-by-step explanation:

have a great day

Answer 2

The sum of the all even factors of 64800 is 236313 by using sum of the even factors is $a^{m}\times b^{n}$.

Given that,

We have to find the sum of the all even factors of 64800.

We know that,

What is factor?

Writing all numbers as the product of primes is a procedure known as prime factorization. Consider the case of the number 20, for instance. We can divide that into two elements. Well, that's four times five, we can say. Furthermore, 5 is a prime number.

So,

The sum of the even factors is $a^{m}\times b^{n}$

Which is (a⁰+a¹+a²+.....$a^{m}$)×(b⁰+b¹+b²+.....$b^{m}$)

Factor of 64800 = 2⁵×3⁴×5²

=(2⁰+2¹+2²+2³+2⁴+2⁵)×(3⁰+3¹+3²+3³+3⁴)×(5⁰+5¹+5²)

=63×121×31

=236313

Therefore, The sum of the all even factors of 64800 is 236313 by using sum of the even factors is $a^{m}\times b^{n}$.

To learn more about factor visit:

https://brainly.in/question/9426948

https://brainly.in/question/49135457

#SPJ2