Question

Prove the divisibility:

45^45·15^15 by 75^30

Answer 1

`3^(75)`.

Explanation:

We have been an division problem:
`(45^(45)*15^(15))/(75^(30))`.

We will simplify our division problem using rules of exponents.

Using product rule of exponents
`(a*b)^n=a^n*b^n` we can write:

`45^(45)=(9*5)^(45)=9^(45)*5^(45)`

`15^(15)=(3*5)^(15)=3^(15)*5^(15)`

`75^(30)=(15*5)^(30)=15^(30)*5^(30)`

Substituting these values in our division problem we will get,

`(9^(45)*5^(45)*3^(15)*5^(15))/(15^(30)*5^(30))`

Using power rule of exponents
`a^n*a^m=a^(n+m)` we will get,

`(9^(45)*5^((45+15))*3^(15))/(15^(30)*5^(30))`

`(9^(45)*5^(60)*3^(15))/(15^(30)*5^(30))`

Using product rule of exponents
`(a*b)^n=a^n*b^n` we will get,

`((3*3)^(45)*5^(60)*3^(15))/((3*5)^(30)*5^(30))`

`(3^(45)*3^(45)*5^(60)*3^(15))/(3^(30)*5^(30)*5^(30))`

Using power rule of exponents
`a^n*a^m=a^(n+m)` we will get,

`(3^((45+45+15))*5^(60))/(3^(30)*5^((30+30)))`

`(3^(105)*5^(60))/(3^(30)*5^(60))`

`(3^(105))/(3^(30))`

Using quotient rule of exponent
`(a^m)/(a^n)=a^(m-n)` we will get,

`(3^(105))/(3^(30))=3^(105-30)`

`3^(105-30)=3^(75)`

Therefore, our resulting quotient will be
`3^(75)`.