Using division, please find the square roots of the following numbers: ( 1 ) 10404 ( 2) 11025

1) Using the division method, we'll notice that if we add the digits from the number 10404 = 1+0+4+0+4=9 which is divisible by 3, so we could write as:

10404 = 3*3468

We'll verify if the sum of the digits of the number 3468 is divisible by 3:

3468 = 3+4+6+8=21

10404 = 3*3*1156

Because the sum of the digits of the number 1156 is not divisible by 3, but the number is ending in a digit which is divisible by 2, we'll divide the number 1156 by 2:

10404 = 3*3*2*578

Again, 578 is divisible by 2:

10404 = 3*3*2*2*289

We notice that 289 = 17*17

So the number 10404 could be written as:

10404 = 2^2*3^2*17^2

sqrt10404 = sqrt(2^2)*sqrt(3^2)*sqrt(17^2)

sqrt 10404 = 2*3*17

sqrt 10404 = 102

2)  For the number 11025, we'll verify first if the sum of it's digits is divisible by 3:

11025 = 1+1+0+2+5 = 9

So the number could be divided by 3:

11025 = 3*3675

Again , we'll add the digits form the quotient 3675:

3675 = 3+6+7+5 = 21

We'll divide the number again, by 3:

11025 = 3*3*1225

We'll divide 1225 by 5:

11025 = 3*3*5*245

Again, we'll divide 245 by 5:

11025 = 3*3*5*5*49

11025 = 3^2*5^2*7^2

sqrt 11025 = sqrt3^2*sqrt5^2*sqrt7^2

sqrt 11025 = 3*5*7

sqrt 11025 = 105