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

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1) 10404

First we notice that the endign number (4) is even, then the number is divisible by 2:

10404= 2* 5202

Now also, 5202 is ending with an even number (2), then it is diisible by 2:

10404= 2*2*2601

Now 2601 is not divisible by 2, we will try and find the sum 2+6+0+1= 9, then it is divisible by3

10404= 2*2*3*867

Now 8+6+7= 21 which is also divisible by 3:

10404= 2*2*3*3*289

Now we have 289 which is the (17)^2

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

sqrt(10404)= 2*3*17= 102

2)11025

The ending number is 5, then the number is divisible by 5:

11025= 5* 2505

Also 2505 is ending by 5, then it is divisible by 5.

11025 = 5*5*441

Now the sum of 441 =4+4+1=9 then it is divisible by 3:

11025= 5*5*3*147

Also 1+4+7=12 which is divisible by 3

11025= 5*5*3*3*49

but 49= 7^2

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

sqrt(11025)= 5*3*7= 105

To find the sqrt10404) by division method:

1)1 04 04(102

1

----------

20) 04

00

------------

202)404

404

-----------------

Explanation. The number 10404 is gruoped by 2digits from unit place towards left.: 1 04 04.

The first group from left is 1 whose square root is 1. So 1-1*1 = 0. Now go to the next step take 04 . Double 1 the earlier found square root and we get 2. Try to get x such that (2*10+x)x = 04. So we get x= 0. So 04-20*0 = 04.

Burrow next 04 and the number becomes 404. Try for getting (200+x)x = 404 . Now x=2 solves this.

Thus 102 is the square root of 10404 by division method.

2) To find sqrt 11025.

1) 1 10 25(105

1

------------

20)0 10

00

-----------

205)10 25

10 25

----------------------

00 00

So sqrt(11025) = 105

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**