What is S.E.?

S.E. stands for Square Edging, an invented term to describe the “non-conventional” method of taking the square root of a number based on the popular algebraic equation (A + B)² = A² + 2AB + B².

Square Edging is divided into three formats;

1) S.E. “Pairing” Method (MSM-1 Format) – it is the easiest among the three and the very ‘fundamental’ of square edging. Learning it makes the principles behind MSM-2 and MSM-3 much easier to grasp. ( See: http://easysqrtsforkids.blogspot.com/2010/08/five-digitsix-digit-square-edging.html )

2) S.E. “Grouping” Method (MSM-2 Format) – it deals with perfect squares ending in 25, where MSM-1 failed. The purpose of MSM-2 is to complete the function of S.E., as a short cut method of taking the square roots of perfect squares in huge numbers. MSM-2 is a little bit tricky but once you become familiar with the rules and procedures, it becomes an enjoyable fun-game.

3) S.E. “Universal” Method (MSM-3 Format or U.S.E.) was developed through the manipulation of numbers using the two original S.E. methods. It is the ‘true’ method of taking the square root of any number, big or small, perfect or non-perfect. It makes the process direct and simple, avoiding the painstaking trial and error method and long processes of divisions and subtractions that the traditional way of taking the square root of numbers is well known (also called as the “long-hand” division, See: http://www.itl.nist.gov/div897/sqg/dads/HTML/squareRoot.html )

Variations of MSM-2

Initially, the S.E. “Grouping” Method seems to be difficult to grasp (or to understand). But the only tricky part is the creation of the sets of possible square roots

Rule 1: If the given problem ends with …225

1) N = 1, 3, 6 and 8

2) There are always two sets (or 2 groups), each having five (5) elements

Rule 2: If the given problem ends with …025

1) N = 4, 5 and 9

2) There are always two sets (or 2 groups), each having four (4) elements

Rule 3: If the given problem ends with …625

1) N = 2 and 7

2) There is always one set (or 1 group), having five (5) elements

Case No. 2

Given:

√66,341,025

Step 1:

(Left Column, 1^st Row)

√66’34’10’25

... 8 ........... 5

N : 4, 5, 9

Step 2:

(Right Column, 1^st Row)

P : 66’34

85 : 72’25 ↓

4↓ M : 0, 1, 2, 3, 4

Step 3:

(Left Column, 2^nd Row)

1^st Square Root Locator

.. 72’25

.. 68’06 \ ↓

.. 64..... /

.136’25 / 2

.. 68’12

Modify Step 2:

P : 66’34

85 : 72’25 ↓ /8250/

4↓ M : 0, 1, 2, 3, 4

Step 4:

List down the two sets of possible square roots

(Right Column, 2^nd Row)

... ↓ .............. ↑

8145 ......... 8245

8095 ......... 8195

8055 ......... 8155

8045 ......... 8145

Step 5:

(Left Column, 3^rd Row)

2^nd Sq. Rt. Locator

.. 68’06 \

.. 66’03 / ↑

.. 64 .....

.132’06 / 2

.. 66’03

Modify Step 4:

As an option, draw a square, enclosing the right set of possible square roots

(Right Column, 2^nd Row)

8245 \ ↑

8195 /

8155 \

8145 / ↓

Step 6:

(Right Column, 3^rd Row)

3^rd Sq. Rt. Loc.

.. 68’06

.. 67’04 \ ↓

.. 66’03 /

.134’09 / 2

.. 67’04

Modify Step 4:

As option, cross out 8245 and 8195, leaving 8155 and 8145

8245 ← Eliminated

8195 ← Eliminated

8155

8145

Step 7:

(Left Column, 4^th Row)

4^th Sq. Rt. Loc.

↓ 8145 ........ 8155 ↑

.. 67’04

.. 66’53 \ ↓

.. 66’03 /

.133’07/ 2

.. 66’53

Step 8:

(Right Column, 4^th Row)

Therefore:

√66,341,025 = 8,145

Case No. 3

Given:

√54,390,625

Step 1:

(Left Column, 1^st Row)

√54’39’06’25

....7 ........... 5

N : 2, 7

Step 2:

(Right Column, 1^st Row)

P : 54’39

75 : 56’25 ↓

4↓ M : 0, 1, 2, 3, 4

Step 3:

(Left Column, 2^nd Row)

1^st Square Root Locator

.. 56’25 \

.. 52’56 / ↑

.. 49 .....

.105’25 / 2

.. 52’62

Modify Step 2:

P : 54’39

75 : 56’25 ↓ /7250/

4↓ M : 0, 1, 2, 3, 4

Step 4:

List down the set of possible square roots

(Right Column, 2^nd Row)

7475

7425

7375

7325

7275

We can re-group them as follows:

.. ↑

7475

7425

7375

.. ↓

7375

7325

7275

Step 5:

(Left Column, 3^rd Row)

2^nd Sq. Rt. Locator

.. 56’25

.. 54’40 \ ↓

.. 52’56 /

.108’81 / 2

.. 54’40

Modify Step 4:

The group with an up arrow is eliminated.

(As an optional, draw a square and enclose 7375, 7325, 7325)

7375

7325

7275

We can separate them into two groups:

.. ↑

7375

7325

.. ↓

7325

7275

Step 6:

(Right Column, 3^rd Row)

3^rd Sq. Rt. Loc.

.. 54’40 \

.. 53’48 / ↑

.. 52’56

.106’96/ 2

.. 53’48

7275 is eliminated

Step 7:

(Left Column, 4^th Row)

4^th Sq. Rt. Loc.

↓7325 ......... 7375↑

.. 54’40 \

.. 53’94 / ↑

.. 53’48

.107’88 / 2

.. 53’94

The ‘up arrow’ indicates that the ‘true’ square root is 7375

Step 8:

(Right Column, 4^th Row)

Therefore:

√54,390,625 = 7,375

S.E. "Grouping" Method

DEALING WITH SQUARES OF FIVE

Table of Multiples of Five Squares (M5.Sq)

05² = 0’25

15² = 2’25

25² = 6’25

35² = 12’25

45² = 20’25

55² = 30’25

65² = 42’25

75² = 56’25

85² = 72’25

95² = 90’25

If you look at the table above, you may notice that there are some “common things” about these following groups;

Group 1

15² = 2’25

35² = 12’25

65² = 42’25

85² = 72’25

Group 2

05² = 0’25 (sometimes, not included)

45² = 20’25

55² = 30’25

95² = 90’25

Group 3

25² = 6’25

75² = 56’25

Take Note:

1) The squares of 15, 35, 65 and 85, all end up with …225

2) The squares of 5, 45, 55 and 95, all end up with …025

3) The square of 25 and 75, end up with …6’25

Always remember that all squares of numbers having a last digit of 5, will all, end up with …25.

But the underlined numbers will give us a ‘hint’ of what the missing digits might be.

Example:

√ 403,225

Step 1:

Use the basic instructions of Square Edging (Review SE Telegram: http://easysqrtsforkids.blogspot.com/2010/08/se-telegram_13.html )

√ 40’32’25

... 6 .._ .. 5

.. N : 1,3,6,8

Step 2:

Create a Parameter Checker (P-Chk)

P: 40’32

65: 42’25↓

4↓ N : {0, 1, 2, 3, 4}

You will notice that the middle missing digit (N). could either be 1, 3, 6 or 8 (The given problem belongs to Group 1 as indicated by the third to the last digit of √40’32’25).

Take note too, that the P-Chk also give us a hint that the ‘true square root’ is below 650 (Due to the notation 4↓),

Step 3

Underline the 1 and 3, indicating that 6 and 8 are eliminated:

P: 40’32

65: 42’25↓

4↓ N : {0, 1, 2, 3, 4}

Step 4

Create A Square Root Locator

↓ 615 ..... 635 ↑

42’25 \

39’06 / ↑

36 .....

78’25 / 2

39’12

Therefore:

√ 403,225 = 635

MSM-2 seems to be simple when dealing with “squares of 5”in five or six digits. All you have to do is to determine which group the missing “middle-digit” belongs and slim down the possibility by using the P-Chk.

“SQUARES OF FIVE” IN EIGHT DIGITS

Case No. 1

Given Problem:

√ 21,949,225

Step 1:

Use the SE telegram procedure as initial instructions:

√ 21’94’92’25

.... 4 ........... 5

N : 1, 3, 6, 8

Step 2:

Create A P-Chk

P : 21’94

45: 20’25 ↑

5↑ M : 5, 6, 7, 8, 9

The N notation is telling us that ‘the second to the last digit’ of the possible square root (which is still unknown), could either ‘any’ of the indicated digits on its right side (Only one of them, 1, 3, 6 or 8 is the correct digit).

The M notation also tells us, that the second digit of the possible square roots could be either be 5, 6, 7, 8 or 9. So we have “4x5” combinations, starting with:

4515, 4535, 4565, 4585, 4615, 4635, 4665, 4685, 4715, 4735, 4765, 4785, 4815, 4835, 4865, 4885, 4915, 4935, 4965 and 4985 and only one of the 20 possible square root is the true square root

Step 3:

To slim down the possibilities, we must use the 1^st Square Root Locator .

25 .....

22’56 \ ↓

20’25 /

45’25 / 2

22’62

Take note that the ‘down arrow’ ↓ indicates that the possible square roots are within the M: “5, 6 and 7” range.

Modify Step 2:

1) As a rule, underline 5, 6, 7(or draw a box, enclosing 5, 6 and 7 of the P-Chk )

2) Insert the notation “/_750/” on the second line of P-Chk . The purpose of this is determine the limit of set of possible square roots

P : 21’94

45: 20’25 ↑ /4750/

5↑ M : 5, 6, 7, 8, 9

Step 4:

On the next column, write down the set of possible square roots

... ↓ ........ ↑

4615 ..... 4735

4585 ..... 4715

4565 ..... 4685

4535 ..... 4665

4515 ..... 4635

This is the “tricky part”:

1) Starting with 4515, write it down by starting from the bottom part, going upward up to 4615 (five elements at each set). Put a ‘down arrow’ symbol on top of the first set

2) Write down the next batch of possible square roots on the next column, starting from 4635 up to 4735, following the same procedures. This time, put an ‘up arrow’.

3) Adding the notation /4750/ on the second line of the P-Chk give us the idea that the ‘upper limit’ only include 4735. Therefore, 4765 and above, are eliminated.

4) Indicate a down arrow for the left group and an up arrow for the right group

Step 5:

Use the 2^nd Square Root Locator, to determine which one of the two groups should be eliminated

22’56 \

21’40 / ↑

20’25

42’81 / 2

21’40

The ‘up arrow’ indicates which group should remain. Enclose inside a square the second group, indicating that it is where the true square root belongs. So the left group is eliminated.

4735

4715

4685

4665

4635

Important:

Enclose and put an ‘up arrow’ ↑ for {4685, 4715, and 4735} while a ‘down arrow’↓ for {4635, 4665 and 4685}

... ↑

4735

4715

4685

... ↓

4685

4665

4635

Step 5:

Continue the process of averaging the square values ( 3^rd Square root locator)

22’56

21’98 \ ↓

21’40 /

43’96 / 2

21’98

The down arrow ↓ eliminates - 4685, 4715, and 4735. There are still 3 remaining square roots:

..↑ / 4685

.... \ 4665 \

...... 4635 / ↓

Step 6:

Continue the averaging and elimination process (4^th square root locator)

21’98 \

21’69 / ↑

21’40

43’38 / 2

21’69

4635 is eliminated.

↓4665 ....... 4685↑

21’98 \

21’83 / ↑

21’69

43’67 / 2

21’83

The up arrow indicates that the true square root is 4685

Therefore:

√ 21,949,225 = 4,685