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