A New Method of Squaring Numbers

Common Way of Multiplying Numbers

Squaring a number is the same as multiplying two numbers having identical values.

Example:

Square the number 743 = 743x 743.

1) Multiply 743 by 3. Put the carries above 743 and the partial product = 2229

2) Then multiply 743 by 4. The partial product = 2972. Put the last digit 2 on the tens decimal place.

3) Lastly, multiply 743 by 7. the partial product = 5201. Put the last digit 1 on the hundreds decimal place.

4) Add the partial products and what we get is = 552049 or 552,049

That is how we commonly get the square of a number.

Systematic Squaring Method (SSQ)

This time, I’ll teach a new way of getting the squares of numbers in an easier and orderly manner. But first, you must also know some new things.

Digit Number

A digit (what I’m talking about here is the numeric digit), is either any of the following;

0, 1, 2, 3, 4, 5, 6, 7, 8 or 9

A number such as 743 has three digits, 7, 4 and 3. Sometimes it is called a three-digit number. All you have to do is to count the digits. Counting the digits of 4,569,742, we can then, name that number, as a seven-digit number. In SSQ, the “count of digits of a number is important”. Later, you will realize the reason why it is important. But for now, giving you the idea of what a digit of a number is all about, would be enough.

Meaning of SSQ

SSQ stands for Systematic Squaring. It is based on a popular algebraic equation, (X + Y)². It is much different from the common method of multiplying two identical numbers.

SSQ has four main parts, namely:

1) PSL (Partial Squares Line)

2) Sub-products (SP1, SP2, SP3…)

3) Sub-totals (ST1, ST2, ST3…)

4) Total Sum (TSum)

Index Squares

Always remember that there are only ten basic digits (numeric digits) and these are;

0, 1, 2, 3, 4, 5, 6, 7, 8 and 9

An index square is a product of a ‘basic digit’ multiplied by itself:

0x0 = 0 / 1x1 = 1 / 2x2 = 4 / 3x3 = 9 / 4x4 = 16

5x5 = 25 / 6x6 = 36 / 7x7 = 49 / 8x8 = 64 / 9x9 = 81

It is safe to call 0, 1, 4, 9, 16, 25, 36 , 49, 64 and 81 as index squares but in SSQ, an index square must be expressed as “two-digit square”. So the proper way of writing them as follows;

Table of Index Squares

0² = 00

1² = 01

2² = 04

3² = 09

4² = 16

5² = 25

6² = 36

7² = 49

8² = 64

9² = 81

Two-Digit Systematic Squaring (2D.SSQ)

Let start by squaring a two-digit number, using the SSQ method

Question: What is the square of 23?

23² = ?

Step 1: Create a PSL

Partial Squares Line (PSL)

The PSL is simply, the “two-digit squares” representation of each, individual digits of a certain number. In 23², the two-digit squares representation of the digits, 2 and 3 are 04 and 09, respectively. So we simply write it this way:

23² = 04’09 ← PSL

But don’t forget to also include this sign - ’ (a special character called single close quote). It will easily give us a clue of how many index squares are there in a PSL.

Step 2: Solve the sub-product

Sub-Product (SP1)

Don’t think that the value we’d taken from the PSL is already the correct answer. The value 04’09 is still incomplete. We must add a sub-product to come up with the ‘true’ square value of 23. But to get the sub-product of 23, we must cross multiply the digits 2 and 3 in a certain kind of pattern.

General Rules in Dealing with Sub-products

Rule 1: Cross multiply the individual digits of a given number using the R.A.R. multiplication pattern

Rule 2: Don’t forget the DTP reminder, “Double The Product”

Rule 3: Follow the decimal place of the reference digit

R.A.R. Multiplication Pattern

R.A.R. stands for “Reference Digit and All the Digits to the Right”. It is a multiplication pattern that is effective in solving the sub-products. How it works?

First Pattern:

If we pick 3 from 23 as our reference digit, ‘the all to the right’ of 3 is nothing, null, empty or zero. Multiplying 3 by 0 is futile, so, we can skip this pattern.

Second Pattern (SP1):

If we pick 2 as our reference digit, ‘the all to the right’ of 2 is 3.

Rule 1: 2 x 3 = 6

Rule 2: DTP reminder 6 x 2 = 12

If you wish, you can skip rule 2 as long as you directly multiply 2 x 3 by 2

2x3x2 = 12 ← SP1

Rule 3: Our reference digit 2 is in the tens decimal place, therefore, the last digit of SP1 must be also, in the tens decimal place

2x3x2 = 12 ← SP1 (provided that 2 of 12 is aligned to the tens decimal place)

Step 3: Get the total sum, add the sub-product to the PSL

Total Sum

The total sum is the final answer. It reflects the ‘true’ square value of a given number. If you multiply 23 by 23, using the common method of multiplying numbers, you will discover that the total sum of SSQ method is exactly equivalent to the product of 23 x 23.

T-Sum = PSL + SP1

... 23² = 04’09 ← PSL
+2x3x2 = 1'2 ← SP1

..............05’29 ← T-Sum

Squares of Three-Digit Numbers

Multi-Digit SSQ

SSQ is applicable to any number. But SSQ becomes a little bit complicated when the digits of a number increases. The 2D.SSQ is the easiest but when the digits of numbers increases, such squaring 54,675², it involves many sub-products and many sub-totals and the process of multiplying and adding the set of numbers also become very complex. BUT DON’T WORRY, we will never to do that 5D.SSQ. But I will just give some new general rules of SSQ so that you will have an idea, how to tackle our next topics, the 3D.SSQ and 4D.SSQ.

General Rules of SSQ

General Rule No.1:

If you square a number, make sure to count the digits. The count of digits doubles as you square a number.

35² = 1,225

(Take note, a from two-digit number it becomes a four-digit number)

But in real life, this rule seems to be not true.

12² = 144

Well in SSQ, it always agrees because we write 12² as 01’44.

12² = 01'44

General Rule No. 2

The number of sub-products increases as the digit of numbers increases. But the number of sub-products depends on the count of digits. The sub-products are always less than one to the count of digits.

In 2D.SSQ, there is only one sub-product (SP1). But be prepared, in 3D.SSQ, you will not only be dealing with one sub-product but two sub-products (SP1 and SP2). While in 4D.SSQ, you need three sub-products (SP1, SP2 and SP3), to be added to the PSL, to get the square of a four-digit number.

General Rule No. 3

The sub-totals are always less than one, the number of the sub-products


Sub-Totals

A ‘sub-total’ is simply, a temporary sum, in between, each time you add a sub-product. In squaring a two-digit number (such as, 23²), there is no sub-total because you directly add SP1 to the PSL. A sub-total is optional; you can either include or ignore it. But personally, I highly recommend you to practice including it because it is important in our study of Square Edging.

Three-Digit SSQ (3D.SSQ)

In 3.D SSQ, the same rules you learned in 2D.SSQ still work. But of course, there are new added features.

Given Problem:

743² = ?

Step 1:Create a PSL

This is the easiest part. Simply write down the index squares of 7, 4 and 3.

743² = 49’16’09

Step 2:Solve the first sub-product SP1

First Sub-Product (SP1)

By the rules in dealing with sub-products;

Rule 1: Multiply the digits 4 and 3

4x3 = 12

Rule 2: Double the product (Optional, if you wish, you can skip this one)

12 x 2 = 24

Rule 3: Take note, the last digit of SP1 must be in the tens decimal place.

4x3x2 = 24 ← PS1

Step 3: Get the sub-total by adding the first sub-product to the PSL

Sub-Total (ST1)

Take note, this sub-total is just simply like a stop-over. It is not yet our final destination.

......743² = 49’16’09 ← PSL
.+4x3x2 = .......2'4 ← SP1
................. 49’18’49 ← ST1

Step 4: Solve the second sub-product


Second Sub-Product (SP2)

Notice that here in 3D.SSQ, the two added features are the sub-total ST1 and this second sub-product (SP2). In dealing with SP2, GIVE EXTRA CARE, in cross multiplying the digits.

Again, by the rules dealing with sub-products using the R.A.R. Multiplication Pattern, the all digits to the right of 7 are 4 and 3. We should not consider 4 and 3 as individual and separate digits. The A in R.A.R. stands for the word “All”. So don’t forget this – consider ALL to the right as ONE GROUP. So the digits 4 and 3, as a group, will become 43. You must read it as forty three.

Rule 1: Multiply the reference digit 7 by 43

7 x 43 = 301 Correct

7 x 4 x 3 = 83 Incorrect

Rule 2: Just to remind you, DTP, “always double the product”

301 x 2 = 602

Rule: The last digit 2 of 602 should be in the hundreds place

7x43x2 = 602 ← SP2

Step 5: As a final step, get the total sum by adding the second sub-product (SP2) to the sub-total (ST1)

.................. 49’18’49 ← ST1

...+7x43x2 = 6'02 ← SP2 (provided that 2 of 602 aligned to 8 of the ST1)

.................. 55’20’49 ← TSum

Summary:

......743² = 49’16’09 ← PSL

........ +4x3x2 = 2'4 ← SP1

.................. 49’18’49 ← ST1

...+7x43x2 = 6'02 ← SP2

.................. 55’20’49 ← TSum

Square Roots of 3-Digit/4-Digit Numbers

What is Square Edging?

The S.E. Method, in its original form, was intended as an alternative method beside the traditional long hand division method (search in NIST Square Root

http://www.itl.nist.gov/div897/sqg/dads/HTML/squareRoot.html

).

The only limitation of Square Edging is that it is only applicable in taking the square roots of numbers called perfect squares. The square roots of perfect squares are always in a “whole numbers”, never as fractions or decimal numbers. In fact, from 1 up to 100, there are only 10 perfect squares. While from 101 up to 10,000, there are only 90 perfect squares. The rest (9,900 other numbers) are useless in dealing with this format of S.E. But TRUST ME, it will WORTH A LOT.

I divided the topics into three;

1) 3D/4D.SE

2) 5D/6D.SE

3) 7D/8D.SE

Four Digit Square Edging (3D/4D.SE)

WARNING:

To easily understand this method of Square Edging, I highly recommend that you first study the SSQ Method.

Let us start by taking the square root of a four-digit number.

Given Problem: What is the square root of 2,304?

√2,304 = ?

Step 1

Count the digits of the given number. Re-group them by twos, starting from the last digit.

√23’04

Step 2

Find an index square, equal to or nearest to but less than the first group of digits of the given number. Write down the equivalent square root, below this first group of digits.

√23’04

4

Step 3:

Find a pair of index squares ending with the same last digit, as to the last digit of the given number. Write down below the last group of digits, their corresponding square roots.

04 is the last group of 23’04. The last digit of 04 is 4. There are two index squares that end with 4, these are 04 and 64. Their equivalent square roots are, 2 and 8.

√23’04

4 2

_ 8

It would be helpful, if you memorized the pairs of index squares having the same last digits. I provided one for you.

Table of Complementary Index Squares

1² = 01

9² = 81

1 + 9 = 10

2² = 04

8² = 64

2 + 8 = 10

3² = 09

7² = 49

3 + 7 = 10

4² = 16

6² = 36

4 + 6 = 10

0² = 00

5² = 25

NO PAIRS

Step 4

Copy the first digit of the upper square root to complete the lower square root.

√23’04

4 2 ← first possible square root

4 8 ← second possible square root

Now, you determined the two possible square roots, only one of them is the ‘true’ square root of 2,304

Final Step

One way to find out which of the two is the true square root, apply the 2D.SSQ

... 42² = 16’04 ← PSL

+4x2x2 = 1'6 ← SP1 (provided that 6 of 16 aligned to 0 of PSL)

............. 17’64 ← T-Sum (provided that T-Sum aligned to PSL)

....48² = 16’64 ← PSL

+4x8x2 = 6'4 ← SP1 (provided that 4 of 64 aligned to the second 6 of PSL)

............ 23’04 ← T-Sum (provided that T-Sum aligned to PSL)

Comparing the results, the second equation matches the given number. We are now sure that 48 is the correct answer.

√2,304 = 48

Helpful Tips:

There is another way of knowing which of the two square roots, is the true square root.

Tip 1: Check the first digit of any of the two possible square roots. Multiply it to a number next to it, higher by 1. Consider the product as our “square root indicator”.

The first digits of of both possible square roots of 42 and 48 are the same and that is 4

The number next to 4 is 5. 4 x 5 = 20

Sq. Rt. Indicator = 20

Tip 2: Compare the ‘square root indicator’ to the first group of digits of the given number

First Condition:

If the first group of digits is less than the square root indicator, pick the square root with lower value as your final answer

Second Condition:

If the first group of digits is greater than the square root indicator, pick the square root with higher value as your final answer.

The first group of digits of the given number is 23. It is greater than 20. So, pick 48 as the final answer.

√2,304 = 48

Three-Digit Square Edging (3D/4D.SE)

Now, let’s try a three digit number

Given Problem: What is the square root of 729?

√729 = ?

Step 1:

Re-group by twos

Important:

Take note, that if we re-group 729, it will appear as 7’29. But in the general rules of SSQ - in the process of squaring a number, the count of digits of a number must be doubled. So, a two-digit number must become four-digit number. To obey this rule, we should write 729 as 07’29.

√07’29

Step 2

√07’29

2

Step 3

√07’29

2 3

_ 7

Step 4

√07’29

2 3

2 7

Final Step

Next to 2 is 3.

2 x 3 = 6

Sq,Rt, Indicator = 6

07 > 6 (or 7 > 6)

27 > 23

√729 = 27

Author’s Note:

The truth is, this 3D/4D.SE in not really new. There are similar ideas that were posted in google and you.tube (this is one good example by Z-Math

http://www.ehow.com/how_2322332_square-root-number-mentally.html

).
I wish to give the credit to a certain Prof. Barbosa, for the Helpful tips (watch:

http://www.youtube.com/watch?v=WNJ2dCavUrA&feature=related

). I will admit, I got that from him.

But still, most of these ideas presented were limited only in taking the square roots of perfect squares in three or four digits. I will extend this method. I will show you how to take the square roots of perfect squares, even up to eight digits.

Square Roots of Five-Digit/Six-Digit Numbers

Check This Out: A Much Easier Explanation, A Revised Version

 
The ideas behind 3D/4D.SE are very important because they are the ‘basics’ of Square Edging (MSM-1 Format). In dealing with 5D/6D.SE and 7D/8D.SE, the procedures you learned in 3D/4D.SE are always included.
The 5D/6D.SE is the start of the “real square edging”. In this topic, you will soon begin to appreciate the things that you learned from the past (specially, the SSQ Method).

Five-Digit Square Edging
Let us start taking the square root of a five-digit perfect square number.

Given Problem: What is the square root of 73,441?

√73,441 = ?

Using what we learned from 3D/4D.SE, we can proceed as follows;

Step 1

√07’34’41

Step 2 :

√07’34’41

Step 3:

√07’34’41

2 _ 1

_ _ 9

Step 4 :

√07’34’41

2 _ 1

2 _ 9

Now, our problem is how to get the middle digit.

Step 5

Instead of leaving the middle of each possible square roots blanks, simply put a letter “N” between 2 and 1. In the same way, put also an N between 2 and 9.

√07’34’41

2 N 1

2 N 9

Now, to find out the missing digits, apply the SSQ Method

Looking For the Missing Digits

Square Edging is not really the opposite or reverse process of SSQ. In fact, they are the same. The only difference is that in SSQ, what we are looking for is the total sum. While in S.E, the total sum is already given and what we are looking for is the missing digit or digits.

To make our search for the missing digits easy and simple, let us first consider the first incomplete square root.

2N1² = ?

Step 6

Create a PSL

2N1² = 04’nn’01 ← PSL

The capital letter ‘N’, represent the missing digit and the small “nn”, its equivalent index square. We do not really have an idea of what these ‘N’ and ‘nn’ are, but including them in our equations will make things easier for us.

Step 7

Cross multiply N to the digit on its right and multiply again by 2. Just leave it that way. It temporarily represent SP1.

Nx1x2 = ? ← SP1

Total Sum

I mentioned before that the difference of SE from SSQ is that in SE, the total sum is already given. Use the digits in the given number as the total sum. Do not include the square root sign “ √ "

√07’34’41 becomes “ 07’34’41 ”

Step 8

Create a T-Sum

07’34’41 ← TSum

Step 9

Put them all together - the PSL, temporary SP1 and T-Sum. Put a plus sign + to the temporary SP1, to indicate that you wish to add the temporary SP1 to the PSL.

.... 2N1² = 04’nn’01 ← PSL

.+Nx1x2 = ? ← SP1

………… 07’34’41 ← TSum

Link to the Missing Digit

If you look at it closely, the last two-digit of the PSL is 01, while the last two-digit of the T Sum is 41”. This is the link for us to know, the missing middle digit. If the last digit of SP1 must be in the tens decimal place, we should "focus our attention" to the "tens decimal places of the PSL and the T-Sum".

0 ← at tens decimal place of the PSL

4 ← at tens decimal place of the PSL

In SSQ, to get the total sum, we must add the sub-product SP1 to the PSL.

0 + ? = 4

What number we must add to 0 to have a sum equal to 4?

The answer is 4.

So the last digit of SP1 is 4

Nx1x2 = ..4 ← SP1

Use the notation “..” (two dots), to indicate that maybe, there are digits before 4 but we do not know yet, what they are. The important thing is that, we have now an idea of what the last digit of SP1.

Knowing the Middle Digit

In N x 1x 2, we can multiply 1 by 2, but not N. Simply multiply 1 x 2.

N x 1 x 2 = N x 2

N x 2 -= ..4

N represents a single digit number. So what is that single-digit number, when you multiplied by 2, give us a product that ends with 4?

There always two numbers

N = 2, 7

If we replace N by 2, 2x2 = 4

If we replace N by 7, 7x2 = 14

Complementary Multiplicands

Always remember that all sub-products are in even numbers. In multiplication, if the multiplier is an even number, there is always a pair of multiplicands that will give products that ends with the same last digits.

Table of Complementary Multiplicands

(Note: Each pair of digits when multiplied by the same even number give products that ends with the same last digits

0 and 5

1 and 6

2 and 7

3 and 8

4 and 9

Step 9

Instead of N, write down the its equivalent digits

221

272

We have now the first set of possible square roots.

Step 10: Repeat the process for 2N9, start again in step 6.

Check if we have the same data for 2N9

PSL of 2N9 = 04'nn'81

Temporary SP1 = Nx9x2

N x 18 can be written as Nx8

N x 8 = ..6

N = 2,7

Our second set of possible square roots is;

229

279

So now we completed the four possible square roots

221

271

229

279

But only one of them is the true square root of 07’34’41

Which among the four possible square roots is the true square root? We can use the 3D.SSQ, squaring them, one at a time.But it will take a lot of time and effort for you to do that. Don’t worry, there is a better way.

Squares Ending in 25

I think it would be helpful, if I’ll teach you first, a “trick” in squaring two-digit numbers ending in 5 (or numbers in the multiples of 5).

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

But actually, there is no need to memorize the table above. If you will notice, all squares in the multiples of 5 always end in 25. But check this out;

0 x 1 = 0

05² = 0’25

1 x 2 = 2

15² = 2’25

2 x 3 = 6

25² = 6’25

3 x 4 = 12

35² = 12’25

.

.

.

9 x 10 = 90

95² = 90’25

Easy Trick

Example:

85² = ?

Step 1

Check the first digit of the given number

The first digit in 85² is 8

First digit = 8

Step 2

Multiply that digit to a number next to it, higher by 1

The digit next to 8 is 9

8 x 9 = 72

Step 3

Write down their product. Also write next to it, the digits 25 with the sign ’

72’25

Therefore,

85² = 72’25

That’s how easy to do that trick.

Parameter Check

Prof. Barbosa’s Method (remember that helpful tips in Four-Digit Square Edging?) is only effective (or accurate), in dealing with two possible square roots. But this time, we have four possible square roots.

All the possible square roots (221, 729, 771 and 771) are, in-between 200 and 300. We can use that limitation as our “parameter”, to correctly guess the location of the true square root of 07’34’41. But the limit provided by 700 and 800, is just not enough. We need the middle-half of 700 and 800 to make our search, more precise and accurate.

The ‘actual’ parameter checker, if we consider the possible square roots in their actual values (3-digit numbers) can be represented this way:

Parameter Checker

300² = 09’00’00

250² = 06’25’00

200² = 04’00’00

Looking at the right side values of the parameter checker, the given number, 07’34’41 is less than 09’00’00 but greater than 06’25’00. So, 07’34’41 is in-between 09’00’00 and 06’25’00. It is in the higher limit area (H.L.A.).

Looking at left side of the P-Chk (parameter checker), the ‘true square root’ should also be, in-between 250 and 300 (or in the higher limit area). If we arrange the four possible square roots, from lowest to highest, it will appear this way;

221, 229, 271, 279.

The P. Chk is giving us a clue that the ‘true square root’ is higher than 250 but less than 300. 221 is lower than 250, so we must exclude it. The same way, 229 is also lower than 250. So the only remaining possible square roots are 271 and 279. So, we reduced our possible squares from 4 into only 2 possible square roots.

271

279

The Parameter Checker only gives us an idea, in which area (higher or lower limit area), the true square root might be located. But it is not enough. In H.L.A. (higher limit area), where we assumed the location of the true square root of 7’34’41, two possible square roots still remained (271 and 279). Which among the two is the correct answer? One way to know it is by using the 3D.SSQ. But again, it would be a long process. Don’t worry, there’s another way.

Square Root Locator

We need to further sub-divide the HLA into two parts – upper portion and the lower portion. To do that, we need a middle number between 250 and 300. That middle number will become a boundary that will separate the upper portion from the lower portion of the HLA.

How to know the middle number between 250 and 300? Just follow these instructions:

First: Add 250 and 300

250 + 300 = 550

Second: Divide the sum by two

550 ÷ 2 = 275

The Middle-Half of 250 and 300 = 275

The first group of data for our Square Root Locator will appear as;

300
275

250

Next, square them and write down their equivalent square values;

300² = 9’00’00

275² = ?

250² = 6’25’00

But how we can get the square of 275? It can be done by SSQ but there is an alternate way. Just follow again these similar instructions:

First: Add 9’00’00 and 6’25’00

9’00’00 + 6’25’00 = 15’25’00

Second: Divide the Sum by two

15’25’00 ÷ 2 = 7’62’50

Third: Subtract the quotient by 6’25

7’62’50 – 6’25 = 7’56’25

275² = 7’56’25

Now, we have a complete data for Square Root Locator

300² = 9’00’00

275² = 7’56’25

250² = 6’25’00

Let us sub-divide them into two parts;

(A)

300² = 9’00’00 \ Upper portion, which is between 275 and 300

275² = 7’56’25 /← Middle Boundary that separate the upper and lower portions

250² = 6’25’00

(B)

300² = 9’00’00

275² = 7’56’25 \← Middle Boundary that separate the upper and lower portions

250² = 6’25’00 / Lower portion, which is between 250 and 275

Looking at the right side data of the Sq. Rt. Loc., the given value, 7’34’41 is below 7’56’25 but above 6’25’00, so it belongs to the lower portion of the HLA.

The two possible square roots have also their respective locations.

279 is above 275 but below 300 (ignore their square signs “ ^{2 ”} ) or in the upper portion.

While 271 is between 275 and 250 or in lower portion.

If 7’34’41 is in the lower portion, it must also follow that the corresponding data on its left side might be its equivalent square root. 271 is also in the lower portion.

Therefore, the true square root is 271.

√7’34’41 = 271