NUMBER SQUARES (in a different way)

BIO – SKETCH

I , Apurva Mehrotra is a complete Math’s enthusiast and I fell in love with numbers around 8 years back , currently residing in New Delhi (INDIA) . I have been working with Microsoft India Ltd. For the past 1 year . I am a two-time 100 percentiler in Quantative Aptitude Section of C.A.T (arguably the toughest exam in the world) both in 2009 as well as 2010, here I would like to share with you some unknown realms from the planet of numbers, which will actually help all to understand things in a way as never understood before, through these modest blogs of mine I’ll try to make the most feared subject in the world that is Mathematics a bit interesting for everyone. I beseech all to kindly go through the blogs and provide your invaluable feedback so as to help me make it better every time I make this trivial effort of mine. All individuals who would like to share something with me regarding absolutely anything are most welcome to mail me at

mehrotra.apurva@yahoo.com

You can also find me on Facebook at

http://www.facebook.com/profile.php?id=100000020233388

Believe me whether you are a 9^th grade student or a PhD in mathematics I sincerely hope that all of you will find this blog to be highly enriching in terms of knowledge. Students preparing for various competitive exams such as CAT , GMAT , CSAT (I.A.S) , BANK EXAMS , CIVIL SERVICES , INTERNATIONAL MATH OLYMPIAD etc. will really benefit in a big way after going through this blog  the tiresome never ending calculations will becoming a child’s play for all.

I hereby humbly request you to be a bit patient while going through this blog as it is a bit on the lengthier side , I would advice people who are not that comfortable with numbers not to read this in 1 go and fragment it into parts and go reading it part by part in order to understand it in a better way.

INTRODUCTION

Frankly speaking I was a bit confused about the fact as to on which topic should my first Blog be, after much deliberation I finally decided to initiate this journey with a topic that caused some real problems to me as a kid, every time I was given a 2 or 3 digit number and was required to Square it up my face had a glum expression. But with this Blog I would to like to share some snapshots as to how Squares of the numbers can b related to some of the most common number patterns available to us and some alternative techniques which will provide with a much easier way of finding the squares of even relatively large numbers. Though there are numerous ways of doing so, I will be explaining some of my personal favorites here.

SQUARES

As we all very well know that a square of a number is actually the result obtained by multiplying the number with itself. Let the number be “n” then Square=n×n. Even if I talk about the geometrical figure The Square the area of it is obtained by multiplying the length of side by itself (side×side) that is the Square of its side. Now whether the geometric figure was named on the arithmetic calculation or vice-versa I’m yet to learn J .

Two Important Number Patterns

·        In order to find the square of any number we can add the sequence of that many consecutive odd natural numbers.

For Example –

In order to find the square of say a number “6”

Add the first 6 odd numbers in order to obtain your answer

1+3+5+7+9+11 = 36

Similarly for any number “n” in order to arrive to n², you need to find the sum of the first “n” natural  numbers that is

1+3+5+7+……….. till “n” terms

This can be understood by a simple fact that if we take an A.P such that

First term =1

Common Difference=2  and

no. of terms = n

and you will arrive to the ans. As n²

another method to do the same is by applying the formula ∑₁  ⁿ  (2n-1) , because any odd number can be represented by the general formula (2n-1) and if we calculate the sum of first “n” odd numbers the  sum comes out to be n².

·        The next number sequence that can be used to find the square of any number is given by adding  numbers from 1 to “n” in increasing order and then again adding them from (n-1) to 1 in decreasing order.

For example

In order to find the square of  “7”

First add all natural numbers from 1 to 7

i.e 1+2+3+4+5+6+7

now add numbers starting from (n-1) till1

i.e 6+5+4+3+2+1

Now on combining both the sequences you will get

1+2+3+4+5+6+7+6+5+4+3+2+1 = 49 = 7²

Again the explanation can be found out by a simple algebraic formula i.e by adding the series ∑n +∑(n-1)

[n(n+1)/2] +[n(n-1)/2]

[n²+n+n²-n]/2

2n²/2

n²

So simple na ! J

TRIANGULAR  NUMBERS

I have a special liking for these numbers in the number theory , algebraically speaking these numbers are those that belong to the series [n(n+1)]/2 , if you substitute values of “n” as natural numbers that is 1,2,3,4,5,6,7………

Geometrically speaking it is the number of beads that are required in order to form a triangle, number “1” being an exception.

A Special Relationship exists between Square nos. and Triangular nos.

·        Two consecutive triangular nos. added together always form a square no.

For example –

3+6 = 9

Algebraically Speaking –

The first ∆ number can be denoted as [n(n+1)]/2

The next consecutive ∆ number can be denoted as [(n+1)(n+2)]/2

SUM of  both of these numbers

[n(n+1)]/2 + [(n+1)(n+2)]/2

[n(n+1)+(n+1)(n+2)]/2

[(n+1)(n+n+2)]/2

[(n+1)(2n+2)]/2

[(n+1)2(n+1)]/2

[(n+1)(n+1)]

(n+1)²

Thus we finally get the whole square of the number (n+1)

·        The second relationship between triangular nos. and square nos. can be shown as

Any odd square number = [(a triangular number)×8]+1

Again let us use some variables in order to get something new

Any odd square no. = {[n(n+1)/2]×8}+1

[4n(n+1)]+1

Since “n” is a natural number hence “n(n+1)”  has to be a natural number  say “m” , hence we can say that any odd square number is always in the form of “4m+1”

Where m=n(n+1) , now substituting n=0,1,2,3,4,5,6,7,8,9,……

We get the sequence 1,9,25,49,81……

That is the sequence of odd squares

PYTHAGOREAN  TRIPLETS

As we all know that the study of square nos. can never be complete until and unless we discuss about a very special type of relationship between three square numbers that is popularly known as the Pythagoras theorem and it actually helped all of us in solving majority of Geometry questions while at school, though I’ll be discussing  the varied properties of Pythagorean Triplets at later stages here I’ll just be discussing about a few of my favorite methods as to how to find them in a quicker way . Now as you all know that three numbers a,b,c are known as  Pythagorean Triplets if they follow the relation

                                           a²  +  b²  =  c²

This relation as we all know geometrically holds true for all right angled triangles where the square of the length of the hypotenuse of the triangle is equal to the sum of the square of the other two sides.

Now we’ll discuss about the methods as to how to find three numbers that exhibit the above stated  relationship , here I’ll be describing the one which I actually found to be the easiest of all I’ve seen or heard . In order to get three numbers which are Pythagorean Triplets :

Take any Integer “n”

Now find the number (n+2)

Now add the fractions 1/n and 1/(n+2)

The numerator and the denominator obtained by adding these two fractions constitute the two smallest numbers of the Pythagorean Triplet desired to be obtained , the largest can be obtained by summing up there Squares.

Let’s take an example in order to clarify the technique in a better way

Let n = 4

Hence (n+2) = 6

Now ¼ + 1/6 = (3+2)/12 = 5/12

Hence 5 and 12 form the smaller pair of the required triplets and we can obtain the third number by adding there sums that is  5²  + 12²  = 169 = 13²

Hence we obtain a Pythagorean Triplet that is :

5²  +  12²  = 13²

Similarly by taking any value of “n” we can find out a new Pythagorean Triplet.     

 Interesting naa !!! J

Now taking this Concept of Pythagorean Triplets a bit further we can also obtain a new relation that has been occurring quite frequently in all the competitive exams that is:

                     a²+b²+c² = d²

Hence we can now see that we will have to find three numbers such that their squares add up to the square of a fourth number.

For this take any number “a” , now you need to take the other two numbers “b” and “c” such that b=(a+1) and c=ab , this sequence will help you determine the three numbers a,b and c and then finally you can find out “d” by squaring them up and then adding them.

Let us again take an example:

Let a=3

Hence b=(a+1) = 4

c= ab = 12

Hence  3²+4²+12² = 9+16+144 = 169 =13²

Hence we get the relation as 3²+4²+12²=13²

Taking this a little further we can obtain the equation as

3² + 4² = 13² – 12²

9+16 = 13² – 12²

 25 = 13² – 12²

13+12 = 13² – 12²

Hence we can deduce from here that the difference of the squares of two consecutive natural numbers is always equal to their sum.

Algebraically Speaking

b² – a² = (b+a) (b-a)

now since “a” & “b” are consecutive numbers it implies that b=(a+1) or

          (b-a)=1

Substituting this we get b²-a² = (b+a)(1)

= b²-a² = (b+a) (if a & b are consecutive numbers)

Remaining Topics of this sort will be dealt in The Blog on “Arithmetic & Geometric Progressions” because they actually carry some of their properties.

SOME SHORTCUTS FOR FINDING OUT THE SQUARES OF VARIOUS NUMBERS

(Math is a subject for lazy people; we always tend to find SHORTCUTSJ)

For Numbers Ending With 5

This is the most famous short cut when we talk about finding the Squares

The statement goes as “in order to find the Square of any number ending with 5 just write 25 as it is because the last two digits of the square number will always be 25, now in order to obtain the remaining digits just take all the digits except the units digit and multiply the number formed by its immediate successor and the result will give you the remaining digits of the required square.”

For Example:

In order to find the square of 65

65²

Now write 25 as it is to get the last two digits of the solution, now in order to get the remaining digits take all the numbers except the unit’s digit and multiply the number obtained by its successor.

Number obtained (except the unit’s digit) = 6

Successor = 7

Product = 6×7 = 42

Hence last two digits = 25

And the first two digits = 42

Hence the required answer = 4225

Let’s take another example in order to make the method even clearer

145²

Last two digits of the required solution = 25

Number obtained (except the unit’s digit) = 14

Successor = 15

Product = 14×15 = 210

Hence the first three digits of the solution = 210

Hence the Required solution is = 21025

So people, do we really require a pen in order to find the squares of numbers ending with “5” ?

A Projection of this Technique

The same technique can be applied in order to find the squares of numbers which begin with the same digits except the unit’s digits and have the unit’s digits in such a way that they add up to 10

Let’s take an example

26×24

Here again just multiply the ten’s digits of the given numbers with its successor to get the first digit, that is 2×3 = 6

Now multiply the unit’s digits of both the given numbers to get the remaining digits, that is 6×4 = 24

Hence the required solution is 624

Three digit multiplications by the same method

Let’s multiply 137 & 133 by the same method

Again leaving the unit’s digit multiply the number obtained by its successor, that is 13×14 = 182

Now multiply the unit’s digits of the given numbers to get the remaining digits = 3×7 = 21

Hence the required solution is 18221

Many of you now might be thinking, are calculations really as tough as we thought them to be ? J

For Numbers Ending With 1

Now as stated above finding squares of numbers ending with 5 is an easy job, in a similar way it’s equally easy to find squares of numbers ending with the digit 1 as well

The statement goes as “As the square of all numbers ending with1 also ends with 1 hence the unit’s digit of the solution also has to be 1 only, now in order to obtain the other digits of the solution you just need to find the successor of the given number and multiply that successor with the number obtained by eliminating the unit’s digit of the original number in order to get the remaining digits of the solution, (quite similar to the method used above, isn’t it ??)

Let’s take an example in order to illustrate method a bit further

Let’s find out 81²

As we already know that the unit’s digit of the solution has to be 1 only

Successor of the given number = 82

Number obtained by eliminating the unit’s digit = 8

Remaining digits of the solution = (82×8) = 656

Hence the required solution = 6561

One more example in order to make it Crystal Clear J

Let’s find out 121²

Again the unit’s digit has to be 1

Successor of the given number = 122

Number obtained by eliminating the unit’s digit = 12

Remaining digits of the solution = (122×12) = 1464

Hence the required solution = 14641

Now here we actually discussed some methods which were quite precise in a way , as there had to be a prerequisite for each , now let’s discuss some methods that are more generic and can help you find the square of any number till 125 within a couple of seconds in your head .

The below stated method is based on multiples of 25, many other methods also exist but I found this one to be the easiest of the lot but (as they say there’s always a “but” in life L) the thing that you need to keep in mind is that in order to use this method you need to be thoroughly well versed with the square of all numbers till 25

For Numbers from 26 to 75

Take any number in the given range, let us take a prime number (numbers which do not have any other factors except 1 and itself ) , because we always find it difficult to deal with prime numbers, Let the number be 67

Now we need to find 67²

Now in order to find the first two digits of the solution subtract 25 from the given number, hence the first two digits of the solution is (67-25) = 42

Now in order to obtain the last two digits of the solution subtract 50 from the given number and Square the answer that is obtained i.e (67-50)² = 289

Now since we know that the square of a 2 digit number has to be a 4 digit number , but here we are getting 5 digits , so we’ll carry over the leftmost digit obtained in the second part and add it to the first two digits obtained

That is, we obtained 42 in the first part and 289 in the second part

Now the leftmost digit of the second part that is 2 will have to be carried on to the first part and added

Hence the first part becomes 44 and the second as 89

Combining both we get the ans. as 4489

Let’s take another example

This time let the number be 53

So we find 53² by the same method

First two digits = (53-25) = 28

Last two digits = (53-50)² = 09

Hence the required solution = 2809

Here we see that no carrying over is required as we obtain only two digits in the second part as well

For Numbers from 76 to 125

The method of finding the squares of numbers in this range is even easier 

We’ll divide this method in two parts, 1^st for numbers lesser than 100 and the 2^nd for numbers greater than 100

For Numbers lesser than 100

Let’s again start with an example, let’s take a number between 76 and 100, how about  97 ??

So we have to find 97² in such a way that it hardly takes 5 seconds to solve it mentally

For this take the difference of the given number from 100 (with sign)

That is 97-100 = -3

Add this difference to the original number 97+(-3) = 94

These are the first two digits of your solution

In order to obtain the last two digits just square the obtained difference, that is  (-3)² = 09

Hence on combining the obtained results we get 9409 (which is actually the required solution)

Believe me, it cannot get simpler than this !!!! J

Again if there’s a carryover we’ll add it to the first two digits, for example let us take 79²

So first we need to find out (79-100) = -21

Add this to the original number, that is, 79+(-21) = 58

Now find the square of the obtained difference = (-21)² = 441

Now since we are obtaining 3 digits in the second part we’ll again carry over the leftmost digit and add it to the first part

So now, the first 2 digits = 58+4 = 62

And last two digits = 41

Hence the required ans. = 6241

For Numbers Greater than 100

Again the whole framework of the solution remains exactly the same as before

Let’s again take an example 107²

 First, take the difference of the number from 100 = (107-100) = +7

Now add the difference obtained to the original number = (107+7) = 114

These are the first 3 digits of the solution, as now the solution consists of 5 digits

Now again Square the obtained difference (+7)² = 49

These are again the last 2 digits of the required solution

Hence the required solution is 11449

now 1 for the carryover J

117²

Difference from 100 = +17

Adding this difference to the original number = 117+17 = 134

Now Square of the number obtained = (17)² = 289

Now since the obtained square has three digits carryover the leftmost digit to the first part, hence the first part becomes 134+2 = 136

Hence the required solution is 13689

Believe me try 3 questions this way and the 4^th one you will be able to solve even before your friend is able to write down the question ;)

A WORTHY APPLICATION (most important)

The technique explained above has a great application which can helps us in solving the seemingly tough multiplication problems within a blinking of an eyelid.

The above explained method can help us solve multiplication problems when they involve either (even×even) numbers or (odd×odd) numbers because in either of the two both the numbers add up to an even number.

It’ll be better off to explain this technique by taking an example, let the question be 79×73

Now as the question involves two prime numbers the methods that state the breaking down of the given numbers into their factors is not an option. Hence we’ll try to do something else.

Firstly, find the difference between both the numbers and divide it by 2

That is (79-73)/2 = 6/2 = 3

Now you will observe that if the above obtained number is either subtracted from the higher of the two numbers or added to the lower of the two numbers we obtain the same answer (it is actually the arithmetic mean between the two numbers).

(79-3)=76

(73+3)=76

Hence the given question can be stated as

(76+3)(76-3)

Now using the algebraic identity (a+b)(a-b)= a²- b²

We get (76+3)(76-3) = 76² - 3²

So now we have to calculate 

76² - 3²

Now it’s a child’s play , isn’t it

76² can be calculated by the above explained method in a matter of few seconds

We get 76² = 5776 and we all know that 3² = 9

So finally we just have to calculate 5776-9

Do you think it’s actually tough?

5767 is your answer

Again 3 questions by this method and the 4^th one actually won’t take more than 10 seconds mentally J

Let’s take another example and this time let it be an even×even problem

Let us take two numbers say 84 & 58

So we have to calculate 84×58 using the same technique as above

Again find the difference between the numbers and divide it by 2

That is (84-58)/2 = 26/2 = 13

Now again subtract the obtained result from the higher number and add it to the lower number

(84-13) = 71

(58+13) = 71

So again we get the same identity that is

(71+13)(71-13)

That is 71² – 13² needs to be calculated , so how much time do you require , I guess 8 seconds will be enough , isn’t it ??

71² - 13²

71² = 5041

13² = 169

Hence 5041- 169

= 4872

So did you find it easy ??

(But this method actually has a bit of limitation and that is that it cannot be applied in (even×odd) type questions)

Now an application that is even easier J

The above stated method becomes as easy as a 3^rd standard mathematics question if both the given numbers add up to a multiple of 10

Let’s take an example, let’s calculate 62×38

Again we’ll have to employ the same method, the difference of the numbers divided by 2

(62-38)/2 = 24/2 = 12

Now again the obtained result has to be subtracted from the higher number and added to the lower number

 62-12 = 50

38+12 = 50

Hence 62×38 = (50+12)(50-12)

Again using the identity (a+b)(a-b) = a²-b²

62×38 = 50²-12²

2500-144= 2356

Now tell me do you actually need a pen for this !!

Now another even a bigger example by the same method

Let’s multiply 137 & 63

Oh My God !! this is tough , it’ll take about 2-3 mins. To solve it on paper

Really ?? do you feel so ?? I require just 2 mins. To break your perception and believe me you will solve it in 10 seconds and that too mentally

Again applying the same technique we get (137-63)/2 = 74/2 = 37

137-37 =100

63+37 = 100

Hence 137×63 = (100+37)(100-37)

= 100²-37²

= 10000-1369 (C’mon we know how to calculate 37²)

Now, I have seen that people find it a bit tough to subtract  from numbers containing a lot of zeroes, so here is yours truly with another shortcut for you , see I told you Humans are Lazy by nature ;)

Remember this mantra for all subtractions from such numbers involving a high number of zeroes

All from 9 and the last from 10

The above stated rule states that in order to subtract any number from a number that contains a large number of 0s starting from the left hand side subtract all digits from 9 and the last one from 10 in order to get your solution (provided the difference between the number of digits of both the numbers is 1)

10000-1369 = (9-1)(9-3)(9-6)(10-9) = 8631

VEDIC MATHEMATIC METHODS

The next two methods that I’ll be describing here have been taken from the Ancient Indian Method of VEDIC MATHEMATICS which actually makes calculations extremely simple.

The first method described here is the DUPLEX METHOD, though it requires memorization of 3 simple formulas, but once these formulas are memorized the calculation of even the most complex squares becomes simple.

For this method we first have to know what a Duplex of a number is, the Duplex of numbers are given as follows

For a single digit number (a), Duplex =a²

For a two digit number (ab), Duplex = 2ab

For a three digit number (abc), Duplex = 2ac+b

Here I would like to suggest that we already have the simplest technique for finding out the squares of numbers till 125, this method based on Duplex of Numbers will be suitable for numbers above 125.

Let’s take an example

Let’s Calculate 352² using the same method

Now here you need to remember that

Square (abc) = duplex(a)/duplex(ab)/duplex(abc)/duplex(bc)/duplex(c)

(here the “/” signs do not represent division , they just represent demarcation)

So now we have Square(352)  =duplex(3)/duplex(35)/duplex(352)/duplex(52)/duplex(2)

= (3)²/(2×3×5)/[(2×3×2)+5²]/(2×5×2)/(2)²

=9/30/37/20/4

Now everywhere we are having a two digit number we will carry over the leftmost digit to the next formed duplex

So the last digit will be 4

Then 0 will be the second digit from right and 2 will be carried over to make the immediate left duplex as (37+2)=39 , hence the third digit from right will be 9 , now 3 will be carried over to make the next duplex as 33 , so the fourth digit from right will become 3 and the remaining 3 will again be carried over to make the leftmost digits as 12

Hence we’ll finally obtain the solution as 123904

I am not discussing this method in detail here but leaving it here itself , It seems to be a little complicated here but believe me try it once on paper and you will never ever use any other method to find the Squares of larger numbers (that is , above 125)

The Brahma–Gupta Identity

This Identity again has its roots in Ancient Indian Mathematics, this Identity states that “The Product Of Sum of Two Distinct Squares Taken Two at a Time yields the Sum of the Squares of two distinct numbers”

That is : (a²+b²)(c²+d²) = (x²+y²)

Where x=(ac-bd) , y=(ad+bc)

Let’s again take an example

Let a=2, b=4, c=6 & d=8

(a²+b²)(c²+d²) = (2²+4²)(6²+8²)

= (4+16)(36+64)

=(20)(100)=2000=400+1600

= (20²+40²) = [(ac-bd)²+(ad+bc)²]

This Brahma-Gupta identity actually has some simply amazing applications and I can tell it you from my own personal experience that this identity actually helps us solving a majority of questions across various examinations at an unimaginable speed and accuracy. 

The 10^x Trick

This trick is actually based completely on common sense , here we find squares of numbers which are close to multiples of 10 , the formula that we’ll be using here will be : x² = (x²-y²)+y²

Where y = (100-x)

Let’s take an example here, let’s calculate the Square of 98 with this method

98² = (98² – 2²) + 2²

= (98+2)(98-2) + 4

= (100)(96)+4

=9600+4

=9604

Simple naa !!! J

Again applications of this method can be found out across numerous examinations.

Squares from 51 to 59

I know you are now starting to think that now you actually know quite a lot of short-cuts as far as the Squares of numbers hence finally this is the last one that I’m putting up

If a number is represented in the form of 5n (where n is the unit’s digit of the number)

The Square of such number can be found out in a very simple way

The first two digits of the solution = 25+n

The last two digits of the solution = n²

Let’s find the Square of 56 by this method

The first two digits = 25+6 = 31

The last two digits = (6)² = 36

Hence the required solution is 3136

Now toh even I’m bored of telling you how simple it actually is !! J

Finally , Some Important Properties of Square Numbers

·        The difference between any square number and its predecessor is given by the formula (2n-1)

Algebraically

n² – (n-1)²

(n+n-1)(n-(n-1))   [applying identity a²-b² = (a+b)(a-b)]

(2n-1)(1)

(2n-1)

From the above stated method we can easily deduce the fact as to why the difference between two consecutive square numbers is always an odd number , if we substitute n= 1,2,3,4,5,6,7….. in (2n-1) we actually obtain the well known series of odd numbers , that is , 1,3,5,7,9,11,13…

Which again proves the fact as to why the sequence 1+3+5+7+9….. always yields square numbers J

·        The square of a number can also be obtained by adding the last square number, root of the last square number and root of the present square number, now since two consecutive numbers will always be even and odd hence their squares will also be even and odd respectively which will always add up to give an odd number, this again shows that the difference between two consecutive squares is always odd.

Algebraically

n² = (n-1)² + (n-1) +n

Solving the Right Hand Side

=n²+1-2n+n-1+n

=n²-2n+2n+1-1

= n²

Hence L.H.S = R.H.S

CONCLUSION

At the end of what is just a beginning I would first and foremost like thank all of you who read this trivial effort of mine in which I tried my best to unravel some of the mysteries of  numbers and reduce the generic perception that regards mathematics and calculations as very complicated topics . Believe me this is just the tip of the iceberg and I could’ve actually written more about Squares of Numbers, but haven’t I bored you enough already ! I promise to post more on other topics such as Cubes, Arithmetic/Geometric Progressions, Quadratic Equations and My Favorite of all Number theory and many more topics on Mathematics.

Finally it’s my humble request to all who have gone through this endeavor of mine to kindly post your comments regarding  the same and enrich me with their valuable feedback so as to improve my future undertakings . So , until next time GOD BLESS JJJ !!!