Square of a number (shortcut method)
The square of a number means the multiplication of the number by itself. Let us keep our discussion to two digit numbers. However idea can be extended for higher digit numbers too.
For two digit number the shortcut method goes like this. Let the number be 10a+b type where unit place digit is b and tens place digit is a then by theorem
(10a+b)^2 = 100×(a^2)+b^2+2x10xaxb
= 100x (a^2)+(b^2)+ 10x(axbx2)
==> do square of a (tens place) , put it after two positions meant for keeping for square of b (unit place). To this add the multiplication of a, b, and 2 after keeping it at place leaving the first place.
This strategy solves problems fastly like, take example of 72^2
72^2, unit place 2, tens place is 7》square them as 4, 49 respectively, 》 keep them properly we get 4904 because 49 is to be kept leaving two places and square of unit digit is to be kept in two places format.》 Find additional 2ab by 2x7x2= 28 》 add the 2ab to number earlier arrived at, which is 4904 after placing it leaving the first digit (or make first digit 0), so as like 4904 + 280 = 5184》72^2 = 5184.
Some more examples :::
Ex 1.1 | To find | 56^2 |
Unit and Tens | 5 | 6 |
Square them | 25 | 36 |
Combine properly | 2536 | |
Find 2ab | 5x6x2=60 | |
Add 2ab by keeping it properly | 25 | 36 |
6 | 0 | |
Adding | 31 | 36 |
Ans | 56^2= | 3136 |
Ex 1.2 | To find | 42^2 |
Unit and Tens | 4 | 2 |
Square them | 16 | 4 |
Combine properly | 1604 | |
Find 2ab | 4x2x2=16 | |
Add 2ab by keeping it properly | 16 | 04 |
1 | 6 | |
Adding | 17 | 64 |
Ans | 42^2= | 1764 |
Special case : if a number is having unit place 5, then first step would be same, that is find square of unit place number which will be only 25. Second number which is to be concatenated to left of 25, will be multiplication of the tens place digit and its successive number. Thus 45^2 = (4x5) concat with 25 = 2025.
We have "concat" function in Excel defined like this concat("a","b","c")= "abc".
Then 45^2 = concat (4*5, "25") =2025
Or generalizing
(10a+5)^2 = concat ( a*(a+1),"25")
Same way there is an "&" function also which does the same work. Thus (10a+5) can be written as "a"&"5", thereby, we could formulate
("a"&"5")^2 = concat ( a*(a+1),"25")
If we want to redefine for general "a" and "b" , "a" being tens place and, "b" being unit place number, we get
("a"&"b")^2
= concat(a^2, if(b^2>9,b^2,"0"&(b^2))) + concat(2*a*b,"0"),
Where if(b^2>9,b^2,"0"&(b^2)) means
If b^2 > 9 then its value will be b^2, else its value will be b^2 concatenated in the left by "0".
The concat () can also be written as cat () as we generally do in UNIX operating systems commands.
Thus having a number which is having unit place as b and rest digits as a, that is number is a&b or cat (a,b), then
(a&b)^2 = (cat(a,b))^2
= cat (a^2, (b^2,\nn)) + cat(2*a*b,"0")
Another Method is based on identity
a^2 - b^2 = (a+b) x (a-b)
In this shortcut method, we find that identity a^2 - b^2 = (a+b) x (a-b)
==> a^2 = (a+b) x (a-b) + b^2
That shows that square of a number could be found by introducing another external number "b" of our choice.
Take a number for our example say 53, which is very near to 50, then if we want to find sqaure of 53, we get an identity like
53^2 = (53+b) x (53-b) + b^2
= (53+50) x (3) + 2500 taking b as 50, because it is near to 50, or we can call this number as base.
= (50+3+50) x 3 +2500
= (100)×3 +3x3 +2500
= 300 +2500 +3x3
= (25+3) x 100 + 3×3
= cat((25+3),"0",3^2)
="28"&"0"&"9" = 2809
this gives a hint to solve such questions in shortcut mode.
We go like this.
In the identity a^2 = (a+b) x (a-b) + b^2, we should define a base number B and d is difference that is there for "a" with respect to base B, so as
a = B + d, and take b = B
a^2 = (a+b) x (a-b) + b^2 will become as
a^2 = (B+d)^2
= (B+d+B) x (d) + B^2
= 2B x d + B^2 + d^2
= (B^2 +2Bd) + d^2
= (B/2+ d) x (2B) + d^2
Particularising the identity for B =50, we get
a^2 = (25+d) x100 +d^2
= cat ((25+d), if(d^2>9,d^2,"0"&(d^2)))
Where if(d^2>9,d^2,"0"&(d^2)) means
If d^2 > 9 then its value will be d^2, else its value will be d^2 concatenated in the left by "0".
Thus, for near 50,
(50+d)^2 = cat (25+d, (d^2,\nn))
Where if(d^2>9,d^2,"0"&(d^2))) is written as (d^2,\nn).
Here more elegant expression is used for putting d^2 in two digits format, i.e. if(d^2>9,d^2,"0"&(d^2))) is written as (d^2,\nn).
Example:::
2.1) Take 56^2 = (50+6)^2
=cat(25+6, 36) = 3136
Similarly 47^2 = (50-3)^2=cat(25-3,09) =2209.
2.2) 49^2 = (50-1)^2
= cat (25-1, "0"&"1") = 2401
2.3) 57^2 = (50 +7)^2
= cat (25+7, "49") = cat(32,49)
= 3249
Now change base to 100, that means if number is near to 100 like 102, 109, 98, 96 etc.
a^2 = (B/2+ d)x (2B) + d^2 reduces to (B+ 2d)x (B) + d^2 implies
a^2 = (100+ 2d)x100 + d^2
= cat (100+2d, (d^2,\nn)) or
(100+d)^2 = cat (100+2d, (d^2,\nn))
Thus, some examples are
Ex3.1) 101^2 = (100 + 1)^2
= cat (100+2*1, (1^2,\nn))
=10201
Ex3.2) 103^2 = (100 +3)^2
= cat( 100+2*3, (3^2,\nn))
= cat( 106,"09")
= 10609
Ex3.3) 97^2 = (100-3)^2
= cat(100+2*(-3), (3^2,\nn))
= cat(94, "09") = 9409
Ex3.4) 95^2 = ? We can find it by two methods. Due to 5 in unit place, we have 95^2 = cat(9*10,25)= 9025.
Another method, due to its proximity with 100.
95^2 = (100-5)^2
= cat(100+2*(-5),(5^2,\nn))
= cat(90,25)=9025.
We also have some methods due to shortcut methods of multiplying two different numbers. If in those methods, we assume that both the numbers are same, then that will also provide square of numbers. Let us explore.
If there are two numbers near to base 100, say 100+d1 and 100+d2, then their multiplication can be written as
(100+d1)*(100+d2)
= 100*100+ 100*d2+ 100*d1+ d1*d2
= 100*(100+d1+d2) + d1*d2
= cat(100+d1+d2, (d1*d2,\nn))
= cat (first number +(second number -100),(d1*d2,\nn))
Thus,
(100+d1)*(100+d2)
= cat(100+d1+d2, (d1*d2,\nn))
Ex4.1) 102*103
= cat(100+2+3, (2*3,\nn))
= cat (105,"06") = 10506
Ex4.2) 105*107
Elegantly doing like
105 5 cross addition 107+5 =112
107 7 cross addition 105+7 =112
---------------------------------------
112 35
---------------------------------------
So 105*107 = 11235
Ex4.3) 98*93
98 -2 cross addition 93-2 = 91
93 -7 cross addition 98-7 = 91
---------------------------------------
91 14
---------------------------------------
So 98*91 = 9114
Rewriting the formula near 100, is as follows. Say "a", "b" be two numbers near 100, then
a*b = cat ( cross addition, product of differences from 100 in \nn format )
Or
a*b = cat ( a+(b-100), ((a-100)*(b-100), \nn) )
Ex4.4) Now let us have two numbers near to 100, but one is greater than 100 another is less than 100 like 103 multiplied by 97 then
103 3
97 -3
Cross addition 100, product of difference in \nn format is -09. If we concatenate such numbers it will give "100"&"-09", which means in our case taking 1 from 100, reducing it to 99 and giving 100 to -09 to make it 100-09= 91, thus "100"&"-09" = 99&91 =9991. This is a important step when (d1*d2, \nn) becomes negative and it is to be made positive. // here it develops the arithmetic of cat() //
Corollary of above method is to take a number near to 100, say 100+d and find its square. The square of 100+d will be
(100+d)^2
= 100+d d
100+d d
---------------------------------------
Cross addition 100+d+d
Product of differences d*d
So (100+d)^2 = cat(100+d+d,(d*d,\nn))
Or more elegantly
a^2 = cat (a+d, (d^2,\nn)) near 100, where d = a-100.
Ex5.1) Find square of 103, 107, 112, 98, 96, 93 ‽
Ex5.2) 103^2 = cat(103+3,(3^2,\nn))
=10609
Ex5.3) 107^2 = cat(107+7,49) = 11449
Ex5.4) 112^2 = cat(112+12, (12^2,\nn) )
= cat(112+12, (1\44)) this is important as it gives carry. // here it develops the arithmetic of cat() for (x^2,\nn), when x^2 spills over the format \nn //
= cat( 112+12+1,44) = 12544
Ex5.5) 98^2= cat(98-2,(2^2,\nn))
=cat(96,"04") = 9604
Ex5.6) 96^2= cat (96-4, 16)=9216
Ex5.7) 93^2= cat(93-7,49)= 8649
Recalling that if Base is B for a number "a". Then, say "d" is difference from base "B", we have
a^2 = (B+d)^2
= 2B*d + B^2 + d^2
= B*(B+2d) + d^2
= B*(B+d+d) + d^2
= B*(a+d) + d^2
Example: we know 56^2 = 3136, now going by above equation
56^2 = 50*(56+6)+6^2
= 50*62+36 = (100*62/2) + 36
= (62/2)*100+6^2
= cat((62/2), (6^2,\nn))
= cat (31, 36) = 3136
This gives the idea that if we make B as multiple of 100, then concatenation can be done.
a^2 = (B+d)^2
= B*(a+d) + d^2
= (B/100)*(a+d)*100 + d^2
= cat ( (B/100)*(a+d), (d^2,\nn))
// here it develops the arithmetic of cat() like
x*100 +y = cat(x,(y,\nn))
= x&(y,\nn)
Thus we can also relate like
10x+y = cat(x,(y,\n))
= x&(y,\n) //
Or
a^2= cat((a+d)*(B/100), (d^2,\nn))
=cat((that# +d)*(base)/100,(d^2,\nn))
The formulae works with ease for numbers, whose base chosen is greater than or equal to 100.
Ex6.1) 108^2 = (100+8)^2 //base 100//
= cat((108+8)*100/100, (8^2,\nn))
=cat(116,64)=11664
Ex6.2) 209^2 = (200+9)^2//base 200//
= cat((209+9)*200/100, (9^2,\nn))
= cat (218*2,81) = cat(436,81)
=43681
Ex6.3) 502^2 = (500+2)^2 //base 500//
=cat ((that #+d)*(base)/100, (d^2,\nn))
=cat(504*5,(2^2,\nn))
= cat(2520,04) = 252004
Ex6.4) 496^2 // take base 500//
=(500-4)^2
=cat((#+d)*base/100,(d^2,\nn))
=cat((496-4)*500/100,(16,\nn))
= 492*5&16 = 246016
Ex6.5) 256^2=(250+6)^2 //base=250//
=cat((256+6)*250/100,(6^2,\nn))
=cat((262)*10/4,36)
=cat(2620/4,36)=65536
Ex6.6) 256^2=(260-4)^2 //base=260//
=cat((260-4)*260/100,(4^2,\nn))
=cat((256)*26/10,16)
=cat(6656/10,16)
=cat(665+(6/10),16)
=cat(655+10+(6/10),16)
=cat(655,..+16) // here there is a need to develop arithmetic of cat() so that we get correct answee 65536 //
// However, one lesson, we get here that, if elegant base is chosen, we get results easily //
Ex6.7) 256^2=(200+56)^2
//base=200//
// let us do adventure of taking a base far off//
=cat(a+d)*B/100, (d^2,\nn))
=cat((256+56)*200/100, (56^2,\nn))
=cat(( 312*2, 56^2\nn) ... Eq(1)
//representing (56^2,\nn) as 56^2\nn. //
Further, 56^2 = (25+d)&(6^2)\nn =(25+6)&36 = 3136 ... Eq(2)
From Eq(1) and Eq(2), we have
256^2 = cat(624,3136\nn)
=cat(624+31,36)
=cat(655,36) = 65536
Though it is lengthy process, but we could find out 256^2 by assuming a base 200, instead of 250. // In the process, here it develops the arithmetic of format function \nn, which says in this particlar case 3136\nn = 31&(36\nn)
= cat(31,36\nn) //
// However, another lesson, we get here that, if elegant base is chosen nearer to number, we get results easily and in fewer steps //
// Once again an expression of square of "a" as
a^2 = (B+d)^2, where B is chosen base
= (B/100)*(a+d)*100 + d^2
= cat ( (B/100)*(a+d), d^2\nn)
= cat ((B+2d)*B/100, d^2\nn)
=cat((base+2d)*(base)/100,(d^2,\nn))
This form of formula is in terms of only base and difference, easy to remember and applying. If chosen base is divisible by 100, then it further eases the competition.
//
Ex7.1) 409^2
=cat((base+2d)*base/100, d^2\nn)
=cat((400+2*9)*400/100,9^2\nn) // base is taken 400//
=418*4&81 = 167281
Ex7.2) 898^2 // take base 900, implying d= 898-900 =-2, that is given number is 2 less//
= cat ((900-2*2)*900/100, 2^2\nn))
= (896*9)&4\nn
= 8064&04 = 806404
When two numbers are multiplied and one of them is one of such type of number like 9, or 99, or 999, or 9999 ... then a concept of complementary digit arises. Complementary digit is defined as like complementary of 1 will be c(1) = 1' = 9-1 = 8, c(2) = 9-2=7 thus for any unit digit a, complementary of a will be 9-a. Having defined so, let us take a multiplication 68*99. What is its value? 6732 by the fundamental methods. Also by general cross method which goes like this:
68
99
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Vertical product = cat(6*9,8*9\nn) =5472
Sum of cross product = 54+72 = 126
vertical product + cross product *10 = 5472 +1260 = 6732
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But being 99 as one of the multiplicand this method reduces to following shortcut process like
68*99 = (68-1)&c(6)&c(7)
=67&6'&7'=67&(9-6)&(9-7)
=67&3&2=6732
The only precaution is to take make digits of both the multiplicands same or be made same.
Ex8.1) 57*99 = 56&5'&6' =56&4&3 = 5643
Ex8.2) 83*99 = 82&8'&2' =8217
Ex8.3) 86*999 = 086*999 // this is the catch if digits mismatch, provided 9 repetitive multiplicand has more digits, for vice-versa see Ex8.4 //
= 085&0'&8'&5' = 085914 =85914
Ex8.4) When a question like 768954*999 is given to evaluate, it doesn't follow the pattern which we devised earlier, because it will be equal to 768954*999
= 768954*000999
= 768(953&231)046 ???
= 768(953231)046 is not equal to correct value, which is 768185046, where 768(185)046. The head 768 and tail 046 matches but within inside it does not match. Therefore in such situations, where 9 repetitive multiplicand has lesser digits than the other multiplicand, we can rearrange them differently like, in this case, it can be rewritten as 768954*999 as 768000*999+954*999 which then can be solved as
= 1000*(768*999)+(954*999)
= 1000*(767&232)+(953&045)
= 1000*(767232)+953046
=767232000+953046
=767&(232+953)\nnn&046
= 767&(1185\nnn)&046
= 767&(1\185)&046
=(767+1)&185&046
=768185046
Well after discussing this the square of any 9 repetitive digited number can be found like
Ex9.1) 9^2 = 9*9 = (9-1)&8' =81
Ex9.2) 99^2= 99*99 = 98&01=9801
Ex9.3) 999^2 = 999*999 = 998&9'9'8'
// let us abridge the & and complementary digitization like this one for faster calculations, as each digit is unit one. //
= 998001
Further if we further assume a notational shortcut that a,b,c,d all unit digits like a&b&c&d as abcd and (abcd)' be as a'b'c'd' then we have
(999...n-times)^2
= (999...n times - 1)&(999...n times - 1)"
Ex9.4) So (99999)^2 = 9999800001
999999
Ex9.5) 99999999^2
= 99999998(99999998)'
= 9999999800000001
