How to solve the following equations

(1)a²  + b² =  x and a + b = y 

(2) a²  + b² =  x and a - b = y

(3) a²  - b² =  x and a + b = y and

(4) a²  - b² =  x and a - b = y

If a+b = x and a-b = y then a = (x+y)/2 and b = (x-y)/2

(1) a²  + b² =  x and a + b = y ; (a+b)² = y² = a²  + b² + 2ab = x + 2ab ; y² – x = 2ab or b = (y² – x)/2a

      a + b = y; a + (y² – x)/2a = y or 2a²  - 2ay + (y² – x) = 0 ; a = (y/2) + ½ √ (2x –y² ) and b = (y/2) - ½ √ (2x –y² )

(2) a²  + b² =  x and a - b = y; (a-b)² = y² = a²  + b² - 2ab = x - 2ab ; x – y² = 2ab or b = (x- y² ) /2a

       a - b = y; a - ( x - y²)/2a = y or 2a²  - 2ay - (x - y²) = 0 ;    a = (y/2) + ½ √ (2x –y² ) and b =- (y/2) + ½ √ (2x –y² )

(3) a²  - b² =  x and a + b = y ;  a²  - b² = (a+b)(a-b) = y (a-b) =  x or a-b = x/y ; a+b= y

      a = (y² +x)/2y and b = (y² - x)/2y

(4) a²  - b² =  x and a - b = y; a²  - b² = (a+b)(a-b) = y (a+b) = x or a + b = x/y ; a-b = y

      a = (x + y² )/2y and b = (x - y²)/2y

 8

The addition of three two digit numbers gives a three digit number  which can be a cube where the single digit numbers 1 to 9 exit once only . one such example is 

83 + 54 + 79 = 216 = 6³   Can you work out for few more examples 

The cube numbers can end with any digit ,however its number digit will be 1. 8,9 only. By simply interchanging the ending digit of the two digit numbers more examples can be obtained.

84 +59 + 73 = 89 +53 + 74 = 216,

A variant of this problem is  the addition of two three digit numbers with a resultant of  a three digit number

146  + 592 = 738

154  + 782 = 936

159 +327 = 486

182 +763 = 945

183 + 492  = 675

193 +275  = 468

218 + 349  = 567

317 + 529 = 846 

By interchanging the equivalent digits in the numbers  added together , one can get more solutions but  there are very few solutions with resultant being a square. One example is here  

143 + 586 = 729 = 27² Find your own solutions.

194  + 382 = 576 = 24²

369  + 415 = 784 = 28²

 Dr.M.Meyyappan, Thavakkai Academy of Creative Thinking and Skill Development 

 5

In an addition/subtraction with two numbers where all the digits (excluding /including zero) are different, no repetition.  Find one or more possible solutions?

 There are many solutions

Subtraction                     addition                                                 Both

236  -149   =  87             183 +492 = 675                           273 + /-195 = 468/78      

263  -194    = 69              319 + 248 = 567                          275+/- 193 = 468/118

419    326     =93             391 + 284 = 675                         293 +/- 175 = 468/82           

491 - 362    =129                                                                   295 +/- 173 = 468 /122

914  - 623    =291                                                                  318 +/- 249 = 567/69

941   -632   =309                                                                   517+/-329 = 846/188

                                                                                                519 +/- 327 = 846/192

                                                                                                 527 +/- 319 =846/208

                                                                                                529 +/- 317 = 846/212 

                                                                                               782 +/- 154 = 936/628

                                         

 Dr.M.Meyyappan, Thavakkai Academy of Creative Thinking and Skill Development 

 6

Find an addition of two numbers with any number of digits where the additive numbers and its resultant have all the single digit numbers from 0 to 9 once only.

Example

         437 + 589 = 1026

Try for more solutions

        347 + 859 = 1206

        489 + 573  = 1062

       849  + 753 = 1602

The possibility is more when we add two three digit numbers to get a four digit number as its resultant. It is noted that the last digit of one of the numbers must be 9, otherwise a number may exist twice in the problem and its solution.

 4

2^a + 2^b + 2^c = 148 .Find the value of a,b,and c.

Conventional method . All the three terms are some power of two  When x (=a,b,c)  = 1,2,3.4.5,6,7 , 2^x = 2,4,8,16,32,64,128. .The sum of three numbers among these numbers must  equal to 148. ie. 148= , 128 + 16 + 4.i.e., a=2,b.=4 and c = 7

Alternative method.  2^a + 2^b + 2^c =   2^a [ 1 + (2^b + 2^c)/ 2^a ]= 148. That is 148 is divisible by 2^a

2^a [ 1 + (2^b + 2^c)/ 2^a ]= 2² x 37; a = 2 and 2^b + 2^c = 144 ; 2^b [1+ 2^c-b] = 2⁴ x 9 ; b = 4; 2^c-b = 8 or c = 7

Other examples 

3^a + 3^b + 3^c = 759;    a = 1, b = 3, c = 6

If the base numbers are different method-II will not be suitable as the factor for one of the terms is the factor for the sum of the two other terms . e.g.,  2^a + 3^b + 4^c = 531;    a = 5, b = 5, c = 4 and 2^a + 3^b + 5^c = 300;    a = 5, b = 5, c = 2

 3.If ab = xy , then (a +b) ≠ (x+y) ,which smaller ? what will be its difference ?

     If the two numbers in the pair are widely separated, its sum will be larger  .  e.g., 8x 3 = 6x4 = 24

8 and 3 are widely separated than 6 and 4, hence the sum  8 + 3 =11 is larger than the sum 6 + 4 = 10.

Let ab = xy

(a+b)²  =  a² + b² + 2ab  or  2ab = (a+b)²  -  a² - b²  and

(x+y)²  =  x² + y² + 2xy  or  2xy = (x+y)²  - x² - y²

Invariably It satisfies the relation (a+b)²  -  a² - b²   = (x+y)²  - x² - y² or (a+b)² -  (x+y)² = a² + b² - x² - y²

If  ab = xy ,then  (a+b)²  + (x-y)² =  (a-b)²  + (x+y)²

 2. If a + b = x + y  then ab ≠ xy. If ab  ≠ xy which is smaller ?  what will be its difference ?

 If the two numbers in the pair are widely separated, its product will be smaller.  e.g., in 7+11 = 5 +13 , 5 and 13 are widely separated than 7 and 11,  hence  the product 5 x 13 = 65 is smaller than the product 7 x 11= 77. 

Let a + b = x+y = 2k

a = (k-n) and b = (k+n) , ab = k² – n²

x = (k+m) and y = (k-m), xy = k² -  m² 

which shows that n= {(b-a)/2] and m = [(x-y)/2]

(a+b)²  =  (x+y)²

 

a² + b² + 2ab =  x²  + y²  + 2xy

(k-n)²  + (k+n)²  + 2ab = (k +m)²  + (k – m)²  + 2xy

 

2(k² + n² ) + 2ab =  2(k² + m² ) + 2xy

Or ab – xy = m²  -  n²  = [(x-y)/2]² – [(b-a)/2]²

 1.  A----------------------------------------------------------------------B

                                                  <-L->

    A rope AB with length L is given. With which one can construct many different rectangles ,but all with same perimeter P. These different rectangles have different area A.

                                  

                                                                           

                                                                           

 

Which rectangle will have maximum area ?

These rectangles have same perimeter P = 2(l+b)

Area of a rectangle is the product of its length and breadth A = l x b

This geometrical figures predict an arithmetical fact – If the  sum of  any two numbers equals with sum of two different numbers  then their product will not be equal and vice-versa.

If a + b = x +y , then ab ≠ xy

If ab = xy then a+b ≠ x + y

          a + b = x+y = 24     ab              ab =  xy =24    a+b                                                                                     .        ………………………………………………

             1 +23                  23                   1x24               25

             2+22                   44                    2x12               14

             3+21                   63                    3 x 8               11

             4 +20                  80                    4 x 6               10

              5 + 19                 95   

             6 + 18               108 

             7 + 17               119

            8 +  16               128

            9 + 15                135

          10 + 14               140

          11 + 13               143

          12 + 12               144  

  ………………………………………………………………

From the table we find that when the difference between the pair of two numbers giving fixed sum is smaller greater will be its product and when the difference between the pair of two numbers giving fixed product is smaller will be its sum. 

Any increase in the length will have equal decrease in the breadth of the rectangle to keep up its perimeter same.To have maximum area, the length of the rectangle must be equal to its breadth, l =  b, where the rectangle becomes a square.  If P = L, then l = b = L/4 for a square and l =(L/4- x) , b = (L/4+x) for any rectangle .The area of the rectangle  A = (L/4 – x)( L/4+x) =( L² /16)  - x²   . To be maximum x = 0 .For a given perimeter, the area of a rectangle will be maximum when it becomes a square, 

Logical answer: The area of the rectangle depends upon two parameters  length and breadth. If length is greater ,breadth will be smaller ,if length is shorter, the breadth will be greater. Which make the area smaller than the maximum. To have maximum are, both the parameter must have optimum value that is they must be equal.

  Thavakai Academy of Creative Thinking and Skill Development

 In the mechanized world, everyone has to spend more time to improve their quality of life, even if they earn and provide the expenses of the family, personal care and individual attention towards the family is decreasing. The time spent with children is reducing . Children are not being taught morals and life education by parents, the first teacher to all children . Most of the children spend more time on their cell phones than their parents. Children are not only wasting their time but also they get some irrevocable  losses which affect the later part of their life.  They are prone to eye sight defect even in the childhood stage. They fail to do the other important works and neglect to develop skills that are required to protect life in the future. What best a parent can do in this issue? .

Stopping the habit all of a sudden is a reactive action and usually  our children don’t accept and follow our words.  Our approach must make our children to get some benefits without banning the unwanted habit once allowed freely. The reactive mode must become proactive for the continuous practice of children.  Children should be trained to use the cell phone and computer for blended learning. When they are introduced to improve problem solving skill through mathematical puzzles they use digital devices not only just for entertainment but also   to stimulate creative thinking with self-interest. With this idea in mind I would like to start a you-tube channel with a self organization called 'Thavakai Academy of Creative thinking and Skill Development' for the welfare of children worldover. After knowing the solution of a problem  its variants are discussed  in successive issue.