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Dear all readers
I request the readers to visit my you tube channel - Thavakkai Academy of Creative Thinking and Skill Development , which I hope will be useful to students of all age group. The posts are in different subjects like recreational mathematics, chemical elements, Problems with chess board, problems with Magic square, Research in Tamil language, moral guidance to have peaceful life etc.,
Thanking you
Dr.M.Meyyappan
Sunday, August 4, 2024
Saturday, July 20, 2024
Friday, July 19, 2024
Monday, July 15, 2024
Sunday, July 14, 2024
Thursday, July 11, 2024
Monday, July 8, 2024
Tuesday, July 2, 2024
Find the value of 217 -1 , 218 -1 and 219 +1
Recreational
Mathematics
Find the value of 217 -1 , 218 -1 and 219
+1
When
the exponent is even 218 -1
=( 29)2 – 12
= (29 -1) (29 +1)
29
= 512; 218 -1 = (511) (513) = (500+11) (500+13 ) = 5002 +
500(11+13) + 11x13 = 250000 + 12000 + 143 = 262143
When
the exponent is odd
217
-1= 2 .216 -2 +1 = 2 (216 – 1 ) + 1
= 2[(28
-1)(28 +1)] +1 = 2[(256-1)(256+1)] +1
= 2(255)(257) +1
= 2[(250+5)(250+7)] +1
=2[2502 +250 x 12 +35] +1
= 2[62500 + 3000 + 35]+1
= 2(65535) +1=
131071
219
+1 = 2.218 -2 + 3 =2(218 – 1) + 3
= 2[(29
-1)(29 +1)] +3 = 2[(512-1)(512+1)] +3
= 2(511)(513) +3
= 2[(500+11)(500+13)] +3
=2[5002 +500 x 24 +143] +3
= 2[250000 +12000 +143]+3
= 2(262143) +1= 524286 + 3=
524289
Sunday, June 16, 2024
Solve (a+b)/(a+c) = (a2+b2)/(a2+c2)
.Find the whole integer solution for a,b and c.
(a+b)
(a2+c2) = (a+c) (a2+b2)
On
simplification we have a(c + b) = a2 – cb
Give
any value to a and find the values of b and c in accordance with the equation.
When
a = 8 ; 8(c + b) = 64 – cb or c = (64-8b)/
(8+b)
If
b = 1 ,c = 56/9; (8 +1)/(8+56/9) = (72 +
9)/(72 +56) = (722 + 92)/(722 +562)
If
b = 2 ,c = 48/10 = 24/5; (8+2)/(8+24/5) = (40 +2)/(40+24) = (402 +22)/(402+242)
Tuesday, May 28, 2024
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
Saturday, May 25, 2024
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
Friday, May 24, 2024
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
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.
Wednesday, May 22, 2024
4
2a + 2b + 2c
= 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 , 2x =
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. 2a
+ 2b + 2c =
2a [ 1 + (2b + 2c)/ 2a ]=
148. That is 148 is divisible by 2a
2a [ 1 + (2b
+ 2c)/ 2a ]= 22 x 37; a = 2 and 2b +
2c = 144 ; 2b [1+ 2c-b] = 24 x 9 ;
b = 4; 2c-b = 8 or c = 7
Other examples
3a + 3b + 3c
= 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.,
2a + 3b + 4c = 531; a = 5, b = 5, c = 4 and 2a + 3b
+ 5c = 300; a = 5, b
= 5, c = 2
Friday, May 17, 2024
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
Thursday, May 16, 2024
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
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.
Friday, April 12, 2024
பக்காத் திருடன்
செட்டிநாட்டில் உள்ள ஒரு பழங்கால பெண்மணியின் உருவப்படம் அருங்காட்சி யகத்தில் வைக்கப்பட்டுள்ளது. அதன் சட்டகம் 32 விலையுயர்ந்த கற்களால் அலங்கரிக்கப்பட்டுள்ளது, இதனால் ஒவ்வொரு பக்கத்திலும் 12 கற்கள் உள்ளன. ஒரு இருண்ட இரவில் ஒரு புத்திசாலியான திருடன் அருங்காட்சியகத்திற்குள் நுழைந்து நான்கு கற்களை எடுத்துச் சென்றான். மறுநாள் காலை காவலர் வந்து ஒவ்வொரு பக்கத்திலும் உள்ள கற்களை எண்ணினார். அதே 12 கற்கள். 4 கற்கள் திருடப்பட்டதை தாமதமாகஅருங்காட்சியக அதிகாரிகள் கண்டறிந்தனர். அது எப்படி சாத்தியமாக இருக்கும்?
சட்டத்தில் பாதிக்கப்பபெற்ற ரத்தினக் கற்கள்
4 4 4
= 12
4 4
4 4 4
= 12
12 12
ஒவ்வொரு பக்கத்திலும் மையத்தில் பாதிக்கப்பட்டுள்ள இரண்டு ரத்தினக் கற்கள் திருடப்பட்டு , காண்பவர்களால் உடனடியாக கண்டுபிடிக்கமுடியாதவாறு மீதியுள்ள கற்களைச் சற்று மாற்றி திருடனால் பதிக்கப்பட்டன
2 2
5 2 5 = 12
12 12