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MHajduk
An illusionist asks a randomly chosen spectator to pick (secretly) one of the natural numbers smaller than 100 and calculate its cube and give the result speaking it aloud. The illusionist immediately gives an answer, i.e. the original number chosen by the spectator. This trick doesn't require the magician to know all cubes of the numbers between 1 and 100. It's enough to remember only cubes of the ﬁrst ten numbers from 1 to 10 because the method of determination of the original number is based solely on this information.
http://www.slideshare.net/fullscreen/mikhajduk/cuberoot49631334 

02 Jul 2015, 16:54 

Tyler
Kaprekar's routine is another cool trick.


06 Jul 2015, 17:56 

MHajduk
Tyler wrote: Kaprekar's routine is another cool trick. 1. Split the given (positive integer) number into separate digits. 2. Calculate the square of each digit. 3. Sum up all numbers calculated in the second step. 4. Go to the first step. will lead us either to 1 or to the cycle 145, 42, 20, 4, 16, 37, 58, 89 M. Szurek, a Polish mathematician, wrote in his book that this kind of process had been called Steinhaus's problem although in Wikipedia numbers that eventually give 1 as a result of the aforementioned procedure are named "happy numbers". D. R. Kaprekar was not only Ganit ("Mathematics" in Sanskrit) guru worth to mention here. I think that we cannot forget about Shakuntala Devi, who also wrote a few books on the theme of recreational mathematics. Revisiting Numbers with Shakuntala Devi 

08 Jul 2015, 20:47 

MHajduk
shoorick wrote: the A upside down impressed me so much! I remember how confused I was when I saw it for the first time in 1995 during math lectures at university. I don't know is the "new" notational convention truly better than the previous one but we all have to unify our math language to be better understood worldwide. 

10 Jul 2015, 11:56 

shoorick
now I see to say true, if you were place there that lambda ("wedge"), i would not note the difference (but now i will, sure )


10 Jul 2015, 12:32 

Tyler
MHajduk wrote: The "A" symbol written upside down (∀) is a "Western" counterpart for the symbol Λ ("wedge") used previously (before 90's) in Poland and other Central European and Eastern European countries as an abbreviation for the universal quantifier "for all". 

12 Jul 2015, 06:33 

MHajduk
Tyler wrote:


12 Jul 2015, 09:42 

Tyler
MHajduk wrote: It was quite logical because "for all" quantifier may be seen as a generalized conjunction for all elements of the given set whereas existential quantifier may be understood as a generalized alternative. 

14 Jul 2015, 23:21 

MHajduk
Thank to Tyler who mentioned Kaprekar's routine, I've heard for the first time about this Indian mathematician and my searches led me to his article about socalled Copernicus magic squares. Finally, I've made an animation that contains some considerations on the theme of Copernicus magic square and I want to present it here (I strongly recommend you to watch this video in a fullscreen mode):
http://www.youtube.com/watch?v=uMp6NwoEHVQ&feature=youtu.be 

17 Jul 2015, 00:04 

Tyler
That's cool. Nice video. How'd you make it? Also, any clue how he came up with that? Surely he didn't brute force it. My first guess would be some weird application of group theory, but he's way, way before group theory.


17 Jul 2015, 05:00 

MHajduk
Tyler wrote: Also, any clue how he came up with that? Surely he didn't brute force it. My first guess would be some weird application of group theory, but he's way, way before group theory. Also, there exist some methods of creation of magic squares explored and explained very well by French mathematicians of the 17th century  La Loubere, Bachet and La Hire. 

17 Jul 2015, 22:45 

MHajduk
When I'm fed up with foggy autumn days I create sunny landscapes that serve me as a background for mathematical problems and their solutions.


03 Nov 2015, 13:47 

MHajduk
Just for the sake of completeness (every question should have corresponding answer ):
Use of inequality between weighted arithmetic mean and weighted geometric mean: 

03 Nov 2015, 13:47 

MHajduk


03 Nov 2015, 13:48 

MHajduk
YONG wrote: To keep the proof simple, I would start from BTW, in the middle of the proof we get the inequality "(a + b)^2 >= 4ab" for every real a,b that is frequently used in solutions of other inequalities. 

03 Nov 2015, 13:49 

MHajduk
A graph presenting a kind of "genetic relationship" between basic inequalities.
The symbols of a form "/<operator> f(a,b)" next to the arrows symbolize the operation that is to be done on the both sides of the given inequality to transform it into another inequality. For example, symbol "/ + 4ab" means that we have to add 4ab to both sides of inequality. 

03 Nov 2015, 13:50 

MHajduk
YONG wrote: Yeah, your proof gives some spinoff inequalities which are very useful. 

03 Nov 2015, 14:38 

MHajduk


18 Nov 2015, 22:38 

tthsqe
(x1)/(x+1) is increasing for x>1
(a/b)^p > (a/b)^q > 1 by hypothesis 

19 Nov 2015, 00:48 

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