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Problem Solvers
Registrace 20. 06. 2021
A channel for people who love to ponder about challenging Mathematical puzzles and problems.
I post solution videos to Math Olympiad and various challenging Math problems on a regular basis. Occasionally, I also post challenge videos and Math trivia, just to mix things up a bit.
I post solution videos to Math Olympiad and various challenging Math problems on a regular basis. Occasionally, I also post challenge videos and Math trivia, just to mix things up a bit.
Seemingly tough question cracked through clever substitution
Substitution, if cleverly done, could help instantly solve or simplify a lot of tough Olympiad Math questions. Today, we take a look at such instance, where the choice of substitution arises quite naturally from familiar factors we identify in the question.
We also touch briefly about factorisation techniques, revisited the quadratic formula method for quadratic polynomials of two variables, and place particular emphasis on the importance of checking solutions through back substitution once we have 'solved' the question.
We also touch briefly about factorisation techniques, revisited the quadratic formula method for quadratic polynomials of two variables, and place particular emphasis on the importance of checking solutions through back substitution once we have 'solved' the question.
zhlédnutí: 4 236
Video
Pretty geometry question - Sunset over square cities
zhlédnutí 244Před 2 lety
Today's geometry question is quite pretty. You could easily visualise the three concentric semicircles as halos of light as sunset descended over cities of squares. The question itself is pretty elementary and accessible as well. We will see how the usage of Pythagoras theorem allows us to make quick deductions which lead to the solution of the question. This question is taken from the blog 'Ma...
What if the 1/x key on your calculator is broken ?
zhlédnutí 99Před 2 lety
Do you know that the trigonometric functions of sin, cosine, tangent and their inverses could be used to simulate the function of the reciprocal key on the calculator ? Believe it or not, this is a valid Math question posed to students of trigonometry! Enjoy the walkthrough of this fun question, although I don't think you will be putting this into practice anytime soon...
Why can this never be a prime number ?
zhlédnutí 2,5KPřed 2 lety
Today, we will show why 14^n 11 can never be a prime number. Leveraging on modulo arithmetic allows us to draw very quick and straightforward conclusions, which would otherwise be hard to achieve using other methods. This question serves as a good use case for modulo arithmetic and also illustrates the technique of 'divide-and-conquer'- a problem solving technique which considers different case...
MISSION IMPOSSIBLE: Area of the square
zhlédnutí 1,4KPřed 2 lety
The power of a point theorem, or intersecting chord theorem is a handy and versatile tool when it comes to dealing with intersecting chords in one or multiple circles. Given just one length, how do we figure out the area of the square in the diagram ? Join me as I unravel this nice geometry puzzle, and show you how the power of a point theorem could be used to tackle similarly styled geometry q...
One Diophantine equation, two different approaches!
zhlédnutí 5KPřed 2 lety
A Diophantine equation admits only integer solutions, and do not come with a ready formula which allows us to immediately obtain the solution. In this video, I will show you two approaches to solving Diophantine equations: factorisation and the method of discriminant. These two methods are particularly powerful if you are dealing with Diophantine quadratic equations, since the factorisation met...
Amusing Math cryptogram puzzle - Strangely satisfying to solve!
zhlédnutí 407Před 2 lety
A cryptogram is a Math puzzle where each digit is represented by a single alphabet. The cryptogram we will be solving today is fun, yet tough! I will be showing you how we can systematically eliminate wrong cases and keep track of our progress as we work through the cryptogram, so please join me on this fun journey!
Telescoping series sum: Beautifully done!
zhlédnutí 2,8KPřed 2 lety
Telescoping series could be a powerful tool when it comes to evaluating series sum and product. However, oftentimes, some work is needed to reveal their existence. In today's example, we illustrate the key ideas and concepts of telescoping series using a nice series sum.
Sum of squares of Fibonacci numbers
zhlédnutí 655Před 2 lety
We take a look at this lovely geometry proof for the identity involving sum of squares of first n terms of the Fibonacci series. The geometry proof arises so naturally that it could be a little hard to believe... Diagrams are taken from 'Proofs without Words- Exercises in Visual Thinking' by Roger Nelson of Mathematical Association of America.
Trisection of a line segment
zhlédnutí 463Před 2 lety
In today's proof without words series, let's take a look at how the trisection of any straight line segment can be achieved by using compass and straightedge (ruler without markings) construction. You will see that we make good use of the property of the centroid of triangle, which divides each of the medians in a lengths of ratio 2:1. This useful geometrical fact will come in handy for tacklin...
Learn a Math theorem through watching anime!
zhlédnutí 556Před 2 lety
In today's video, we take a look at a curious Math question posed by Magata Shiki from the anime Subete ga F ni naru, or perhaps better known as "The Perfect Insider' among the English speaking anime community. We will see how was can use the Fundamental Theorem of Arithmetic (FTA) to confirm some of her claims made in the anime. You might or might not have heard about the Fundamental Theorem o...
Cauchy Schwarz inequality - A visual proof
zhlédnutí 456Před 2 lety
In today's proof without words series, we take a look at the Cauchy Schwarz inequality, and see how we could deduce a nice visual proof through geometry. The best part of it ? The geometric proof allows a very natural and satisfying establishment of equality conditions, as opposed to algebraic methods for the usual Cauchy Schwarz proofs. Diagrams are taken from 'Proofs without Words- Exercises ...
Pokemon Emerald Nuzlocke: Only Prime Number Pokedex allowed!
zhlédnutí 23Před 2 lety
I will attempt a Nuzlocke challenge of Pokemon Emerald, but with a slight twist that I could only catch pokemon whose pokedex number is a prime number. Given that no two consecutive (except for 2 and 3) numbers are prime, this means I couldn't evolve any of my Pokemon at all and my choice of Pokemon to build a team would be severely limited. Will this Nuzlocke be a success given these stringent...
Pythagoras theorem- 5 visual proofs!
zhlédnutí 159Před 2 lety
In today's video, I present 5 different visual proofs of the fabled Pythagoras theorem, which relates the relationships of lengths in a right angled triangle. The proofs all have different flavour, but all of them are stunningly elegant and never fail to amaze me every time I look at them. This establishes the Pythagoras theorem as perhaps the single theorem with the most number of verified pro...
Solving a cute Math puzzle inspired by the Clannad anime!
zhlédnutí 181Před 2 lety
In this video, we look at a class of Math puzzle - the cryptogram - which requires the solvers to possess both good logical deduction and problem solving skills. This question is especially cool for anime lovers since it is directly inspired by the anime Clannad and Clannad After Story. We will see how the cryptogram could be solved by drawing upon the techniques of case-by-case analysis, as we...
Ukraine Math Olympiad 1999: A clever number theory trick!
zhlédnutí 206Před 2 lety
Ukraine Math Olympiad 1999: A clever number theory trick!
Vertex angles in a star sum to 180 degrees
zhlédnutí 287Před 2 lety
Vertex angles in a star sum to 180 degrees
The Crazy Math Problem in Madoka Magica
zhlédnutí 383Před 2 lety
The Crazy Math Problem in Madoka Magica
China’s hardest Math exam- for high schoolers!
zhlédnutí 387Před 2 lety
China’s hardest Math exam- for high schoolers!
[Proof without words]: AM-GM Inequality proven in 3 different ways
zhlédnutí 12KPřed 2 lety
[Proof without words]: AM-GM Inequality proven in 3 different ways
The domino model - a better way to learn math induction
zhlédnutí 104Před 2 lety
The domino model - a better way to learn math induction
Trisecting an angle in infinitely many steps
zhlédnutí 212Před 2 lety
Trisecting an angle in infinitely many steps
'Family tree' of all mathematicians from the past, current and beyond - Math Genealogy Project
zhlédnutí 379Před 2 lety
'Family tree' of all mathematicians from the past, current and beyond - Math Genealogy Project
Pythagoras theorem proof by the US President!
zhlédnutí 646Před 2 lety
Pythagoras theorem proof by the US President!
[STEP Series]: Logarithms and inequalities! (2015 STEP II - Q1)
zhlédnutí 79Před 2 lety
[STEP Series]: Logarithms and inequalities! (2015 STEP II - Q1)
[STEP Series]: Are you careful enough to avoid the careless mistake? (STEP II 2018 - Q6)
zhlédnutí 61Před 2 lety
[STEP Series]: Are you careful enough to avoid the careless mistake? (STEP II 2018 - Q6)
what about the hairy ball theorem (differential topology)
quite helpful and hope your channel would grow at an unprecedent speed
😊
anime about math got 0 comments? not unexpected but lemme fix that!
Why construct that parallel dotted line? The answer is staring you in the face! (1+3) + (2+4) + 5 = 180
Really nice, but this is infact identical to the standard proof with the diagram halfed ( original proof uses 2 more copies of the original triangle, creating a side-c square inside a side-(a+b) square, and counting areas is easy )
好生硬
thank you!
it was very helpful thank you.
czcams.com/video/NMkwmsljFns/video.html
Yuyushiki teaches mathematics 0.o
I swear....that is exactly how I did it without first looking....nice problem!
Good, thanks for sharing, here is an exercise: ABCD × 1.5 = DCBA Enjoy.
Believe ABCD is 4356, thanks for the question ! Will try and do a solution video if there’s time
@@ProblemSolversInMaths Well done, this cryptogram is a difficult one, you are worthy of being named <Problem Solvers>, In fact, there is a beautiful pattern, 1089 × 9 = 9801 × 1 2178 × 8 = 8712 × 2 3267 × 7 = 7623 × 3 4356 × 6 = 6534 × 4 multiples of 1089 😎
This is a great problem and I am surprised no one else has given any comments. I did it different. From the center of the circle you can draw radius r to the north eastern point, eastern point and western point on the square. Connecting the north eastern point with the western point gives an obtuse triangle and the obtuse central angle I call theta. Based on this diagram, you can use sohcahtoa in the right triangle with adjacent r-1 and hypotenuse r to write cos(180-theta) = (r-1)/r. Using Pythagorean theorem you can find the longest side of the obtuse triangle to be SQRT(10r-5). Then using the Law of Cosines in the obtuse triangle: 10r-5=r²+r²-2r²cos(theta). But cos(180-theta) is known in terms of r and using sum/difference formula for cosine you can now eliminate the cosine term altogether to arrive at the quadratic equation 4r²-12r+5=0 which gives the desired r=2.5 (the other solution is extraneous) from which the rest follows.
x=1 x-=-1
Nice step by step solving ❤️
Odd primes > 5 is+/-1 mod 6. Taking mod 6, things are much easier
very nice explanation
Уравнение сводится к однородному. Ответ: 1
x=±1
x>0!
@@user-qy8re3yx3d Х может принять любое значение
nice
Thanks!
Thanks.
You are welcome :)
0:18 Why do you say "greater" when it's in fact "greater or equal"?
Yes, you are right, it should have been greater and equal, with equality if a=b. Thanks for pointing it out!
Wow the discriminant method is super
Indeed, especially when you are dealing with quadratics! Glad you find the video helpful :)
Amazing
Thanks, and glad you have enjoyed it :)
Namaste🙏.
Namaste🙏.
Your checking out special cases shows its value in motivating a route to a solution. It suggests moduli 3 and 5 as good candidates for a starting point.
Indeed, I like to play around with a few cases to see if there are patterns I can exploit. It proved helpful in this question !
Same proof as above!
Your 1st solution is easy and straightforward, but I like your discriminant method better.
Glad you enjoyed it , and thanks for the comment! :)
Well, that solution was over my head. I started with a mental graph with (0,0), and found that it worked. Also (0,1) and (1,0) worked. Anything beyond that starts to not work, but I didn't know how to prove it. After this video, I still don't know how to prove it 😉.
Perhaps if you could tell me which part of the solution you didn’t understand, then I could try and elaborate more ? :)
Namaste🙏.
:)
Infinite steps is not possible.
Exactly, it’s more like an approximation , if you let the number of steps tend to infinity, then you can get as close to a trisection as you want (within 1/2^n error) Thanks for commenting ! :)
Thanks..
You are welcome :)
YO MOMMA
Quack
Cute puzzle indeed! I needed some hints for G and K but it was fun.
Thanks, glad you have enjoyed it :D
Hey, I was myself curious about this question when I came across your channel. Though I didn’t really get a clear answer, as I was thinking about why is the principal of distribution is placed that way itself! Anyways I’m glad I found this video , it was fun to investigate from different viewpoints as to how and why it might be placed that way! Keep up the good work! 😄👍
Thanks, as for why the principle of distributivity holds, it has to do with Axioms of ring and field theory, but I won’t elaborate more here since I will just raise more questions ;) maybe it will be a video topic for next time Feel free to stick around and have a look at the other videos and thanks for your kind support again!
@@ProblemSolversInMaths ohhh i didn't know that thanks for the the info 😄
👍 great, solving integrals is so much fun 👍
czcams.com/video/2edpYfNsBa0/video.html
Right. hope human will be presist.
Thank you sir
Thank you
Beyond!!!!!
Ha and they say you can't learn anything from anime ha!
I love you
Amazing, we weebs are truly capable of anything.
What teachers expect us when going to war: **calculating size of m1 garand and calculating 400x50 german soldiers**
😎👍