Відео

@@ald6980 thx i had seen that done on unit circle, but had forgotten. I guess the general proof is straightforward?

Vieta's theorem for quadratic equations :)))) By the way - 2-nd order curve may has no rational points. For example x^2+y^2=3.

Not clear to me why that would follow from Vieta result. Will look at it when I get time. I constructed this hyperbola problem based on the “almost prime” constant 73*137

Just express one variable in terms of another. It would be a linear relationship. And substitute into your 2-nd order curve. You will get a quadratic equation with rational coefficients and one rational root. Second root = coeff/first root is rational.

The given information in the instructions circumvents the need to find a fourth degree polynomial and list potential rational zeros/roots.

@@theupson thanks for the different approach. Your usage of the term simple is gratuitous if not immodest. Binomial expansion appears to have been used to establish your recurrence relation where students would prefer to see the “simple manipulation.

Both approaches involved expanding x+y to the 5th and 3rd and substituting. I like this one because x and y are complex numbers with irrational coefficients. Raising them to 5th power is messy as far as I know. The method we used avoids having to directly solve for x and y.

@@als2cents679 nice way to do it that does not rely on that fifth power algebraic factoring identity used in video. In math contests, the competitors are sometimes looking for ways to speed up the computation due to the time constraints.

Your method relies on 20^5 and 70^5 having a "nice" common power of 10, right?

@@MyOneFiftiethOfADollar Yeah, just like your solution relies on the fact that both of them are to the power of 5, no?

@@als2cents679 what I meant to convey is any quotient in the form (x^5 + y^5/(x^4 -C + y^4) = x+y whether x and y have common factors or not. The way you did it is purely arithmetic. Try it your way for x=23 and y=74. I am interested in computational efficiency and appreciate your help and interest.

@@MyOneFiftiethOfADollar Yeah, it does not work as well for that case. What I would have done I guess is similar to what you have, viz. knowing that x^5 + y^5 is divisible by (x + y), do the long division to get the other number, then realize it is same as denominator, to get the answer.

@@aiueohehe the possible factors of 589 must sum to (n-x) + (n-y) + (n-z) = 3n -x -y - z.

@@MyOneFiftiethOfADollar will it only apply to natural numbers or will it apply to any condition?

@@aiueohehe it applies to all integers. The prime factorization of any natural number is the key to this process being effective. There may be some sort of analog problem type for real numbers. Would be worth exploring.

Conclusion?

Right!, My way was definitely banal algebra bashing. Thx for the more elegant practiced competitor solution. I still expect someone to claim it is obvious, by inspection, that x^18 is not present in the expansion of f(f(x))

Your line faintly reminds me of a fairy tale, but I don’t remember the title. Hard to believe it has anything to do with math 😃

@MyOneFiftiethOfADollar the connection is this: some have said and proven that 0.999... equals 1. Many people tend to believe that if something is unconventional, it must be incorrect, and thus, they assert that 0.999... =1, even though it's suggested to be incorrect by many mathemation. This resistance to unconventional ideas is not new; for many years, mathematics did not accept the existence of negative numbers, then complex numbers, and now hyperreal numbers. The moral of that story is that confidently repeating what most say can sometimes lead you to mistake a naked emperor for one who is fully clothed. 😉

Thanks for the detailed explanation. The irrational numbers were once unacceptable if not ostracized in Ancient Greece I believe. In fact, the term surd has a pejorative connotation I believe. I think the ancient Greeks were close to horrified when they had to consider the diagonal of the unit square could NOT be written as the ratio of two positive integers!

Glad you enjoyed it. Was the topic something any of your students might enjoy or more of a personal interest?

Thx, I wasn’t certain. Did you get same value as I did for largest Fibbish integer?

​@MyOneFiftiethOfADollar yup 112369

Just a hunch, but I doubt fibbish numbers will ever reach perfect number stardom.

31 is a prime divisor, but not the largest prime divisor. 33!-31!= 31!(32*33 - 1)=31!(5*211). So 211 is largest prime factor

What did you like about this "puzzle"?

@@MyOneFiftiethOfADollar 33!-31! looks easy at first glance, and hard at second glance, but it turneed out not to be hard after all. A real world application of 33! and 31! would have a division sign between them, not minus. I love puzzles that either look easy but is hard, or looks hard but is easy. This was both.

Thx, will try to find an instance of the difference of two factorials having a “real world” application. There is bound to be some type of combinatorics usage for the difference.

Quick? I noticed you had to edit your solution . How would you know to use your method without first knowing the remainder was 38?

​@@MyOneFiftiethOfADollar The edit was minor (adding a space after a minus sign for esthetics). On the other hand my comment is not intended as a full solution from scratch, it is just a shortcut for an already known solution. I think the factorization trick has its own interest.

@@mlerma54 I certainly think your approach is flawless for different instructions, namely “prove the remainder for this quotient is 38”. However, the problem in video is “Find the Remainder” Will try to compose a problem that uses your, what I would call, verification technique. Thx for your interest and the approach you shared.

I coded in assembly language long ago and have reverent respect for binary and hexadecimal. Love the thought of reducing any problem down to the most rudimentary number system known to man. I think they inscribed binary on an unmanned spacecraft believing binary to be the most “universal” way to count if other forms of life happened upon the craft!

Ah, interesting...yeah, lol, perhaps it would universal...aliens might, even if not recognizing 0 and 1, guess the two symbols might represent binary, lol, even if perhaps not in the direction (our horizontal right lower to left higher) they might use...

My understanding of why the scientists chose 0,1 is that the notion of night and day on earth might be the closest thing to something “universal “ given that about everything in the universe revolves around a source of light, but one could argue that light is just a waveform in the electromagnetic spectrum that helps humans see. Other forms of life might not depend on it the way we do 😂

I suppose so, lol...yeah, interesting...but I was talking about the symbols themselves, 1, 0, etc...they are not even universal to all humans, or at least were not...2000 years ago the Romans were using different symbols (I guess they didn't have one for 0, but still, lol)...so, what I mean is that aliens are very unlikely to be using the same symbols we use, etc...

@@archangecamilien1879 understood, but the usage of only two symbols irrespective of their appearance may be what is important in conveying on/off, is/ain’t, night/day, absence/presence, true/false blah blah. Even though, it’s highly improbable that strings of binary digits would mean anything to other life forms AND we haven’t even broached what it means to be alive 😂

That was my original attempt, but then I realized modular arithmetic was essentially “one step”. How many iterations was your long division? If you have time, will you show your work?

@@MyOneFiftiethOfADollar Four iterations. On the final step, all the powers of 2 cancel, leaving 37 minus -1.

I tried that way and messed up about half way through it. I'm not very neat when writing and was too lazy to track my error. Will try the long division again in the airport next time I get that inevitable late flight.

Thank you. I modeled it after a similar AMC problem or I would have spent a lot of time on it too! The lemma at beginning of problem is a big time saver.

You need to take responsibility for your own learning. It was explained sufficiently for the purposes of this problem. Phi function counts the number integers less than n relatively prime to n. So Phi(6)= 2 since 1 and 5 are the only two integers less than 6 which are relatively prime to 6. You will be better at math if you can read math textbooks rather than expecting to be spoon fed by YTube videos.

20^20 = (2^80)(5^20) which has 81x21 divisors >> 720. 18^18 = (2^18)(3^36) which has19x37=703divisors < 720. Oh I see! Is 20 the smallest you could find? I was just using single occurrences of two primes. My mistake. thanks! What do you mean by prime signatures? prime factorization.

I get 20^20 = 2^40 * 5^20 so is has 41*21 = 861 divisors. And 20 is the smallest positive integer with that property. Prime signatures are found by listing the exponents with multiplicity in a prime factorization. So primes have signature (1) an 20 has signature (2,1). But maybe 20 is small enough to check positive integers until 20 and don't use prime signatures.

Maybe also see OEIS sequence A062319 to verify n = 20

@@DavidCorneth right another mistake on my end. Thx for explanation regarding signatures. I had heard of omega function which just counts the distinct primes in in the prime factorization . It seems like it might be related to this……

You're welcome! Maybe omega could be used for this, like one could know for any number k with omega(k) >= 3 we have k^k has more than 720 divisors. But it might get tricky when it produces false negatives.

Your n has 16 divisors only if k=1. Your answer would hold if we dropped the requirement that n possess 16 divisors. I wanted a unique answer and I think 2744 = (2^3)(7^3) is only answer. Let me know if you find another.

@@MyOneFiftiethOfADollar True. I misinterpreted the "16 divisors" at some point so k=1 is the only solution.

Thanks for reporting that. It should be 2.414 since sqrt(2) is approximately 1.414. So answer of 483 is correct. 2.14 is a fingerfehler on my part😀 Glad you liked the linear algebra explanation of Cauchy Schwarz inequality

The a,b,c,d were specified as distinct in the instructions or what you wrote would lead to additional solutions. I don’t think I missed any others, but possibly.

@@MyOneFiftiethOfADollar I think these are the solutions: m = 18 via (18 - 19) * (18 - 29) * (18 - 7) * (18 - 5) m = 42 via (42 - 43) * (42 - 41) * (42 - 55) * (42 - -79) m = 48 via (48 - 49) * (48 - 47) * (48 - 59) * (48 - -95) m = -12 via (-12 - -11) * (-12 - -13) * (-12 - -25) * (-12 - 109) m = -18 via (-18 - -17) * (-18 - -19) * (-18 - -29) * (-18 - 125) m = 12 via (12 - 11) * (12 - 23) * (12 - 1) * (12 - 25)

Thx for this. Use your methodology for a+b+c+d+e=9 and the constant on right hand side is 2009. This case is 5th degree rather than 4rth degree that you just solved.

You mean a question that settled ties in competitions?

@MyOneFiftiethOfADollar yes, in the national math team competition in the United States. It had a bunch of logs inside each other and you had to find the domain.

@@justpotatoit thx much for the feedback and best of luck to you in the competition. Were you selected by your school as having the aptitude required to compete at that high level?

@@MyOneFiftiethOfADollar yep, thanks

Whenever time permits, will you share one of the tougher problems you have seen in the national US math team competition?

This is interesting since it involves finding the prime divisors of the smaller number 10^n + 1 where n=3 in this case. Stylistically I would have written 7x11x13 = 1001 after I had discovered that it was useful.

@@MyOneFiftiethOfADollar thank you..This is the method taught to my grand child who is in grade 5.

Great that you are teaching your grandchild arithmetic/mathematics. He/she will remember how you helped when they secure that big salary high tech job! Only drawback to the method you presented is finding the prime divisors without a calculator for expressions of the form 10^n + 1 when n is much larger. With other method, one can just continue multiplying last digit by 4 and adding to what remains until what remains is small enough for the easy computation.

@@MyOneFiftiethOfADollar thank you.Already I taught her tests of divisibility.I gave combined test for 7 and 13( same test).I am going to her level and teaching.She wrote state level pre algebra test and qualified in USA.

I may not have mentioned that a relatively common competition problem is to solve without calculus. Some who uses calculus aren’t even certain why they set derivative = 0 to find mins and maxes. Like you said completing the square is the “obvious” way after the substitution so one can find the vertex. This approach is closer to “first principles” than calculus. There is a way to use Cauchy Schwartz if you think about the linear algebra product of the magnitude of two vectors approach. see if you can figure it out. thx for viewing.

The given trigonometric expression has a vertex with a y coordinate of 11 and 11/50 which means the expression is > 11 for all x. Instructions asked for largest integer that is less than the expression. Since 11 is the closest integer to 11 and 11/50, 11 is the answer. This type of problem is not uncommon in math competitions. It is quite similar to recent Singapore Math Olympiad problem. That may explain why it appears “odd” to you. Hope this clears things up. I think you may have been evaluating the expression at integers only, but it is the sum of two continuous sinusoids defined for all reals.

@@MyOneFiftiethOfADollar thanks for replying. Yes, makes sense once I re-read it and knowing it is a well-known type of question. Was good to look at the problem anyway and see both solutions - the expected way and my way, even if my way is not correct! Cheers PS. The big giveaway is the "for all x" symbol, implying that it's not a maximum but in fact the entire function.

Right the universal(For all) and existential quantifiers can mean everything with respect to problem interpretation. I make many mistakes still in that regard. There is a large class of "no calculus" problems that I like a lot and this was intended to be of that ilk. Quite a few hurried students use calculus without understanding the motivation for the steps they are taking. Are you a math hobbyist or do you use it your profession? thx for your interest

It may make it seem that way to you. How would you interpret f(x)^2 ? Based on your comment, you would interpret f(x)^2 as f(x^2) which is certainly mistaken. Hope this helps.

writing it as sin^4(x) removes ambiguity, since some people write "sin x" instead of "sin(x)" and "(x)^2" instead of "x^2". If we would all agree to always put brackets around the function input ("sin(x)"), your argumentation works. However, some people don't. And sin^4(x) is unambigously interpreted as (sin(x))^4 which is IMO the perfect syntax, since some absurd people write f^3(x) to mean f(f(f(x)))

@@noel.friedrich I guess it's how u have seen things from your school vs how I have seen them I from my school. If u bring this with any other Indian student u will get a response like mine. But I do agree with you on how you say the actual representation must be (f(x)) ^y it's most clear expressions of all the above mentioned ways.

@@noel.friedrich the trig functions and the logarithm functions are the only culprits that come to mind which omit parentheses with respect to function input argument. Would love to know historically how (sin(x))^2 morphed to sin^2x ?! I do understand that practiced IMO competitors do not struggle at all with potentially cryptic notation such as sin^nx. It saves them time. However beginning trig students should write (sin(x))^n in lieu of sin^nx to reinforce the truth they are manipulating functions. For how long I don’t know…..

Tic Tac Toe is an easy "theoretical draw". Have you ever done any coding for checkers or chess? I would expect checkers to be a draw also, but not sure about chess since there are so many more possible chess games. Some think white having the first move allows for an. initiative that will lead to a win. I doubt that though. Close to unprovable. I think Backgammon programs play perfectly since it is all about probability and counting which is fairly easy to code.

Right, that is essentially that same idea as the AM-GM inequality where it can be shown that (x+y)/2 >= sqrt(xy) where x=15 and y=14

@@MyOneFiftiethOfADollar Oh I didnt even notice that. This will be useful for my math compitions! Thanks so much!

What level of math competition are you competing in? Glad this might be help to you.

@@MyOneFiftiethOfADollar I’m doing amc 10 targeting aime

2^12 does divide the specified integer, but it does not divide it fully. 2^13 also divides the specified integer and the video shows that 2^13 is largest power of 2 that does.

Thanks for the alternate proof. I could ask you to prove x^2 is increasing, but you could ask me to prove the AM-GM inequality which is simple for two variable case, starting with (sqrt(x) + sqrt(y))^2 >=0

You are assuming what you are trying to prove the moment you square the given. Directly proving it is the principled approach imo. Many trig students commit this same logical fallacy when trying to verify identities.

@@MyOneFiftiethOfADollar What? You're only 'assuming' that squaring two positive numbers does not affect the ">" which is true. Not sure what else you mean - can you link the stackexchange post?

@@patrickthompson6885 i only remember it was posted by a user named fleablood. Mentioning the squaring function is increasing on interval [0, inf) should be stated in any proof that begins with squaring the very inequality that one is trying to prove. Equivalent inequalities are created via squaring and square rooting. So yes it’s a proof if that is stated, but is the “easy way out” 😀

You are mistaken again. One can refer to the roots of an expression without explicitly setting expression = 0.

@@MyOneFiftiethOfADollar we use zeroes of expression and roots of equation.Different countries use different wording.For example in some countries 1 is called prime number.But we don't.

@@SrisailamNavuluri I understand. So you are suggesting the whole world adopt the conventions of your country, India I gather? 😀 The mere mention of the term root implies the algebraic expression is being set = 0.

@@SrisailamNavuluri also 1 is not considered prime in most of the world since the prime factorization would not be unique otherwise. For example, 14=2x7 = 14=2x7x1=2x7x1x1…..

@SrisailamNavuluri you can “feel” any way you want 😃, but it is not uncommon to reference the roots of an expression or a function. For example, the roots/zeroes of x^2 + 7x + 12 are -4 and -3. What is a "root"? A root is a value for which a given function equals zero. When that function is plotted on a graph, the roots are points where the function crosses the x-axis. So the roots of f(x) are those values of r such that f(r)=0. Put another way, the definition of root implies the expression/function is being equated to zero. Language is our servant, not our master. I don’t understand why this is difficult for you to grasp? In any event, thank you for your genuine interest in this topic.

I have been doing this almost 3 years and you are the second to mention 1/50 of a dollar is 2 cents.

@@MyOneFiftiethOfADollar 😂 My mathematical contribution might actually be worth even less than that. ... 1+1 = 2 ... beyond that I need a calculator.

You are likely a much more well balanced human being than me. I spend an inordinate amount of time thinking about math. At least it is a healthier addiction than alcohol and most drugs. I hope??!!!

The divisibility rule for 7 is similar, but not the same. You subtract 2 times last digit from what remains for the 7 rule You add 4 times the last digit to what remains for the 13 rule.

@@MyOneFiftiethOfADollar 2+7=9 9+7=16 You can multiply last digit by 2 or 9 or 16 to subtract from the remaining part for divisibility test for 7 Try 284256 which is divisible by 7. Take 284256 28425-6×16=28329 2832-9×16=2688 268-8×16=268=128=140=7×20.

I encourage you to produce videos on your own channel. Can you prove anything you just stated in your last comment OR is it just something you have memorized?

@@MyOneFiftiethOfADollar you told to subtract multiple of 2 from the remaining part.I used 16 and did the same for divisibility test for 7.You try with multiple of 23.It is valid.

@@MyOneFiftiethOfADollar you try with multiple of 22 for divisibility test for 13 instead of 9. In the same way we can give divisibility test for 17,19,23,29 etc

Right, there is a limit to how large the constants should be for this to be a fair contest problem. There is a modulo 10 condition for perfect squares stating any perfect square is 0,1,4,9,6,5 mod 10 meaning if a integer's last digit is 2,3,7,8 then is not a perfect square which is a little bit of an aid I guess. As far as substituting 7 back into the sextic rational function, I am sort of on the fence about checking to see if it evaluates to an integer. If it had not in this case, I would have ended up in psych ward somewhere?!?!? I think even on IMO level competition problems, they rarely go past the squares of 2 digit number, but not certain.

@@MyOneFiftiethOfADollar Yes, sure. BTW, I was just comparing how mathematicians do arithmetic calculations vs other people. Sure, there's no need to evaluate 49^2 - 49x7 + 2. However, plugging it into the remainder after long division just shows that no careless mistake crept into the algebraic and arithmetic calculations. Alternatively, we can just re-check those calculations. Plugging 7 into the denominator of the initial fraction (before long division) would of course be very stupid.

Sure, it is an interesting comparison. I would like to know if there is an approach to this problem that does NOT involve long division? Was thinking about possible modular arithmetic schemes, but have not hit on anything yet.

@@MyOneFiftiethOfADollar Why? What's wrong with long division? It should be a mandatory skill for everyone in my opinion, and it's extremely simple and short in this problem. It's definitely the best way by far to solve this problem. If we were to extract the square root of 2 on paper, say, with 12 digits of precision, the long division would be a hundred times more difficult and tedious but still easy as it is only a square root, not a fifth root or sin 20.

Did not mean to intimate there was anything wrong with long division which certainly appears to be the best way to reduce computations for this particular problem, but am genuinely curious if there is another computation reducing scheme we can try that excludes long division.

17/6 = 187/66 is a solution. Been awhile since I did that video. If I wrote 17/16, then probably just a typo. Thx for letting me know.

@@MyOneFiftiethOfADollar Yeah, sorry. I posted that too quickly failing tor realize that 187/66 = 17/6. My brain is glitching sometimes.

@@billmorrigan386 no worries, good to hear from you. Have you been spending much of your idle time doing math?

@@MyOneFiftiethOfADollar No, I'm engaged in other things. Actually, all my life I've been engaged in other things, even in childhood. The upshot I don't even have a PhD in it.

@@billmorrigan386 having a PHD is way overrated nowadays(easier to get one online than it should be) Amateur(labor of love) types frequently know more about specific topics than PHDs who crammed a lot into their brains in a relatively short duration. I have a masters in math, but was lured away from earning a PHD by the lure of big bucks in the “real world” It’s great and mentally healthy that you have varied interests!

Right. Thanks for that correction.

Yep, if this problem was one given in a math competition, the graders might not deduct points if test taker left answer in the form 44*46. The algebraic simplification is the heart of the problem since it reduces computational difficulties.

@@MyOneFiftiethOfADollar true. I just thought that using (a+b)(a-b) here is clever.

Yes definitely. The difference of two squares identity is the basis for a ton of contest problems. For example Simplify sqrt(83^2 - 34^2) is solvable directly without evaluating 83^2 and 34^2

I thought about Newton sums too, but the algebra yielded a^2 + b^2 + c^2 so conveniently, I went with it. Newton sums probably better idea if one has an expression to evaluate with exponents greater than 3. If you have time, will you provide some details about the recurrence approach you mention?

@@MyOneFiftiethOfADollar ohh !! sorry for the late reply , busy with the exaams rn , but just to clarify the recurrence and newton sums is the same approach!! its just like name of the particular recurrence, nd yes since 3 is quite small the algebra bash you did is more elegant here. i can send you newton/recurrence approach but where? email?

@Singularity007 thanks, no worries life is busy isn’t it! I had already read up on how recurrence related to this so no need to send me anything. I think email is listed somewhere on home page though if you have topics of interest to share. Is any of your course work significantly proof based?

It is quite easy to do mentally if you know the fractional part is just a shifted line of slope 1 which creates a triangle. So no need to even find an anti derivative.

That must be devastating for you 😢 So many video games to be played and porn to be watched.

@@MyOneFiftiethOfADollar i didn't mean to hurt you. I genuinely like your content and I have you on bell. Just told you what could be better, because I wish you well. Don't take that stuff personally.

It was a joke aimed at your usage of the word “forever” for video that lasts maybe 6 minutes 😀 Many who have never seen the subject matter need a slower more deliberate pace. A major complaint students have for some college math professors is how fast they rush through the lessons. So I wouldn’t write any faster even if I was using platform where I could. I may start saving the video at 1.5x or even 2x as a novelty so thanks for letting me know.

Thanks, that’s easier than my approach. I Need to find an initial condition that makes problem more challenging 🙄

Good to hear from you! So you don’t like interval notation? Appealing to intermediate value theorem a bit in the direction of overkill in my view, but certainly instructional to point out floor(x)ceil(x) is continuous everywhere except x= an integer. That came up on later video where definite integral had an integrand of floor(x)ceil(x){x} where {x} is fractional part function. Are you doing any videos yourself these days?

@@MyOneFiftiethOfADollar Nice to see another of your videos. I actually really like all sorts of set notation and keep the symbol ∈ (along with a few others) on a Notepad++ tab that I can quickly open to copy-paste into comments. I think I have to use intermediate value theorem if I want a bit more rigor in demonstrating that there has to be a value of x that makes x^2 equal to f*c somewhere near the middle of the range where x ∈ [f, c]. Although I properly should also appeal to the monotonically increasing nature of x^2 to prove that there is only one solution in each range. As usual, I find myself doing too many other things to find a large chunk of uninterrupted time that I know I would need to ever contemplate doing a video. Maybe next week when I've finished campaigning in the local elections ...

As a candidate or as an advocate of another candidate? I don't envy what you are doing as politics/elections have become so polarized/vitriolic/tribalistic all over the world. If you follow what is happening in the US at some of the large campus colleges, you realize how universities have become inculcation centers for political ideology. They always have been to some extent, but it is far more extreme nowadays in my opinion.

Formally, one could not contest your choice to use IVT. However, floor(x)ceil(x) = i(i+1) in the interval [i, i+1) where i is an integer. So you either have y=x^2 intersecting horizontal line y=i(i+1) (count 17) or vertical line x=i (count 18) IVT is more useful in the many cases where a convenient geometric interpretation is not available or easy to determine😀

Thanks, sometimes appealing directly to definitions is not sufficient, but in this case it is as you noted. How long have you been teaching and at what level?

Very true, but it’s often a good starting point. I’ve been teaching lower-level college math (college algebra, precalc, calc) for about 5 years

I’ve done a lot already which you can locate by searching Home Screen.