Math Olympiad 3^m-2^m=65 | Math Olympiad Problems | Algebra

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  • Опубликовано: 21 дек 2024

Комментарии • 3,3 тыс.

  • @tiehoteele874
    @tiehoteele874 Год назад +120

    The comment section is polluted by critics but i learned some tricks from this video.. EXCELLENT

    • @onlineMathsTV
      @onlineMathsTV  Год назад +16

      Wow!!! Thanks and we are glad you gained some values from this video tutorial sir.

    • @thefireyphoenix
      @thefireyphoenix 8 месяцев назад +2

      it isnt critism.. in every maths channel you will see people doing that.. they do it for alternate solutions and writing down their own way

    • @tiehoteele874
      @tiehoteele874 7 месяцев назад +2

      @@thefireyphoenix this video wasn't made for them

    • @Theuserisadick
      @Theuserisadick 7 месяцев назад

      Polluted by critics??
      You're definitely not a student of maths

    • @Maths_withs_NHLaskar
      @Maths_withs_NHLaskar 6 месяцев назад +2

      I learnt something new from this video... And I wanna say something to dear sir please do you job not listen other criticism... Love you sor i am from India ( Bengali) i am also a Maths teacher 🧡🤍💚

  • @danieldavies1829
    @danieldavies1829 Год назад +166

    I worked it out a bit different. My solution was simply determine what of 3 exponent would get me a number greater than 65 that would be an odd number (3^4). I then subtracted that 65 from that number (81) and I got 16 which is 2^4.
    In other words you can rewrite the equation in this instance as
    3^m - 2^m=65
    3^m - 65 = 2^m
    The first exponent of 3 which results in a number greater than 65 is 4
    so 3^4 = 81
    81-65 = 2^m
    16 = 2^m
    16 can be written as 2^4
    m=4

    • @Xhopp3r
      @Xhopp3r Год назад +6

      That's exactly what I did.

    • @Smith_14
      @Smith_14 Год назад +8

      m can be a negative number?

    • @leishajuneja2994
      @leishajuneja2994 Год назад +8

      You cant take any value of 'm'.consider the question is same but with a very large value(instead of 65),probably in crores,it wout take an eternity to reach that number

    • @leoosu
      @leoosu Год назад +4

      I did the same😂

    • @DownhillAllTheWay
      @DownhillAllTheWay Год назад +9

      You're assuming that m is an integer?

  • @alfredomulleretxeberria4239
    @alfredomulleretxeberria4239 Год назад +74

    I was impressed by the analytic demonstration used to figure out that m = 4. Sometimes procedures can be more interesting to follow along than just knowing the result.

    • @italixgaming915
      @italixgaming915 Год назад +5

      Well, you'll be impressed to see that the proof is not valid. If you suppose that m is an even number then x+y and x-y are integers but if it's an odd number then they are IRRATIONALS and you can't use 65=5*13.
      Plus the fact that even in the case where x+y and x-y are integers, 5*13 is not the only way to get 65, you must also look at 1*65...

    • @jarikosonen4079
      @jarikosonen4079 11 месяцев назад

      ​​​​​@@italixgaming915This looks like diophantine method used here, which would work only with the integers.
      In case the solution works like in above case it could prove that no other integer solutions exist... But it seems 65x1 was not checked.
      This could work for 3^m-2^n cases also. Variables 'm' and also 'n' are often used for integers, but not necessarily always.

    • @claudiohermeslima
      @claudiohermeslima 10 месяцев назад +1

      You are the top!😊 You are a great teacher!!!!

    • @elmehdiazzouz7888
      @elmehdiazzouz7888 9 месяцев назад +2

      the proof is not valid, we need to proceed much more cautiously with analytic ways, i d say : some more conditions/discussions needed to be added to the video .... i gree with italixgaming
      the arithmetic way stays safer ..

    • @matsen2c5-1
      @matsen2c5-1 2 месяца назад

      Support

  • @rubikaz
    @rubikaz Год назад +133

    There is a problem here, when you write (x+y)(x-y)=5x13 you can not deduce that x+y=13 because you do not know if x+y is a natural number. If m is an odd number, then x=3^(m/2) and y=2^(m/2) are not natural numbers. So you have prove that if m is an even number, then m=4. it is very easy to check that there is only 1 solution.

    • @onlineMathsTV
      @onlineMathsTV  Год назад +11

      Noted.

    • @nicadi2005
      @nicadi2005 11 месяцев назад +4

      @rubikaz "So you have prove that if m is an even number" - Or, you can assume that m is even (thus making the quantities x+y and x-y (positive) integers etc.) and see whether it pays off - which it does, actually.
      "it is very easy to check that there is only 1 solution." - Indeed. The uniqueness of the solution is a direct consequence of the properties of the exponential function at work here...

    • @babetesfaye1001
      @babetesfaye1001 10 месяцев назад +4

      ❤❤❤
      May I use any number Which have same numerator band denominator ? like 3/3 , 4/4 , 5/5....

    • @with.love.from.siberia
      @with.love.from.siberia 10 месяцев назад +1

      ​@@babetesfaye1001да, потому что оно равно 1

    • @Quasar900
      @Quasar900 10 месяцев назад +5

      @@babetesfaye1001 the function f(x) = 3^x - 2^x where x>0 is strictly growing, therefore with x=6 , f(6) > 65 so x must be less than 6, and so on trying integers until finding x=4 or m=4

  • @georiashang1120
    @georiashang1120 Год назад +22

    65=13•5=(9+4)×(9-4)=9^2-4^2=3^4-2^4
    4 is the m;
    My high school math teacher used to tell me,the easiest way to understand an equation is to make them look the same,that is to say,we should make the brief side more complicated other than simplifing the complicated side for the most of the time.

  • @samuelmayna
    @samuelmayna Год назад +38

    You can also factor 65 into 65 and 1. This gives values of a and b as 33 and 32 hence b=2^(m/2)=32, m= 10 but m will have different value for 3^(m/2)= 33. Using logs(or ln) m=(log 33/log 3)=6.365

    • @eliasgitau7353
      @eliasgitau7353 Год назад +2

      This makes sense and it is mathematically correct

    • @nicadi2005
      @nicadi2005 Год назад +3

      @samuelmayna "You can also factor 65 into 65 and 1." - Yes, you can, but these won't be proper factors for 65, in the sense that ANY NUMBER could be "factored" as itself and one... Also, as you've seen yourself, this breaks the consistency of the original equation by forcing the unknown to take different values simultaneously - which is obviously not possible.

    • @AtefFarrouki
      @AtefFarrouki Год назад

      m is an integer

    • @samuelmayna
      @samuelmayna Год назад +4

      ​@@nicadi2005 but it is mathematically logical.Mathematics is about thinking all cases.

    • @samuelmayna
      @samuelmayna Год назад

      ​@@AtefFarrouki but my approach is sound which shows that 65 and 1 won't work but it can give different answers for some equations.

  • @nicholastergech8525
    @nicholastergech8525 Год назад +139

    Similarly you can as well re-write 65 as 81-16....From there you make the bases of the two numbers to be similar with what you have on the left hand side..From there you take one of the corresponding bases and equate them together,when bases are the same powers will also be the same.

    • @usmanmusa8028
      @usmanmusa8028 Год назад +2

      This is what I actually expected from him

    • @TomJones-tx7pb
      @TomJones-tx7pb Год назад +3

      If you do not notice that 16 = 2**4, you are not a computer geek.

    • @lgmoses3876
      @lgmoses3876 Год назад +2

      I did it,in my braine.

    • @sfqamd
      @sfqamd Год назад

      ruclips.net/video/sIlRHg730CA/видео.htmlsi=ujk2KD_xjoMGqiP5

    • @shivaprasadmallikarjunaiah3751
      @shivaprasadmallikarjunaiah3751 Год назад +20

      you are not mathematically solving the problem, but doing so by trial and error. These were smaller numbers so it is easy for anyone to come to that conclusion ( becausethe solution is "visible" in the numbers in front of you). In other words, how would you solve the same problem with entirely different and larger numbers involved? ...say 2059 for instance.

  • @f-s406
    @f-s406 Год назад +113

    I got ‘m=4’ by mental arithmetic. Because 3^m > 65 and 2^m < 3^m, that’s necessary.
    The value '65' determines that the value range of m must be less than 5 and greater than 0. When m is a positive integer, test the m=5, 4, 3, 2, 1 and finally get m=4.

    • @CBSE24
      @CBSE24 Год назад +3

      I too

    • @tuyu6404
      @tuyu6404 Год назад +2

      Since we don't participate in the Olympics, this short answer is good. But I'm not sure if I would use this method for an Olympics. I think logarithms are simpler than the answer And have a good axiom to condense those complicated answers.❤

    • @lophocthienuc7345
      @lophocthienuc7345 Год назад +2

      I dont think so. 2^m 65 why m

    • @gheorgheneacsu3356
      @gheorgheneacsu3356 10 месяцев назад +2

      Me too! 😊😊

    • @chrissyday67
      @chrissyday67 8 месяцев назад +2

      exactly! if students are good enough to do olympiad problems then they'd know powers of 3, 3, 9, 27, 81, 243 at least and also 2 to even higher powrers, 2, 4 , 8, 16, 32 etc so very easy to work out in less than 20 seconds . However I'm used to doing mental arithmatic as I never had a calculater when I was in school

  • @sofianikolidou5640
    @sofianikolidou5640 2 месяца назад

    Ευχαριστώ πολυ γι αυτες τις λυσεις...ειμαι 65 ετων και μου αρέσει να θυμαμαι τα χρόνια μου στο σχολείο...Παντα μου αρεσαν τα μαθηματικα και τωρα τα θυμάμαι ολα και οτι δεν θυμαμαι το μαθαινω μαζι σας thank you❤

  • @sanmus100
    @sanmus100 Год назад +774

    How can one just assume that x+y = 13, and x-y = 5, respectfully, as there's 65 and 1 as well. Furthermore, this wouldn't really work if 65 had many more factors, making it more complicated, opening up to a lot more possibilities.

    • @onlineMathsTV
      @onlineMathsTV  Год назад +198

      Nice question @Santhosh John. There are principles and rules that govern the operations in mathematics as it is for all other sphere of life.
      Once you are a maths student or tutor you must get urself familiarized with these rules.
      They become part of you and you know what to do once you have a math challenge/problem before you.
      Once a mathematician sees a math problem, his head automatically runs through different means of approaching the problem for a better solution.

    • @thegreathussar9442
      @thegreathussar9442 Год назад +40

      When taking difference of squares if smaller part is 1(assuming it is the smaller part and it is) , m=2 and positive part becomes 5, not 65 therefore the result will be 5, not 65.
      If it has more factors, you just have to make arithmetical inferences and simplify it. That's it.

    • @naharmath
      @naharmath Год назад +103

      3^(m/2) is not necesserly an integer!

    • @thegreathussar9442
      @thegreathussar9442 Год назад +21

      @@naharmath but there is no other solution since both parts are exponential and even if m is a rational number the result wouldn't be integer( they are different primes). So this equation requires to be analyzed numerically first

    • @mariosantangelo9929
      @mariosantangelo9929 Год назад +25

      Il professore ti ha risposto in maniera adeguata. Però io consiglio al professore di spiegare certe regole anche se ciò richiede qualche minuto in più. I fruitori di you tube non sono tutti matematici, ma persone desiderose di capire ed imparare.

  • @НиколайЧуприк-ъ4с
    @НиколайЧуприк-ъ4с Год назад +54

    It doesn't work if m is odd. In this case (x+y) and (x-y) aren't integer and can't be assumed as 5x13.

    • @danielrivera2278
      @danielrivera2278 Год назад +6

      If you do the analysis, and supposing m is an integer, you can conclude m is even.
      That's because 3^m-2^m must be congruent to 0 mod (5). If m is even you get 1 mod 5 or 4 mod 5. But m being even, you have 0 mod 5 always.
      Anyways, it's not proven in the video, maybe it would be amazing to have hows and whys in the video

    • @НиколайЧуприк-ъ4с
      @НиколайЧуприк-ъ4с Год назад +2

      @@danielrivera2278 of cause. I mean that "trick" in the video is not universal.

    • @Change_Verification
      @Change_Verification Год назад +4

      @@danielrivera2278 and who even said that m must be an integer ?

    • @danielrivera2278
      @danielrivera2278 Год назад

      @@Change_Verification exactly. I also think like that, that's because assuming integer was the first thing I said

    • @Eismann1
      @Eismann1 Год назад

      Yes, he didn't do the preliminary work. But this is an important step to make tricks like this one work.

  • @Sapped6
    @Sapped6 Год назад +9

    A matemática é uma língua universal como a música. Parabéns, ótima técnica ❤

  • @airtonreis2675
    @airtonreis2675 Год назад +33

    Parabéns. Não sei falar nada em inglês e mesmo assim consegui aprender com sua aula. Até eu estou surpreso de ter assistido sua aula até o final sem saber se iria entender o seu modo de esnsinar. A matemática pode ser universal, mas o jeito de ensinar é fundamental.

    • @onlineMathsTV
      @onlineMathsTV  Год назад +12

      @Airtonreis, we want to sincerely say you are such a wonderful person and thanks a million for watching our contents despite the language barrier.
      We all @OnlineMathstv deeply cherish and love you from the depth of our hearts sir. ❤️❤️❤️💖💖💖💕💕💕🙋🙋🙋

    • @airtonreis2675
      @airtonreis2675 Год назад +1

      @@onlineMathsTV 🤝

    • @MATHSHEADMASTER
      @MATHSHEADMASTER Год назад +1

      You are welcome to subscribe to our channel for more interesting math problems.

    • @pedrooo13
      @pedrooo13 10 месяцев назад

      Brasil tá em todo lugar não tem jeitoooooo

  • @mitahaubica6498
    @mitahaubica6498 Год назад +159

    I immediately saw that 65 can be decomposed as 81-16, and conveniently 81 is 3^4 and 16 is 2^4, so matching coefficients suggests m is 4.

    • @BrightonMutero
      @BrightonMutero Год назад +3

      You are a genius

    • @nicadi2005
      @nicadi2005 Год назад +18

      @mitahaubica6498 "I immediately saw that 65 can be decomposed as 81-16" - That's not decomposition... You can find an infinity of pairs of numbers that have their difference equal to 65.
      The fact that you selected one such pair that also happens to be powers of the respective bases in the original problem merely indicates you have approached solving this by trial and error...
      *The question would be whether you can do better than finding the solution that way...*

    • @counterpoint9260
      @counterpoint9260 Год назад +3

      that is not the right method..may work here but not al the time

    • @davidmajor1508
      @davidmajor1508 10 месяцев назад +4

      You just got lucky. It was pure luck that you used the right numbers to subtract, and that m is an integer in this case.

    • @danielvazquez6301
      @danielvazquez6301 8 месяцев назад +5

      Always try some values of m to see the behaviour.
      m=0 or 1 or 3 or 4... I've found the solution!
      Obviously it is not an Olympiad problem.

  • @karlm9584
    @karlm9584 Год назад +2

    I like how you used indices to get past that first section. I've never really been good at spotting where to use substitution, the section when you brought in x and y

    • @onlineMathsTV
      @onlineMathsTV  Год назад +1

      Hahahaha....thanks a bunch my good friend and thanks for watching our contents consistently.
      We all here love deeply....💖💖💕💕😍😍

  • @Amy-601
    @Amy-601 Год назад +13

    The way I see it, 3 cubed is 27, less than 65, and 3 raised to 4 is 81. Therefore 65 is between 27 and 81. Upper bound, lower bound or range even. Now 81 would mean m is 4. So 2 should be raised to 4 also which gives us 16. 81 minus ➖ 16 is 65. So m is 4. The other ways are using log or binomial series which is overkill for smaller numbers. - Amy

  • @Bossudeboss898
    @Bossudeboss898 7 месяцев назад +6

    Rewrite the equation as 3^x = 65 + 2^x. We can safely divide by 3^x and then we have that 1 = 65 * (1/3)^x + (2/3)^x. Because the function on the R.H.S. is strictly decreasing, being the sum of 2 other strictly decreasing functions, it means that f(x) = 1 has one solution at max. We notice that x=4 checks, so that is our only solution.

    • @onlineMathsTV
      @onlineMathsTV  7 месяцев назад +2

      Bravo!!!
      You the best sir.
      Maximum respect sir 🙏🙏🙏

  • @dannychown2593
    @dannychown2593 7 месяцев назад

    Your work opens up the horizon of my mind how to approach this type of question. Thank you.

  • @elio9008
    @elio9008 Год назад +48

    If you guess the solution, m=4 and present the equality as 3^m - 81 = 2^m - 16 , then it would be easy to prove that both functions (on the right and on the left) increase and therefore their graphs have only one intersection.

    • @ca1498
      @ca1498 Год назад +2

      You need more than that. Two increasing functions can intertwine and cross each other all the time. But if they have one intersection, and after that one of them grows faster than the other all the time, then it follows that they won't intersect again. It's like two cars racing. They both increase their distance from the start all the time, but they could swap places many times during the race--unless one of them is always faster than the other after the point in which they were even with each other.

    • @ca1498
      @ca1498 Год назад

      @@reginaldocalvo4361Each of the two sides can be a function e.g. y = 3^x - 81. The solution of the equation 'side 1' = 'side 2' is a number x (or m) where the two functions give the same y for the same x. You first find, by guessing, one x (or m) for which the two functions have the same y, which would mean that the particular m is one solution to the equation of function 1 = function 2 for some x. You then show that each of the functions only grows, and that the difference between the two y-s for each x (which is a function of x as well) also only grows. Therefore there can be only one value of x for which the difference is 0, so only one m (the one we already guessed) is the solution to the equation where one of the functions has the same y as the other for a given x.
      If I were participating in this Olympiad and solving this problem, I would have guessed 4, and then I would have argued that the first side 3^m... grows much faster than the other side 2^m for each subsequent m. I wouldn't be using derivatives, as I did not know any calculus in high school. But if they ask about integer solutions, I would talk about growth of y with respect to changes in m by +1. And if they did not limit it to whole numbers, I would probably still try to talk about slope of the graphs of the functions and hope to make an acceptable argument, as I don't see how you can analyze these functions without using calculus.

    • @MeWa-z4g
      @MeWa-z4g Год назад

      @@reginaldocalvo4361
      3^m-81= f(m) this can be considered as a function depending on m
      2^m-16= g(m) also could be considered as function of m.

    • @MeWa-z4g
      @MeWa-z4g Год назад

      @@ca1498 you are right, but
      When you see this function it is easy to notice that the growing path or direction is already know but partially

    • @Vitzyk
      @Vitzyk Год назад

      Consequence is false.Contra-example x and x^3. Both increase but have 2 intersections

  • @jaimeduncan6167
    @jaimeduncan6167 Год назад +24

    Another option: the solution is pretty clear, it's a small number one can calculate with the mind. Then it reduces to show that their solution is unique. One can use calculus to show that the analog continuos function is monotone for x>4 and be done with it, or use induction to show that it grows on the integers for n>4.

    • @onlineMathsTV
      @onlineMathsTV  Год назад +4

      Your approach is superb sir. I find it very fascinating and I will try it out in subsequent math challenges.
      Thanks for sharing this nice and wonderful procedure with OnlinemathsTV.
      You the boss and much respect boss...👍👍👍

    • @GirGir183
      @GirGir183 Год назад

      I think he used a simplified equation for the demonstration. When it's not so simple and the numbers are much larger, then this method can be used as well.

  • @ludmilak9396
    @ludmilak9396 8 месяцев назад

    Очень грамотное изложение, и очень удобно следить на доске. Однозначно плюс!🎉

  • @Quasar900
    @Quasar900 Год назад +5

    the function f(x) = 3^x - 2^x where x>0 is strictly growing, there for if x=6 f(x) > 65 so x must be

    • @onbored9627
      @onbored9627 11 месяцев назад +1

      All I did was think whats the first interger power of 3 that goes past 65, 3 is 27 so the answer is 4. Then just test. and it gave right answer. simple.

    • @Quasar900
      @Quasar900 11 месяцев назад

      @@onbored9627 Where are you from ? Cause how did you know I was still alive after 4 months ? 🙂Here is a secret about me : I 've never studied Mathematics in English !

    • @onbored9627
      @onbored9627 11 месяцев назад

      @@Quasar900 Ah, I apologize I only speak English. I'm from the USA. I wasn't trying to take away from your explanation I should've been more clear on that, I just thought yours was so good I didn't even need to say. Figuring out it's strictly growing is clever as hell. I didn't even think of that. I meant simple, as in, my answer was simply a guess really and it worked.

    • @Quasar900
      @Quasar900 11 месяцев назад

      @@onbored9627 Oh please Sir , no need to Apologise , The fact that I've never studied math in English doesn't mean I don't know English 🙂 It's just I'm not that familar with English terms in math !
      But Thank God I do speak and read : French, English, Arabic, Spanish + some Japanese !
      I did study math in French (after high school) and Arabic (until high school) , but that waaaas 21 years ago !
      What class are you in ? I hope you're safe from those ongoing blizzard storms !
      Greetings From Morocco and Free Palestine 🙂

    • @Quasar900
      @Quasar900 11 месяцев назад

      @@onbored9627 I think you do know these techniques involving the continuity of a function, the intermediate values etc.. to solve equations !
      For example :
      solving in IR set :
      Arctan(x+1)+Arctan(x-1)=Pi/4

  • @FractAlkemist
    @FractAlkemist Год назад +37

    I have been learning Genetic Algorithms in Python; they are good for problems like this.
    The value I get for 'm' is 3.97, 3.99, 4.00, etc.; different each time as there is a random element for the convergence.
    A little intuition is also required;
    If you plug 4.0 into the equation you get correct 65. Program run time ~5 seconds.

    • @onlineMathsTV
      @onlineMathsTV  Год назад +4

      Wow!!! Nice sir.

    • @pierrecurie
      @pierrecurie Год назад +4

      If you're going the numerical route, bisection search is much faster.

    • @mistertwister1015
      @mistertwister1015 Год назад

      Blunt enumeration of roots is not always the best solution)

    • @victorfildshtein
      @victorfildshtein Год назад +1

      Hello. I program in PureBasic. I made this task by binary search method, 30 iterations,
      result 3.9999999991, precision 0.0000000047. Time is almost instantaneous.

    • @wilsonuche9389
      @wilsonuche9389 Год назад

      You need to review the Python solutions cos only 4 is an exsct solution. 3.97 is far from it, 3.99 is just an approximate

  • @teamredx1pro952
    @teamredx1pro952 9 месяцев назад

    Thank you for reminding me that you can't believe everything on the internet. Props to you i almost believed it untill i tested it with calculator. BRAVO YOU SHOULD WIN AN OSCAR

  • @therichcircle.8819
    @therichcircle.8819 Год назад +8

    You tried here tutor Jakes. I have learnt something here. Just keeping on running this channel. More grace, love from Port Harcourt ❤.

    • @onlineMathsTV
      @onlineMathsTV  Год назад +1

      It is our pleasure to serve you sir. Thanks for watching

  • @johnpalagye7036
    @johnpalagye7036 Год назад +15

    Love your video! I got lost half way and I did not know that you could just square the exponents and make them equal

    • @onlineMathsTV
      @onlineMathsTV  Год назад +4

      We are glad you love what is happening here.
      We promise to give more educative contents in the area of mathematics with the help of God.
      Love you.....💖💖💕💕

    • @karlm9584
      @karlm9584 Год назад

      I like this too. I have done it with roots before. Roots are just indices but I wouldn't have thought of doing it.

  • @clownphabetstrongwoman7305
    @clownphabetstrongwoman7305 10 месяцев назад

    Maybe it's a bad argument but I would say for a^m - b^m = X, and a, b, m belong to N, a^m> X > a^m-1.
    Here 3^m>65>3^m-1
    81>65>27 => m=4.

  • @ЛидийКлещельский-ь3х

    It can decide differently. The function (1) f(x)=3^x-2^x -is increasing . { for x>0 (2) 3^x>2^x ; x10 . (3) f(x2)-f(x1)= ……..=3^x1*[3^(x2-x1)-1]-2^x1*[2^(x2-x1)-1 ]> 2^x1*[2^(x2-x1) -1 ]-[ “--“ ]=0 ; (3) f(x2)-f(x1)>0 !!!!! So , it takes all its meanings once a time. f(4)=3^4-2^4=65 . So , x=4 - is the only root of equation ! Respectfully , Lidiy

    • @onlineMathsTV
      @onlineMathsTV  Год назад +3

      The elders in our mist are highly learned.
      Love your detailed explanation sir.
      Thanks for finding our time to watch our content and commenting even at this age of yours sir.
      Much respect and we All @onlinemathstv love you dearly...💖💖💕💕

    • @Rodolfo-Martin-Santos
      @Rodolfo-Martin-Santos Год назад

      There is no X! 😂but however, good job!

    • @andreasandre4756
      @andreasandre4756 Год назад +1

      Pay attention that m=constant not variable, so M must be grater then 1 otherwise solution will not be true because ln1=0 or 3-2=1 which is not equal 65.
      So M>1 and could be anything. So if it is not an integer number? So if it is not equal 65 but 63.5 - ?

    • @ADSemenov_ru
      @ADSemenov_ru Год назад +1

      You took the words right out of my mouth. :)

    • @Marat7973
      @Marat7973 Год назад

      Здравствуйте,Лидий! Не ожидал Вас здесь увидеть)

  • @ravirajshelar250
    @ravirajshelar250 Год назад +6

    I did it in a different way.
    Write 3 as 2+1 and perform binomial expansion, so the 2^m cancels and subtract one from both sides. We have 64 on one side and some series on other side.
    Notice that the series 2 + 2^2 +.....+ 2^m will definitely be smaller than series on left which is equal to 64.
    So make this G.P. sum less than 64, you will get that m should be less than 5, and once you have known this, you have proved that you just need to find a solutions less than 5 and those will be the only solutions.
    Only m=4 works out.

    • @albajasadur2694
      @albajasadur2694 Год назад

      a good method to find the range of m and it makes sensible checking easier with limited number of m

    • @theboss73104
      @theboss73104 8 месяцев назад

      Yeah

  • @tesfayewegderesegn
    @tesfayewegderesegn 4 месяца назад

    Great job, You are an excellent instructor. From an admirer Ethiopian in America

  • @mdmahin7299
    @mdmahin7299 Год назад +9

    Very well solution but the rules are very lengthy or Expensive ☺️
    So if we assume the value of "m" here from 1-4
    We are easily getting the value of m
    Such as
    Let , m=1,2,3,4,... ♾️
    and now,
    3^1-2^1=1≠65
    again
    3^2-2^2=5≠65
    And now if we let, m=4 then
    3^4-2^4=65=65
    So we can easily get m=4😊

  • @pwmiles56
    @pwmiles56 Год назад +6

    In a slightly fancier approach we can make a recursion
    2(3^m - 2^m) + 3^m = (2+1)3^m - 2^(m+1)
    = 3^(m+1) - 2^(m+1)
    Put a(m) = 3^m - 2^m
    a1 = 3^1 - 2^1 = 1
    a2 = 2 a1 + 3^1 = 5
    a3 = 2 a2 + 3^2 = 19
    a4 = 2 a3 + 3^3 = 65, done

    • @onlineMathsTV
      @onlineMathsTV  Год назад +5

      Wow!!! This approach is impressive but a bit obscure sir.

    • @vejayashanker
      @vejayashanker Год назад

      can u explain what u mean by obscure pls😂

    • @soltanchalkarow905
      @soltanchalkarow905 Год назад

      can you help me sir?
      if ab+bc+ca=1
      prove:
      sqrt(a + (1/a)) + sqrt(b + (1/b)) + sqrt(c + (1/c)) >= 2( sqrt(a) +sqrt(b) + sqrt(c) )

  • @danielmelo5609
    @danielmelo5609 Год назад +2

    Excellent explanation. It seemed very complicated, but it turned out to be easier than expected. A hug

  • @fisicamatematicasprofewilliam
    @fisicamatematicasprofewilliam 11 месяцев назад +3

    profe en olimpiadas de las matematicas se aprende mucho. y con usted bastante. felicitaciones

    • @onlineMathsTV
      @onlineMathsTV  11 месяцев назад +2

      Thanks a million sir, we appreciate this comment my good friend.

  • @AndrewUnruh
    @AndrewUnruh Год назад +5

    Clever solution! One thing I did not get is how you knew that m was a positive integer - or was this just an assumption that happened to work? For example, if the problem had been 4^m - 3^m = 65, m would be approximately 3.36.

    • @charleskaruru481
      @charleskaruru481 Год назад

      m can never be a negative otherwise we wont have 65but fraction

  • @jamesmichuki5804
    @jamesmichuki5804 11 месяцев назад +1

    Thanks teacher. Today I've learnt something very new in maths. Am really surprised....

  • @88kgs
    @88kgs Год назад +12

    Sir, we can also do this by hit and trial method, assuming different values for m=1,2,3,4....
    But your way was also very nice 👍👍.
    Thank you for this video sir🙏

    • @onlineMathsTV
      @onlineMathsTV  Год назад +2

      Yes, but that will only work if the working process is not in the examination but this is needed when the examiner wants you to show your procedure step by step...👍👍👍

    • @gregfarnham5651
      @gregfarnham5651 Год назад +1

      Yes, trial and error could work if we assume m is a positive integer. I don't believe that was a given, however.

    • @ralfimuller8948
      @ralfimuller8948 Год назад +1

      @@gregfarnham5651 The solution in the video also made use of the assumption that m is an integer. Otherwise the factorization of 65 would not be unique. Actually, the hit and trial method should be entirely ok.

    • @gregfarnham5651
      @gregfarnham5651 Год назад +1

      @@ralfimuller8948 Agree. Thank you.

    • @danielrivera2278
      @danielrivera2278 Год назад +1

      Also, by trial and error you can't prove that's the obly answer

  • @michaelsidorov5508
    @michaelsidorov5508 Год назад +5

    Остроумно и красиво! Как и вся математика.

  • @ГульнараКожошева-ц2ф
    @ГульнараКожошева-ц2ф 6 месяцев назад +2

    Thank you very much!
    Без перевода мне всë так понятно было👍

  • @Toxa_Azimov
    @Toxa_Azimov Год назад +11

    Слева возрастающая функция при m>0 ( можно найти производную и убедиться ), справа постоянная функция, значит у них может существовать только одна точка пересечения, методом оценки m=4

    • @onlineMathsTV
      @onlineMathsTV  Год назад

      Bravo 👍👍👍
      You the best sir.

    • @onlineMathsTV
      @onlineMathsTV  Год назад +1

      Bravo 👍👍👍
      You the best sir.

    • @PlumbuM871
      @PlumbuM871 Год назад

      Чисто случайно подставил вместо m 4, и всё сошлось! Везёт мне

  • @umeshkhetan
    @umeshkhetan Год назад +13

    If a.b=65, a and b can have infinite values. So, the tutor has just one answer where multiple answers are possible.

    • @onlineMathsTV
      @onlineMathsTV  Год назад +3

      Yes it has multiple answer but for the sake of this tutorial we restricted ourselves to this solution sir.
      Thanks for this observation.
      Much love....💕💕👍👍

    • @mustaphaolunrebi8100
      @mustaphaolunrebi8100 Год назад

      I think you add that, where m is an integer. It makes it complete. The only integer factors 65 has are 5 and 13. Nice solution 🎉

    • @ivandonchev474
      @ivandonchev474 Год назад

      Sorry but this video is the most useless shit I have seen. You solved it by guessing and overcomplicated massively

    • @speedsterh
      @speedsterh Год назад

      @@mustaphaolunrebi8100 No, 65 has 4 factors: 1, 5, 13, 65.
      The equation with 2 other factors should be explored for completeness

  • @EllaBanua-q2u
    @EllaBanua-q2u 4 месяца назад

    Thank you so much for sharing this wonderful learning for us Sir! God bless...❤❤❤

  • @АлександрСтоляров-ю2с

    Сначала поделить обе части на 2^m. Тогда слева будет возрастающая функция, справа убывающая. Тогда уравнение имеет не более одного корня. Подобрать корень не сложно. 9 класс, ничего сложного. За проведенное решение минус: нигде не доказано отсутствие других решений, переход к системе ничего не обосновывает.

  • @Psykolord1989
    @Psykolord1989 Год назад +12

    Before watching:
    Alright, so, we are looking at exponential functions. 3^m - 2^m = 65.
    First, we can rule out m=1 and anything below; the difference between those would be smaller than between 3 and 2, and thus much lower than 65.
    Next, notice that we are dealing with an integer on the right. This heavily implies (but *does not necessarily guarantee* ) that 3^m and 2^m are both integers as well. If both are integers, then m must also be an integer. So we should start with integers. (If we were dealing with a mixed number instead, this would be much more complex; as it stands, we can just plug in integers and see which one works).
    Let's start with the first x that gives us 3^m >65, namely 4. 3^4 = 9^2 = 81, and 2^4 = 4^2 = 16
    So for m= 4, we have 81-16 = 65. Fortunately for us, this checks out, and thus we have our answer of *m = 4*

    • @onlineMathsTV
      @onlineMathsTV  Год назад

      Thanks for this thorough explanation. You the best. 👍👍

    • @RoderickEtheria
      @RoderickEtheria Год назад

      Negative m gets fractions.

    • @Psykolord1989
      @Psykolord1989 Год назад

      @@RoderickEtheria Yes, you are correct, and I see I made a typo in the "so for m=4 we have..." section by putting a negative in front of the 4. Fixed now.
      I don't imagine it was a huge problem since in the section right above it,and at the very end of that section, I used 4, but it still may have confused some people.

  • @OkoroUchechukwu-j8r
    @OkoroUchechukwu-j8r 10 месяцев назад

    I'm impressed with this explicit methodology 😊(12:16am)

  • @franciscodeassisbrandaobra898
    @franciscodeassisbrandaobra898 Год назад +5

    exercicio maravilhoso🥰🥰❤❤❤❤❤❤

    • @onlineMathsTV
      @onlineMathsTV  Год назад +1

      Thanks a millions sir, we love you ❤️❤️💖💖💕💕😍😍

  • @RoderickEtheria
    @RoderickEtheria Год назад +19

    Solved 3^m-2^m=65 by just thinking about the first whole number power of 3 above 65.

    • @dilphek
      @dilphek Год назад

      It is not about finding it. Olympiad is a school competition teaching kids to solve these problems mathematically

    • @bleh-zj1hy
      @bleh-zj1hy Год назад

      ​​@@dilphekyou guys got a different Olympiad or something? Here Olympiads (for the kids, totally different type from the subjective ones) are mcqs and the subjective ones are like 3 questions in 3 hrs and if you're able to do even 1 you're qualified (you can imagine the toughness so it's really not for the kids)

    • @ARTAATIF
      @ARTAATIF 3 месяца назад

      I really think just like that...

  • @michaellockett4044
    @michaellockett4044 8 месяцев назад

    Difference of squares into u-substitution. Excellent methodolgy.

  • @DrMikeE100
    @DrMikeE100 Год назад +4

    It took a fraction of a minute to recognize m = 4, but as a mathematician, I did find this interesting. (Dr. Mike Ecker)

    • @JaroGoraJ
      @JaroGoraJ Год назад

      I think because it was easy question

  • @divonsirlopes5409
    @divonsirlopes5409 Год назад +4

    There is a faster workaround, with the assumption that the number m is integer. The term on the right is less than the term on the left. For simplicity, we can assume that the term on the right is zero. This results in:
    3^m = 65.
    m is greater than 3, because 3^3 = 27.
    m can be 4 because 3^4 = 81.
    Let's test m = 4:
    3^4 - 2^4 = 65
    81 - 16 = 65
    65 = 65

    • @onlineMathsTV
      @onlineMathsTV  Год назад +3

      Wow!!!
      This is fantastic. I love this approach sir. We have gained some values from this procedure sir. Thanks for dropping this sir.
      Respect.....👍👍👍
      Much love....💕💕💖💖❤️❤️

    • @divonsirlopes5409
      @divonsirlopes5409 Год назад

      There is a faster workaround, with the assumption that the number m is integer. The term on the right is less than the term on the left. For simplicity, we can assume that the term on the right is zero. This results in:
      3^m = 65
      m = log(65)/log(3) = 3,8
      m is greater than 3,8
      Let's test m = 4
      3^4 - 2^4 = 65
      81 - 16 = 65
      65 = 65

    • @MONSTER2013
      @MONSTER2013 Год назад +1

      Base on which level you’re at. This question and video are made for yr 10-11? So he gave the solution at that grade. Above yr 12 can use other tools as log/ ln skilfully to solve it.

    • @divonsirlopes5409
      @divonsirlopes5409 Год назад

      Thanks for the info.

    • @foudilbenouci482
      @foudilbenouci482 Год назад

      ^you found one solution doesn t mean you found all solutions

  • @elhassanidnarour672
    @elhassanidnarour672 Год назад

    You must first determine if m is even, or do you also handle the second case of m being odd.

  • @mathtv3982
    @mathtv3982 Год назад +5

    First you must show that m is even positive integer

  • @Curufin1984
    @Curufin1984 Год назад +10

    This is so overly complicated.. Just check low values of m and find that m = 4 works.. Then use a simple analysis tool to show uniqueness of the solution e.g. by showing that function f(m) = 3^m - 2^m is strictly increasing for m>=1.
    Additionally your solution contains errors and missing steps:
    1. If you do the trick with 3^(m/2)^2 - 2^(m/2)^2 = 65 then you presuppose that m is even, because if m is odd then 3^(m/2) is not natural anymore, so you cannot use the natural divisors of 65 in the following steps anymore.
    2. Even if that approach worked (because you somehow proved that m must be even): After you rewrite the equation as x^2 - y^2=65, then you have to consider *two* pairs of solutions 65 = 65 * 1 and 65 = 5 * 13.
    Long story short, lots of mistakes in your video unfortunately.

    • @onlineMathsTV
      @onlineMathsTV  Год назад +1

      Thanks for this keen observation and I really appreciate this comment sir.
      Noted. I will do more detailed work on subsequent videos.
      You the best and much love for this detailed comment....💕💕💕

    • @ivandonchev474
      @ivandonchev474 Год назад

      Yes this video is shit full of errors. The fact that it has so many errors and is hugely overcomplicated at the same time makes it complete and utter shit

  • @horstmueller1000
    @horstmueller1000 2 месяца назад

    The presentation and demonstrated solution is very good. rubikaz comment is nevertheless valuable. If the approach solves, then the uniqueness provides all solutions. Using an exclusive deduction from the start is valid, if a working hypothesis. Thanks for this video.

  • @MathHero24
    @MathHero24 2 месяца назад

    if m0 and the derivative of 3^m-2^m is 3^m*log(3) - 2^m*log(2) which in turn larger than 3^m*log(2)-2^m*log(2) which is clearly positive as both log(2) > 0 and 3^m>2^m hold. Therefore the solution is unique, since 3^m-2^m-65 is increasing. A simple guess shows m=4 is a solution.

  • @bdh9202
    @bdh9202 10 месяцев назад

    You worked hard, I found the result by giving the value of m to 4 in 10 seconds, but it is important how the solution is, but I still think it is too long. Greetings from Turkey.

  • @forgottenlegacy5929
    @forgottenlegacy5929 8 месяцев назад

    Wonderfully explained, very informative.
    But it’s sometimes more convenient to use the easy method

  • @elchinpirbabayev5757
    @elchinpirbabayev5757 10 месяцев назад

    Somehow I am having trouble with assuming x+y=13, and x-y=5... why they can't be 10 and 6.5 respectively? Well if you think about m as a fraction the likelihood of the right side of the equation being a round number (=(65) is low, but can it be ruled out?

  • @antoniogomesfigueiredo7835
    @antoniogomesfigueiredo7835 Год назад

    Gostei muito da resolução dessa questão. Não entendi suas palavras, pois, sou português, todavia, a resolução da equação entendi perfeitamente. OK? Obrigado. Thank you very much.

  • @jwilson4163
    @jwilson4163 Год назад

    Great lesson! Thanks. (From São Paulo/BR)😃

  • @tunein1218
    @tunein1218 10 месяцев назад

    Could you explain more on how you arrived at using 13 and 5

  • @stanleysimon6723
    @stanleysimon6723 Год назад

    Criticizers will be always there just ignore them.Good work brother👍

  • @BrukFikru
    @BrukFikru Год назад +1

    Nice one ...... good teacher ...... stay teaching stay doing more math tricks ...... i liked it

  • @charleskaruru481
    @charleskaruru481 Год назад +1

    in maths olympiad time is very critical...what is important is the answer...so the best way is to solve maths olympiad is to have very basic maths then the rest is analysing to get the pattern...so what i did is just look at indecies of 3 that have the last number such that if we subract an indecies of 2 and get 65

  • @BN-hy1nd
    @BN-hy1nd Год назад

    Following your example, can you do (x+y)(x-y) = 65 multiplied by 1 ie 1*65 = 65 becsuse 1 and 65 are factors of 65?

  • @AbdullahSaHeL
    @AbdullahSaHeL Год назад

    Nice. You used Algebraic manipulation.
    I used my laptop to answer the question with Graphical approach, the answer was 4 too.
    However, the answer was almost 4 when I used Newton-Raphson and linearization methods, why?

  • @mymathtutoreg2628
    @mymathtutoreg2628 Год назад

    3^m>65 use log of both sides to solve the inequality we get m>3.8 so start to check m=4 in the original equation the equation is satisfied and m = 4

  • @rajendrasinghbisht2628
    @rajendrasinghbisht2628 9 месяцев назад

    Nice solution. It makes so easy, the way you explained.

  • @nirmalbatra5651
    @nirmalbatra5651 3 месяца назад

    Excellent ! Solution becomes so easy and understandable

  • @1234garga
    @1234garga 10 месяцев назад

    Браво! Много елегантно решение

  • @Saibatou
    @Saibatou Год назад

    You are just great! A very good approach.

  • @dinlendiricidrtv
    @dinlendiricidrtv 6 месяцев назад

    Thank you very much my dear friend

  • @shubhampoddar8403
    @shubhampoddar8403 Месяц назад

    So helpful . And this video gave me idea how to solve elzebric brobepems ❤ .
    Take more video related to solve many brobepems in polynomial . Logorithimc function and trigonometry
    Nice and very helpful video

  • @MonaKhalil-r6e
    @MonaKhalil-r6e 5 месяцев назад

    Amazing work.. Thank you for sharing...

  • @MbwanaHosseinMbwana
    @MbwanaHosseinMbwana 2 месяца назад

    When we find the factor of 65 is there any condition that considered? If no how how we can use 65and 1as our proposed factor ?

  • @sakshamraj1566
    @sakshamraj1566 10 месяцев назад +1

    Sir but what if like its a mcq or rapid round question can we just apply hit and trial?

  • @agnitank5238
    @agnitank5238 5 месяцев назад

    Nice , solution & was very enlightening. Thanks

  • @TheDotBot
    @TheDotBot Год назад

    Great stuff!
    I went the x²-y³=81-16 route, was that cutting corners?

  • @BipedBD
    @BipedBD Год назад

    But how do you prove m is an even number in order to be able to treat 3^(m/2) and 2^(m/2) as integers for the factorizations of 65?

  • @GautamKumar-wx3sm
    @GautamKumar-wx3sm 11 месяцев назад

    Couldn't have thought of this approach. Thanks for this.

  • @mikeblings1504
    @mikeblings1504 10 месяцев назад

    As a math teacher, this is a plus to me. You are amazing 👏

  • @DeadBoY-z8s
    @DeadBoY-z8s Год назад +1

    Mathematics is all abt observation and understanding thee pattern thx a ton Buddy❤

    • @onlineMathsTV
      @onlineMathsTV  Год назад

      Thanks a million for watching and dropping this wonderfully encouraging comment sir. The true is that mathematics is all about finding a solutions to problems/challenges. We really appreciate the fact that you watch and dropped this comment to clarify some debts in the minds of so many viewers and subscribers here.
      Much love from all of us @OnlinemathsTV....💕💕💕💖💖💖❤️❤️❤️

  • @CleopatraNadesan-kn3jg
    @CleopatraNadesan-kn3jg Год назад

    Excellent tutorial. Thank you very much

  • @zulfqarali2994
    @zulfqarali2994 8 месяцев назад

    Gentleman I appreciate your work.

  • @madhabaranjanrout1215
    @madhabaranjanrout1215 4 месяца назад

    Many Many Thanks to you. ❤❤❤

  • @lakshmikutty2255
    @lakshmikutty2255 3 месяца назад

    Beautifully explained.

  • @Maths_withs_NHLaskar
    @Maths_withs_NHLaskar 6 месяцев назад

    I learnt something new from this video... And I wanna say something to dear sir please do you job not listen other criticism... Love you sor i am from India ( Bengali) i am also a Maths teacher 🧡🤍💚

  • @maheshkulkarni6086
    @maheshkulkarni6086 25 дней назад

    amazing explaination of procedure

  • @tejendramohenbaisya8529
    @tejendramohenbaisya8529 Год назад

    Problem solved in a very intelligent way.

  • @FrenziedSword
    @FrenziedSword 2 месяца назад

    What about moving 3^m to the right hand side and equating the equation

  • @xuanhungnguyen7667
    @xuanhungnguyen7667 9 месяцев назад

    Chính xác.anh biến đổi 65=5×13.đây mới là mấu chốt của bài.và tư duy hàm số mũ...tuyệt.

  • @york-tg4cu
    @york-tg4cu 4 месяца назад

    這題如果想到 "平方差公式" 就很簡單! 前提是右邊的數字剛好是兩個質數的乘積~ 如果不是 設計好~很難用第二種簡單的方法計算!

  • @negativeM
    @negativeM Год назад

    Awesome information on math in this video. But I think I saw somewhere that a problem like this can be solved with log or ln some how. I was wondering if this same problem can be solved using that method instead. I also liked this method in this video. Don't get me wrong here. I just wanted to see a way to do it with log/ln in the equation if that is possible..

  • @ajitkumardas1491
    @ajitkumardas1491 8 месяцев назад

    3^m - 2^m = 65
    > 3^m - 2^m = 81 - 16
    > 3^m - 2^m = 3^4 - 2^4
    comparing both sides, it can be concluded
    m = 4

  • @lophocthienuc7345
    @lophocthienuc7345 Год назад

    I think it's very easy.
    We let m = 4, it's correct
    We will prove "with m > 4 , 3^m -2^m > 65"
    So, only m = 4 will be equation.
    m>4 m = 4+a (a>0)
    3^m - 2^m > 3^4 - 2^4
    Because (3^a-1).3^4 always > (2^a-1).2^4
    Thanks from Việt Nam 🎉

  • @Winz_Breaker
    @Winz_Breaker 8 месяцев назад +1

    Thankyou you,i understand😊

  • @F007-n6y
    @F007-n6y 11 месяцев назад +1

    3ᵐ-2ᵐ=65
    y=3ˣ-2ˣ
    y=65
    если построить оба эти графика, то будет видно, что уравнение имеет единственное решение, поэтому можно попробовать подобрать корень подбором:
    х=1: 3¹-2¹=1; 1≠65
    х=2; 3²-2²=5; 5≠65
    х=3; 3³-2³=19; 19≠65
    х=4; 3⁴-2⁴=65; 65=65
    х=4 корень
    m=4

  • @paraskevitsiourva2794
    @paraskevitsiourva2794 6 месяцев назад

    Απολύτως τέλειο!!!!
    Είστε έμπνευση!

    • @onlineMathsTV
      @onlineMathsTV  6 месяцев назад

      Thanks a million for watching and encouraging us with your wonderful comment sir.
      Much love sir 💕💕💖🙏

  • @user-mb8fu6ci1q
    @user-mb8fu6ci1q 5 месяцев назад

    How do you arrive to the power 2/2

  • @sevgiylehayat
    @sevgiylehayat 10 месяцев назад

    Perfect.Thank you sir.