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

In solving this math Olympiad problem, 3^m-2^m=65, Jakes uses a very unique approach to handle this exponential math challenge with easy.
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[@mitahaubica6498]

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]

You are a genius

[@nicadi2005]

@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]

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

[@davidmajor1508]

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]

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.

[@tiehoteele874]

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

[@onlineMathsTV]

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

[@thefireyphoenix]

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]

@@thefireyphoenix this video wasn't made for them

[@Theuserisadick]

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

[@f-s406]

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]

I too

[@tuyu6404]

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]

I dont think so. 2^m 65 why m

[@gheorgheneacsu3356]

Me too! 😊😊

[@chrissyday67]

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

[@danieldavies1829]

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]

That's exactly what I did.

[@Smith_14]

m can be a negative number?

[@leishajuneja2994]

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]

I did the same😂

[@DownhillAllTheWay]

You're assuming that m is an integer?

[@rubikaz]

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]

Noted.

[@nicadi2005]

@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]

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

[@with.love.from.siberia]

​@@babetesfaye1001да, потому что оно равно 1

[@Quasar900]

@@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

[@nicholastergech8525]

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]

This is what I actually expected from him

[@TomJones-tx7pb]

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

[@lgmoses3876]

I did it,in my braine.

[@sfqamd]

kzbin.info/www/bejne/qXrPg3udbJhjeaMsi=ujk2KD_xjoMGqiP5

[@shivaprasadmallikarjunaiah3751]

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.

[@alfredomulleretxeberria4239]

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]

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]

​​​​​@@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.

[@claudiohagra]

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

[@elmehdiazzouz7888]

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 ..

[@georiashang1120]

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.

[@sanmus100]

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]

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]

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]

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

[@thegreathussar9442]

@@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]

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.

[@samuelmayna]

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]

This makes sense and it is mathematically correct

[@nicadi2005]

@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.

[@user-uy7uu1rm5u]

m is an integer

[@samuelmayna]

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

[@samuelmayna]

​@@user-uy7uu1rm5u but my approach is sound which shows that 65 and 1 won't work but it can give different answers for some equations.

[@Amy-601]

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

[@gabrielschiteanu4963]

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]

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

[@Shirlippe]

Hi there! Here is an alternative solution. The difference of the powers on the rhs increases with m, and it is already greater than 65 for m=5. So m must be less than 5. Assuming m is an integer, and noticing that m=3 doesn't work, the only possible solution is 4.

[@onlineMathsTV]

Ya boss. You are very very correct and I like you approach sir. This is an indication that you are a master in this area. Respect sir...👍👍👍

[@ronitmahawar1193]

but this only works if m is integer which we dont know it would be

[@romanyukvictor]

@@ronitmahawar1193 Exactly! To solve this you have to prove that the equation has single solution. Just shift 2^m to the right side and divide the entire equation on 2^m. As a result you will get the increasing function of m on the left (3/2)^m and decreasing function on the right (1+65/2^m) which can intersect only once. So there is only one root. The only way to find the root is to enumerate the integers and find out m=4.

[@ifomichev]

@@ronitmahawar1193 the solution proposed by the author of the video also works only for integers, because it relies on factorization of 65

[@jccamargo99]

For this reason many people don't like math.

[@therichcircle.8819]

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

[@onlineMathsTV]

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

[@marcio5457]

Solução genial. Parabéns

[@jamesclerkmaxwell8545]

Parabéns. Resolução muito boa.

[@user-oi3iv7oo4z]

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]

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

[@user-oi3iv7oo4z]

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

[@Change_Verification]

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

[@danielrivera2278]

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

[@Eismann1]

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

[@elio9008]

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]

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]

@@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.

[@user-xl3mg3om7s]

@@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.

[@user-xl3mg3om7s]

@@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]

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

[@dinlendiricidrtv]

Thank you very much my dear friend

[@Sapped6]

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

[@airtonreis2675]

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]

@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]

@@onlineMathsTV 🤝

[@MATHSHEADMASTER]

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

[@pedrooo13]

Brasil tá em todo lugar não tem jeitoooooo

[@xgx899]

Function 3^x-2^x is strictly increasing for x>0. Since solution (if any) is positive, it is unique. By inspection, m>2 and m

[@teamredx1pro952]

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

[@ravirajshelar250]

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]

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

[@theboss73104]

Yeah

[@user-pd7js7cy9m]

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]

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...💖💖💕💕

[@Lernen-mit-Rudi]

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

[@andreasandre4756]

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]

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

[@Marat7973]

Здравствуйте,Лидий! Не ожидал Вас здесь увидеть)

[@user-qo6ni5sm5p]

Solução genial. Parabéns. Muy interesante, hoy aprendí un buen método, gracias.

[@user-ne7pu8ib7y]

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

[@johnpalagye7036]

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]

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]

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

[@Psykolord1989]

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]

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

[@RoderickEtheria]

Negative m gets fractions.

[@Psykolord1989]

@@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.

[@clownphabetstrongwoman7305]

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.

[@SALogics]

nice explanation ❤

[@franciscodeassisbrandaobra898]

exercicio maravilhoso🥰🥰❤❤❤❤❤❤

[@onlineMathsTV]

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

[@jaimeduncan6167]

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]

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]

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.

[@GautamKumar-wx3sm]

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

[@user-gj5qq4cn8k]

Nice one sir

[@fisicamatematicasprofewilliam]

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

[@onlineMathsTV]

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

[@RoderickEtheria]

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

[@dilphek]

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

[@bleh-zj1hy]

​​@@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)

[@vijayanand8077]

Instead of solving by equations in eloborate lengthy long time , I simply solve it in the beginning itself my applying m=1 m=2 m=3 Wow.. when I apply m = 4, I got the answer *65* # Instead of 65, if the answer @ RHS is 6584 (or any larger number), we should go for equations as described in the above video .. {But when it is small number less than 100, we can approximately conclude range between by directly applying values for *'m'*

[@dannychown2593]

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

[@karlm9584]

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]

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

[@Quasar900]

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]

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]

@@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]

@@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]

@@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]

@@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

[@mdmahin7299]

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😊

[@onlineMathsTV]

Bravo👍👍👍

[@WhiNikkatherealone]

i mean hit and trial is always the last resort

[@user-rz7ym5ts1e]

I love how the way you work out the question..❤❤

[@user-og4vc7ey2y]

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

[@user-qt9qo3ez9x]

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.

[@pwmiles56]

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]

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

[@vejayashanker]

can u explain what u mean by obscure pls😂

[@soltanchalkarow905]

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) )

[@nnaammuuss]

m≤2 makes it too small and don't work. So, m must be even, as 3^m ≡ 3 (mod 4) for m odd, but 65 ≡ 1 (mod 4). On the other hand, 3^m - 2^m = 3^{m-1} + ... ≥ 3^{m-1} and is strictly increasing. Since, 3⁴ = 81 is already too big, we must have m

[@ludmilak9396]

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

[@owlsschoolofmath9732]

Great! Its a fun problem. I did something kind of similar with difference of 2 squares.

[@onlineMathsTV]

Nice, you the best. Kudos 👍👍

[@FractAlkemist]

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]

Wow!!! Nice sir.

[@pierrecurie]

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

[@mistertwister1015]

Blunt enumeration of roots is not always the best solution)

[@victorfildshtein]

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]

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

[@charleskaruru481]

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

[@ekinodoaruoriwo]

Wonderfully done

[@DrMikeE100]

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

[@JaroGoraJ]

I think because it was easy question

[@Toxa_Azimov]

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]

Bravo 👍👍👍 You the best sir.

[@onlineMathsTV]

Bravo 👍👍👍 You the best sir.

[@PlumbuM871]

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

[@lophocthienuc7345]

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 🎉

[@sevgininbahcesi5222]

Perfect.Thank you sir.

[@mathadict1]

Keep up the great work man happy to see somoen in our beloved African continent devoting a portion of their time into this much love from Morocco 🇲🇦 MA

[@onlineMathsTV]

Much appreciated sir. Thanks for watching our contents and the encouragement sir. Much love from everyone @OnlinemathsTV to you sir 💖💖❤️❤️💕💕💕

[@Quasar900]

@@onlineMathsTV 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

[@Quasar900]

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

[@umeshkhetan]

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

[@onlineMathsTV]

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]

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]

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

[@speedsterh]

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

[@andrevadyaskin164]

m = 4 😊 I just guessed it. And this is the only single solution coz with more "m" the difference will be more.

[@onlineMathsTV]

Bravo!!! You are good at what you do sir. You are the master/boss here sir. Maximum respect boss!!! 🙏🙏🙏

[@bdh9202]

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.

[@michaelsidorov5508]

Остроумно и красиво! Как и вся математика.

[@onlineMathsTV]

Thanks for this wonderful comment sir. Love you....💕💕

[@user-nv9rw7nh5w]

Очень, очень. Надо же как можно. Удивительно!

[@88kgs]

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]

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]

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

[@ralfimuller8948]

@@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]

@@ralfimuller8948 Agree. Thank you.

[@danielrivera2278]

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

[@math_qz_2]

Very instructive task

[@xuanhungnguyen7667]

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

[@AntoninaKa-es8tv]

Очень интересное решение! Спасибо!

[@onlineMathsTV]

You are welcome always even as we look forward to seeking better ways of serving you better in our services to you on this channel. We at Onlinemathstv love you without reservation sir....💕💕💕

[@paulmiddletonphotography4368]

Hello Online Maths TV, I really like your approach with this and how your deliver the proof. It is elegant and uses several skills. Your pace of delivery is really good too. As a suggestion for completeness in your proof, can you please include a determination table of which factors of 65 and their order are valid for consideration for equating to (x+y).(x-y). A table similar to; (x+y) . (x-y) | x | y | m from x=3^(m/2) | m from y=2^(m/2) | 1 . 65 65 . 1 13 . 5 5 . 13 If you calculate x and y and then m from x=3^(m/2) and from y=2^(m/2) for each of these arrangements, only (x+y) . (x-y) = 13.5 provides valid and consistent values for m. So 3 out the 4 arrangements can be mathematically eliminated. This would verify that the only valid arrangement and values of the factors is 13.5. I hope this helps. Cheers, Paul.

[@onlineMathsTV]

What a wonderful guide @Paul Middleton. Very very helpful sir. Respect boss. ❤️❤️❤️👍👍👍

[@onlineMathsTV]

Am compared to drop another comment in respond to your first comment sir. Am speechless and overwhelmed by your systematic construction of this comment. It is so encouraging. This is because you appreciated my little effort in the mist of professors and further suggested a wonderful/a great approach to me on solving this same problem or similar problem in my subsequent videos. Sir, with all humility we are glad to meet with you and have you here. Much love sir....💖💖💖💕💕💕❤❤❤

[@paulmiddletonphotography4368]

@@onlineMathsTV You are very welcome indeed.

[@paulmiddletonphotography4368]

@@onlineMathsTV Thank you for your very kind words of appreciation. It certainly is my pleasure to help you and I am really pleased that my suggestion will assist you and into the future with different proofs. May you increase your subscribers, students and people interested in learning your very clear, easy to follow and forthright teaching of maths. All the best to you for the future.

[@bozobanger]

Lazy solution: Powers of 3 = 3, 9, 27, 81, 243 ... Powers of 2 = 2, 4, 8, 16, 32 ... Trying the minus operation for each "m", we get: 3⁴ = 81 2⁴ = 16 81 - 16 = 65 So, m=4

[@ange-bernardferracci4278]

. I have made 3exp1 -2exp1=1 3exp2-2 exp2 =5 3exp3-2exp3=19 3exp4-2 exp4=65 81 -16 = 65 Différence between two perfect squares

[@AndrewUnruh]

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]

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

[@divonsirlopes5409]

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]

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]

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]

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]

Thanks for the info.

[@foudilbenouci482]

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

[@Luis-zj4dv]

Genial, muchas gracias!

[@Saibatou]

You are just great! A very good approach.

[@mathtv3982]

First you must show that m is even positive integer

[@onlineMathsTV]

Noted sir.

[@Curufin1984]

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]

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]

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

[@danielmelo5609]

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

[@MdSamir-st9jv]

Love and support from India 🇮🇳

[@onlineMathsTV]

Thanks a million, much love from all of us @ Onlinemaths TV...💖💖💕💕 We cherish you more sir...🙋🙋🙏🙏

[@jamesmichuki5804]

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

[@abuomar4995]

Very good approach

[@yklim817]

3 power of m-2 power of m=65 (3*3*3*3)-(2*2*2*2)=65 81-16=65 Because 4 times of 3 and 4 times of 2 So the answer for m is 4.

[@joseph-fernando-piano]

This is explained well, but respectfully I notice a logical flaw in the solution. When we start with the original equation, we can tell by the fact that (3^m - 2^m) starts at zero for m=0 and increases monotonically for all m>0 that there is only one real solution for m. However, after doing the variables transform to x and y, we have x^2 - y^2 = 65, which has infinite solutions. This is because when you did the variables transform, you don’t consider that x and y are not independent, they are both functions of m and therefore you can’t just pick them arbitrarily. I’m picking 13 and 5 as the factors, you were fortunate because 65 had few integer factors, and x and y could have infinite solutions. If you were going to guess at integer solutions, you should start from the original expression and guess values of m, since there are so many fewer options. Just by guessing and checking integer values of m, you would get the right answer in about 10 seconds.

[@user-rk5eh2sh9v]

Bruteforced for 10 secs and x = 4 81-16=65

[@marcelo372]

Excelente. Thank you.

[@Burcuqqoo-Tube]

Hedduu Galatoomi! Furmaata bareedaadha bu'uuraa isaa nuf ibsiitee ❤🎉

[@islamjonibragimov]

Increasing function🔥

[@michaellockett4044]

Difference of squares into u-substitution. Excellent methodolgy.

[@onlineMathsTV]

Thanks a bunch sir.

[@chrisw4562]

Nice job! I got the answer by trial and error, but this way of getting a deterministic solution is really cool.

[@italixgaming915]

Actually this proof is wrong. If m is an even number then x+y and x-y are integers but if m is an even number then they are IRRATIONALS and the unicity of the prime factors works only with integers. Euler did the same kind of mistake when he tried to demonstrate the Fermat theorem (that time he used complex numbers).

[@tsengkunli8461]

1. (3^m-2^m) isincreasing function of m. 2.M is integer,because 65 is integer. 3.f(0)=0,f(5)=243-32=211. 4.we only need to test by 1,2,3,4 5.The answer is 4

[@adityamittal4357]

He still had to compare and guess which is 13 and which is 5 in the factors. Same guess applied directly to the original problem shows the subtraction of 2^m from 3^m so m

[@marcofidelpenaavila2129]

Extraordinario , thanks so much.

[@TheMightyOdin]

I did this in my head in less than a minute. 3,9,27,81(bigger than 65) 2,4,8,16( subtract 16 from 81)…. Works.

[@daintydawn2508]

Yeah, it's a bit nd trial method. But if the value of m had been bigger then, it wouldn't work

[@maheshmalkapuram2409]

Good explain

[@Georgeadrian-21]

Very good video

[@antoniogomesfigueiredo7835]

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.

[@DG_EDITZ17]

I was able to solve the problem in less than 1 minute... I just thought about the Multiple of 3 more than 65 that was 81 and then i subtracted 65 from it and the resulting answer was 16,now we know that 3^4 is 81 and 2^4 is 16,so easily the value of m came out to be 4

[@BrukFikru]

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

[@zulfqarali2994]

Gentleman I appreciate your work.

[@user-qg1xq8vr9v]

thank you sir

[@knguyen3348]

You are the best. Thanks

[@Monitor2023]

Excellent 👍👍👍

[@CleopatraNadesan-kn3jg]

Excellent tutorial. Thank you very much