Hi! I'm the organizer for SMT 2022! I'm glad you enjoyed our tiebreaker problems, and thank you for reviewing them!! For those of you who are high school students we encourage you to participate in this year's iteration of the contest, we're very excited :)
For the second question, removing all of the high level notation you could say, how many numbers less than 40 are perfect squares but not perfect fourth powers.
@@blackpenredpen Thanks. Also to finish the interpretation, it comes down to finding how many perfect squares there are below 40 and how many prefect fourth powers below forty and subtracting the later from the former
@@cosmicvoidtree Alternatively, you can find/generate a list of squares of numbers that are not squares themselves. This works for n≤40 because we don't really have to deal with sixth powers and etc.
Luckily there are no perfect 8th powers less than 40 or what you have said would not hold. Eg. 256 is 4^4 but it would satisfy the condition (f(256) can = 16 as f(4) = 2) perfect 16th powers don't hold as f(65536) != 256 since that is f(16). Will hold for 32nd powers but not 64th and so on.
I am from India , i study in 12th standard. when i study integration, i have some questions having a problem then i search in TH-cam then i show your channels good working teachers. I really motivated your teaching skills , i love mathematics 🥰🥰🥰🥰🙏🙏🙏🙏🙏🙏🙏🙏
That was a fun one!! It's so cool to see problems that I initially have no idea how to approach, solved with such a simple, easy-to-understand technique! Thanks!
For the last one, I just multiplied both sides by x^2 + 3x + 1, which cancels out all the third and first degrees, and gives x^4 -7x^2 + 1 = 0. Rinse and repeat twice more. Basically the same process without dividing by x first.
As a 6th grader, our math club took the questions from the Stanford-Math League Tournament. Even though our group was comprised of 6th, 7th and 8th graders, I ended up winning with a t-shirt which has “Stanford-Math League Tournament”. I’ve already learned logarithms and even some calculus.
I don’t know if I ever got one right but I remember seeing questions like the last one all the time in these extracurricular math competitions lol The idea of dividing by x to get a binomial where 2ab removes the variable… so simple yet so crazy to come up with if you’re not used to these types of approaches
Hi Jay. If you want to see similar videos in math competitions, please consider th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
For the 3rd question, you can also notice that the k = sum of the roots of the equation. Then, sum the two roots of the staring equation (found via quadratic formula), each to the power of 8. This will be equal to k.
Yes you can compute the sum of the 8th power of the roots using a similar method of squaring, adding/subtracting repeatedly. But, bprp's method is quite elegant in a way.
Very fun! I did the first the same! For the second, I said that f(a)^2 = a can only be true if a is a perfect square, so a = 1,4,9,16,25,36. For a = 1, f(1) = 1^2, so we can get rid of that. And then 16 is the square of a square, so f(2) = 4 but f(16) = 4 breaks the bijectivity of f. No other is a 4th power so f(a) ≠ a^2 for a = 4, 9, 25, 36. For the third, I used the quadratic formula! Since x^2 - 3x + 1 = 0, we have x = (1/2)(3 ± sqrt(5)). And since x^16 - kx^8 + 1 = 0 is a quadratic in x^8, we get that x^8 = (1/2)(k ± sqrt(k^2 - 4)). raising (1/2)(3 ± sqrt(5)) to the 8th power (using the binomial theorem) gives (1/2)(2207 ± 987sqrt(5)) so (1/2)(2207 ± 987sqrt(5)) = (1/2)(k ± sqrt(k^2 ‐ 4)), which tells us k = 2207.
In my highschool in Poland we did problems like the first one on daily basis when we had logarythms in 11th grade but instead of doing it by factoring it out we did t=log2(n) then solve delta and substitute t1 and t2 and then solve for log2(n)
x^2 - 3·x + 1 = 0 x - 3 + 1/x = 0 x + 1/x = 3. We know that x^16 - k·x^8 + 1 = 0, so we want x^8 - k + 1/x^8 = 0, hence x^8 + 1/x^8 = k. The idea is to compute x^8 + 1/x^8 in terms of x + 1/x solely. This is doable, by considering the binomial theorem. This is the key to the question.
I see that a lot of these, high school juniors could solve, If they had the knowledge, I tried the problem with all the solutions less than 40, I actually did it and it was more logic than anything in my opinion
The trick of dividing by x is very nice but you end up with something very similar by just doing x^2 = 3x -1 square both sides x^4 = 9x^2 -2.3x + 1 and notice that you have an equation for 3x above: 3x = x^2 + 1 giving you x^4 = 7x^2 + 1 Now do that 2 more times and you get the answer. It's less elegant but doesn't require as big a flash of insight, I think.
Fiat Lux! As you were doing the question I thought in my head to make a common denominator on the left side and see that lon_2{n}=1. I was stunned the answer didn't line up. But then I saw you didn't switch the sign on when you moved 5log_2{n} to the other side. Maybe that will get my answer of n=1 and there is probably another one too. PS I made it to Cal because of you and Professor Leonard. TYSM
Hahaha, I love the method you used in the third problem. Usually, I would do x^2= 3x-1, then square both side X^4= 9x^2 -6x +1, but x^2 = 3x-1 X^4 = 9(3x-1) -6x +1 = 21x -8, square both sides again ...
Hi Peter. If you want to see similar videos in math competitions, please consider th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
That won't give you x^8. However, I did it somehow similar, and that gave me x^8. x^4 = 9x^2 - 6x + 1 and 3x = x² + 1 ==> x^4 = 7x² - 1 squaring gives x^8 = 49x^2 - 14x^2 + 1, and we have from the last stept 7x² = x^4 + 1 ==> x^8 = 47x^4 - 1 squaring again...
Squaring 47 in your head. There is a cool way of working out squares of numbers between 41 and 59. 47 is 50-3 so 47^2 is (25-3)(100)+3^2=2209. Why does this work? (50+n)^2=2500+2(50n)+n^2= 2500+100n+n^2=(25+n)(100)+n^2. So for example 56^2=3136.
Alternative Solution to #3: Assume a and b are the root of x² - 3x + 1 = 0. Then a and b must also be a root of x¹⁶ - kx⁸ + 1 = 0. So, a¹⁶ - ka⁸ + 1 = 0 b¹⁶ - kb⁸ + 1 = 0 --------------------------- - (a⁸ + b⁸)(a⁸ - b⁸) - k(a⁸ - b⁸) = 0 k = a⁸ + b⁸ = 2207
In the first problem: If you simply substitute “x” for “log (base 2) n”, you get (3/x) - (x/2) = (-5/2) Solve the quadratic and plug back in the 2 solutions (x = -1 & 6), setting them each = to “log (base 2) n” Same answer in the end. Just a less complex approach. You can actually solve it just by looking at it. It’s pretty clear what X is equal to. Regardless, more than one way to skin a cat. Per usual with the maths. 🤘😝🤘
@@natebobdog23 “EXCEPT you let x = log_2 (n)”. THATS the difference. Same answer-yet the slightest of variety in approaching and yielding that same solution. Your comment just defeated itself. Congrats. 👏👏👏 You also conveniently neglected the end of my original statement. Conveniently. No wonder you’re struggling over there. But hey, you’re learning! Good for you! I’ll be right here if u ever need me. ✌️
I did the first one another way, probably more complicated: () does not represent the base First, we can write the eqn as 1/log2^3(n) + 1/logn(2^-2) = -5/2 As logn(2^-2) =-2logn(2),,, And log2^3(n) =1/3log2(n) We get 3/log2(n) - 1/2logn(2) = -5/2 Then As logn(2) =1/log2(n) 6/log2(n) =log2(n) -5 Which simplifies down to (log2(n)) ^2 -5log2(n) -6=0 Which yields the solutions n=2^6 And n= 1/2
In the second problem, it's given that f is bijective. But I wonder if it makes sense to think about how you would prove such a function would actually be bijective.
Consider any sequence n, n², n⁴, n⁸..., where n itself is not a square. N is a disjoint union of such sequences, so it suffices to prove that f is a bijection on each sequence separately. On any such sequence the function is given by f(n^(2^k)) = n^(2^(k+1)) if k is even, n^(2^(k-1)) if k is odd. From this it is clear that f(f(x)) = x for each x, so f is its own inverse and it is bijective.
in the second question is there a generalized way to solve if, for example, it asked less than 500 or a larger number? how would we approach it? we surely can't go about counting every such pair...!?
Hey i have a question for you. This is from an IITJEE prep book by RD Sharma. Lo and behold: y'•y'"=3y"² I solved it (somehow lol) but later thought, "well doesn't mean that y', sqrt3 y, and y"' are always in GP?" That didn't quite work out well though. I urge you to do this!
Hi Shreeji. If you want to see similar videos in math competitions, please consider th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
hi i have a calculus question that i really hope you'll answer because it's annoying me so bad when you take the derivative of ln x or ln 2x or any natural log of nx it'll always be 1/x right? so why when we integrate 1/x do we just say the integral is ln x and not some ln ax because it can be any constant multiplied by that x please answer and ty
Let me ask you a simpler question, the derivative of x^2 is 2x, right? But so is the derivative of x^2+1, and x^2+2, so why then is the integral of 2x equal to x^2 and not x^2 +a, the answer? It is, that's we have the +C in indefinite Integrals, similarly as we have the integral of 1/x is lnx +C, where C is any real numbers, which is also equivalent to adding ln(a) for any positive a
let's compute the solutions of 1st equation : x1 = (3+sqrt(5))/2 and x2 = (3-sqrt(5))/2. Then k = x1^8 + x2^8 = 2207. (directly, or by Girard-Newton formula)
#questions For the third problem, could you do a u-sub with let u = x^8? Also, I don’t get why we can’t just do the binomial expansion (x^2-3x)^8 = 1^8
Because you have a ton of terms such as x^15, x^14 etc with a binomial expansion of your expression. Yeah you could set up a system so that all the unwanted coefficients equal zero but at best it will be too messy and time conseuming and at worst it may not even work (I’m not gonna give it a try).
I did get all three, but made it harder than needed for #1 (took common log instead of log base 2, then had to substitute k=log4 and factor a multivariable quadratic expression before back-substituting) and for #3 (actually solved for x and found the 8th power--you don't get more and more terms, but you do get larger and larger coefficients--then compared that to the quadratic equation result for the 16th-degree polynomial).
3rd question lai can\t just ake the two roots of the first expression, alpha and beta. so you get alpha = (3 +sqrt5) / 2 and beta = (3-sqrt5)/2 use substitution y = x^4 so the sum of roots of y = -k or ((3+sqrt5) / 2 )^4 + ((3-sqrt5)/2)^4 = 2207 or k =-2207
Who is the target participant for the SMT? I can see some good high school (grade 11 or so) students being able to do 1 and 3, but I don't think 2 would be suitable - maybe if "bijection" was replaced with "1-1 and onto".
Hi Gill. If you want to see similar videos in math competitions, please consider th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
But for the second question, can we rule out that f(1) =/= 1^2? Since it can go either way, isn't it a "maybe"? I would have written 4, and 5 if you count N = 1. But maybe I'm just stupid.
For the last one, x² - 3x + 1 = 0 => x¹⁶ - kx² + 1 = 0, I just said x = y⁸, which means (y⁸)² - 3(y⁸) + 1 = 0 y¹⁶ - 3y⁸ + 1 = 0 since the symbols arbitrary, we can just replace y with x x¹⁶ - 3x⁸ + 1 = 0 k = 3
The symbols are not arbitrary though, since we have specific values of x that solve the quadratic. You can check that the roots of the quadratic don't satisfy the new equation you wrote.
i dont understand the bijection problem. U assigned 2 to 4, so the 16 couldnt have the 4 again. But why didnt u assign the 4 to the 16 (in blue) and then the 16 to the 4? You u will end up with 5 numbers, not 4
I feel so dumb for not being able to solve 1🤦🏻♂️... I even though about change of bases but I was just like "nah I'll just give up and watch him solve it"
Hello, I was wondering, when is it okay to divide by x? Because my teacher always tells us, never divide by x because it could be 0, and I’ve mentioned to her that you could just plug in 0 for x to find out whether it is before you divide, but she just told me not to do it.
"she just told me not to do it" - that is obnoxious..... that's EXACTLY what you do.....say x²=4x is the equation.....x=0 is obviously a solution.....instead of dividing by x, you do: x²-4x=0 x(x-4)=0 x=0 or x-4=0 --> x=4 Hence, x=0 or 4 .....in equations like x²-5x+6=0 there is no way x=0 is a solution ....so if dividing by x helps you get a special useful form of the equation (like in this video), there is no reason in the world not to do it...... BTW what grade are you in rn that your teacher is so restrictive?
@@aayushdhungana360 Thank you! I am in 9th grade right now, however I am taking algebra 2 honors, so my classmates are almost all 10th graders who are slightly advanced, and the actual class is an 11th grade class. I hope that made sense and thank you for the explanation!
I think most teachers simply get desperate after teaching for years and decades and always having to correct errors like: x² = x is the same as x = 1. So they simply tell _all_ pupils that one should _never_ divide by x. And don't mention that there are cases where this _is_ allowed and even helpful...
Could anyone suggest a method to prepare for this year's SMT? Mainly I struggle in combinatorics so would be nice if someone could suggest a good source.
Hi Hesh. We have a lot of videos from SMT and other countries in our channel. please consider th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
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Hi! I'm the organizer for SMT 2022! I'm glad you enjoyed our tiebreaker problems, and thank you for reviewing them!!
For those of you who are high school students we encourage you to participate in this year's iteration of the contest, we're very excited :)
Wow, great to have you here! As I said in both of my SMT videos, you guys have some really nice problems. Keep up the great work!
@@blackpenredpen amogus
@@theuserings no
@@kepler4192 👽
Yes, we would like to.
For the second question, removing all of the high level notation you could say, how many numbers less than 40 are perfect squares but not perfect fourth powers.
Yes!!! That’s a very simple way to interpret it. The hard part was definitely all the fancy math language.
@@blackpenredpen Thanks. Also to finish the interpretation, it comes down to finding how many perfect squares there are below 40 and how many prefect fourth powers below forty and subtracting the later from the former
@@cosmicvoidtree Alternatively, you can find/generate a list of squares of numbers that are not squares themselves. This works for n≤40 because we don't really have to deal with sixth powers and etc.
Wonderful insight! Nice.
Luckily there are no perfect 8th powers less than 40 or what you have said would not hold. Eg. 256 is 4^4 but it would satisfy the condition (f(256) can = 16 as f(4) = 2) perfect 16th powers don't hold as f(65536) != 256 since that is f(16). Will hold for 32nd powers but not 64th and so on.
Another SMT problem (rotating y=x^2) : th-cam.com/video/gYAQg7xn-Xo/w-d-xo.html
cant you just do the quardtaic formula for the last one and then plug in x?
@@yodaimpostor4781 if I'm assuming what I understood from that
I got (3+-√5) ÷ 2 for the first equation
I am from India , i study in 12th standard. when i study integration, i have some questions having a problem then i search in TH-cam then i show your channels good working teachers. I really motivated your teaching skills , i love mathematics 🥰🥰🥰🥰🙏🙏🙏🙏🙏🙏🙏🙏
That was a fun one!! It's so cool to see problems that I initially have no idea how to approach, solved with such a simple, easy-to-understand technique! Thanks!
lol ever since i picked up dummit and foote the only things that come to mind whenever i see or hear "algebra" are rings, groups, fields, modules etc.
@@shamcallado8947 haha yea
I wouldn’t be able to solve these in a half hour but you just make it looks so easy and make so much sense. Ty!
I really enjoy your channel. I don't understand half of what you are saying, but it stretches my brain in the right direction.
For the last one, I just multiplied both sides by x^2 + 3x + 1, which cancels out all the third and first degrees, and gives x^4 -7x^2 + 1 = 0. Rinse and repeat twice more. Basically the same process without dividing by x first.
My 6th graders would probably wonder why lumber is involved in these 'log' math problems.
As a 6th grader, our math club took the questions from the Stanford-Math League Tournament. Even though our group was comprised of 6th, 7th and 8th graders, I ended up winning with a t-shirt which has “Stanford-Math League Tournament”. I’ve already learned logarithms and even some calculus.
@@thegoldlightning 🏅
@@thegoldlightning wut
@@thegoldlightningwait 8th graders don't learn log and calc yet?
@@windowsxpmemesandstufflol a lot of people there are in the top advanced program at our school. Some do even higher outside.
I don’t know if I ever got one right but I remember seeing questions like the last one all the time in these extracurricular math competitions lol
The idea of dividing by x to get a binomial where 2ab removes the variable… so simple yet so crazy to come up with if you’re not used to these types of approaches
The best almost 15 mins. spent today. Thank you for the video.
Hi Jay. If you want to see similar videos in math competitions, please consider
th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
If i was in a math class and the 2nd question was in the test... I'd have passed out 😂
Edit: how you tackled the 3rd problem was satisfying. No lie.
That is the kinda problems I, as a math major, get asked as homework.
the second one was by far the easiest imo
@@bonjour7209 definitely.
For the 3rd question, you can also notice that the k = sum of the roots of the equation. Then, sum the two roots of the staring equation (found via quadratic formula), each to the power of 8. This will be equal to k.
Yes you can compute the sum of the 8th power of the roots using a similar method of squaring, adding/subtracting repeatedly. But, bprp's method is quite elegant in a way.
Very fun!
I did the first the same!
For the second, I said that f(a)^2 = a can only be true if a is a perfect square, so a = 1,4,9,16,25,36.
For a = 1, f(1) = 1^2, so we can get rid of that. And then 16 is the square of a square, so f(2) = 4 but f(16) = 4 breaks the bijectivity of f. No other is a 4th power so f(a) ≠ a^2 for a = 4, 9, 25, 36.
For the third, I used the quadratic formula! Since x^2 - 3x + 1 = 0, we have x = (1/2)(3 ± sqrt(5)). And since x^16 - kx^8 + 1 = 0 is a quadratic in x^8, we get that x^8 = (1/2)(k ± sqrt(k^2 - 4)).
raising (1/2)(3 ± sqrt(5)) to the 8th power (using the binomial theorem) gives (1/2)(2207 ± 987sqrt(5)) so
(1/2)(2207 ± 987sqrt(5)) = (1/2)(k ± sqrt(k^2 ‐ 4)), which tells us k = 2207.
this man explained the entire paper within 15 minutes time limit lol
I love the first question, it helps my understanding for logs tats coming for my mocks, thanks!
That 3rd problem was amazing!!
I am glad that I was able to do the 3rd problem 😁
Nice approach to solve problems 👍
I tried doing it in 15 minutes and messed everything up with haste xD. Thank you for the solving methods exposed in this video.
Your channel is truly awesome! I just wish you diversify the kind of problems you solve to include more stuff outside calculus :) Rock on!
Amazing
All three are really great problems!
More math competition thanks Prof)
In my highschool in Poland we did problems like the first one on daily basis when we had logarythms in 11th grade but instead of doing it by factoring it out we did t=log2(n) then solve delta and substitute t1 and t2 and then solve for log2(n)
Wah new look 🤘🤘
I don't know why I'm smiling on entire video, especially on the last part. Please help me lol.
Takk!
x^2 - 3·x + 1 = 0 x - 3 + 1/x = 0 x + 1/x = 3. We know that x^16 - k·x^8 + 1 = 0, so we want x^8 - k + 1/x^8 = 0, hence x^8 + 1/x^8 = k. The idea is to compute x^8 + 1/x^8 in terms of x + 1/x solely. This is doable, by considering the binomial theorem. This is the key to the question.
I see that a lot of these, high school juniors could solve, If they had the knowledge, I tried the problem with all the solutions less than 40, I actually did it and it was more logic than anything in my opinion
Satisfying explaination ..
Beautiful, that was quite fun. Well explained, and funny too
The trick of dividing by x is very nice but you end up with something very similar by just doing
x^2 = 3x -1
square both sides
x^4 = 9x^2 -2.3x + 1
and notice that you have an equation for 3x above: 3x = x^2 + 1
giving you x^4 = 7x^2 + 1
Now do that 2 more times and you get the answer.
It's less elegant but doesn't require as big a flash of insight, I think.
Quick tip you can use the asterisk symbol instead of decimal point for multiplication like 2*4
Fiat Lux! As you were doing the question I thought in my head to make a common denominator on the left side and see that lon_2{n}=1. I was stunned the answer didn't line up. But then I saw you didn't switch the sign on when you moved 5log_2{n} to the other side. Maybe that will get my answer of n=1 and there is probably another one too. PS I made it to Cal because of you and Professor Leonard. TYSM
BPRP: *Points at 9*
Also BPRP: "Five"
9:24
😂
I liked the 3rd question and the 1st question a lot.
Proud of you sir i also want to be an mathematician like you👍👍👍👍🇮🇳🇮🇳🇮🇳
Hahaha, I love the method you used in the third problem.
Usually, I would do x^2= 3x-1, then square both side
X^4= 9x^2 -6x +1, but x^2 = 3x-1
X^4 = 9(3x-1) -6x +1 = 21x -8, square both sides again
...
Hi Peter. If you want to see similar videos in math competitions, please consider
th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
That won't give you x^8. However, I did it somehow similar, and that gave me x^8.
x^4 = 9x^2 - 6x + 1 and 3x = x² + 1 ==> x^4 = 7x² - 1
squaring gives x^8 = 49x^2 - 14x^2 + 1, and we have from the last stept 7x² = x^4 + 1 ==> x^8 = 47x^4 - 1
squaring again...
Squaring 47 in your head. There is a cool way of working out squares of numbers between 41 and 59. 47 is 50-3 so 47^2 is (25-3)(100)+3^2=2209. Why does this work? (50+n)^2=2500+2(50n)+n^2= 2500+100n+n^2=(25+n)(100)+n^2. So for example 56^2=3136.
Beautiful!!
that last question was satisfying
UC Berkeley ?? That’s awesome
I looked at the problem and thought u-sub by defining int(f(x)dx)=F(x). Pretty much the same thing you did but you get nice algebraic steps
Alternative Solution to #3:
Assume a and b are the root of x² - 3x + 1 = 0. Then a and b must also be a root of x¹⁶ - kx⁸ + 1 = 0.
So,
a¹⁶ - ka⁸ + 1 = 0
b¹⁶ - kb⁸ + 1 = 0
--------------------------- -
(a⁸ + b⁸)(a⁸ - b⁸) - k(a⁸ - b⁸) = 0
k = a⁸ + b⁸ = 2207
That’s beautiful!
How did you get 2207 in the last step? Do you use quadratic formula to find a and b?
@@topgearfan2596 a and b were by assumption the roots of the first quadratic, so you just have to solve that one
@@Prxwler Yeah I get it. I feel like calculating that is more tedious than the method presented in the video.
Another excellent video! Thanks Professor!
i love knowing every step used in the video but not being able to string them together to get a solution:))))))))))
In the first problem:
If you simply substitute “x” for “log (base 2) n”, you get (3/x) - (x/2) = (-5/2)
Solve the quadratic and plug back in the 2 solutions (x = -1 & 6), setting them each = to “log (base 2) n”
Same answer in the end. Just a less complex approach. You can actually solve it just by looking at it. It’s pretty clear what X is equal to. Regardless, more than one way to skin a cat. Per usual with the maths. 🤘😝🤘
this isn't a different way to do the problem. it's literally the same exact thing except you let x = log_2(n). it doesn't save any time
@@natebobdog23 “EXCEPT you let x = log_2 (n)”. THATS the difference. Same answer-yet the slightest of variety in approaching and yielding that same solution.
Your comment just defeated itself. Congrats. 👏👏👏
You also conveniently neglected the end of my original statement. Conveniently. No wonder you’re struggling over there.
But hey, you’re learning! Good for you! I’ll be right here if u ever need me. ✌️
I solved only first one log question and rest two no idea
But i. Amazed 😮😲 by seeing solution of 3 rd one
I did the first one another way, probably more complicated:
() does not represent the base
First, we can write the eqn as
1/log2^3(n) + 1/logn(2^-2) = -5/2
As logn(2^-2) =-2logn(2),,,
And log2^3(n) =1/3log2(n)
We get 3/log2(n) - 1/2logn(2) = -5/2
Then
As logn(2) =1/log2(n)
6/log2(n) =log2(n) -5
Which simplifies down to
(log2(n)) ^2 -5log2(n) -6=0
Which yields the solutions n=2^6
And n= 1/2
Solved this equation in a similar way to yours. Thought it'd be harder, but that's school grade maths
But notice that solution n=1/2 is not integer, so this equation has only one solution n=2^6=64
Not -1/2, but 1/2
In the second problem, it's given that f is bijective. But I wonder if it makes sense to think about how you would prove such a function would actually be bijective.
Sure. I imagine it would be a pretty basic proof by cases. "Suppose f(a) = b. Then either b = a^2 or b = sqrt(a)." Follow from there.
Consider any sequence n, n², n⁴, n⁸..., where n itself is not a square. N is a disjoint union of such sequences, so it suffices to prove that f is a bijection on each sequence separately. On any such sequence the function is given by f(n^(2^k)) = n^(2^(k+1)) if k is even, n^(2^(k-1)) if k is odd. From this it is clear that f(f(x)) = x for each x, so f is its own inverse and it is bijective.
First was nice bro.😍
in the second question is there a generalized way to solve if, for example, it asked less than 500 or a larger number? how would we approach it? we surely can't go about counting every such pair...!?
yeah just find numbers that have natural square roots but not natural fourth roots
so take the number of squares and subtract the number of tesseracts
for the 3rd question, cant you find out the value of x from the quadratic and then substitute in the 2nd eq?
Love from india🇮🇳🇮🇳🇮🇳🇮🇳
Hey i have a question for you. This is from an IITJEE prep book by RD Sharma. Lo and behold:
y'•y'"=3y"²
I solved it (somehow lol) but later thought, "well doesn't mean that y', sqrt3 y, and y"' are always in GP?" That didn't quite work out well though. I urge you to do this!
I knew there had to be a more elegant solution to 3 than just solving the quadratic and making the second expression an equation for k
Very good
You should try to differentiate the function f(x)=x!
Oh I am lovin' it!
Hi Shreeji. If you want to see similar videos in math competitions, please consider
th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
Awesome!
8:24 this should be a meme! (The pause)
hi i have a calculus question that i really hope you'll answer because it's annoying me so bad
when you take the derivative of ln x or ln 2x or any natural log of nx it'll always be 1/x right? so why when we integrate 1/x do we just say the integral is ln x and not some ln ax because it can be any constant multiplied by that x
please answer and ty
bc of the +C
^ This. I was going to say, "We already do!"
Let me ask you a simpler question, the derivative of x^2 is 2x, right? But so is the derivative of x^2+1, and x^2+2, so why then is the integral of 2x equal to x^2 and not x^2 +a, the answer? It is, that's we have the +C in indefinite Integrals, similarly as we have the integral of 1/x is lnx +C, where C is any real numbers, which is also equivalent to adding ln(a) for any positive a
I am very glad while you change blackpenredpen
let's compute the solutions of 1st equation : x1 = (3+sqrt(5))/2 and x2 = (3-sqrt(5))/2. Then k = x1^8 + x2^8 = 2207. (directly, or by Girard-Newton formula)
#questions For the third problem, could you do a u-sub with let u = x^8? Also, I don’t get why we can’t just do the binomial expansion (x^2-3x)^8 = 1^8
Because you have a ton of terms such as x^15, x^14 etc with a binomial expansion of your expression. Yeah you could set up a system so that all the unwanted coefficients equal zero but at best it will be too messy and time conseuming and at worst it may not even work (I’m not gonna give it a try).
I did get all three, but made it harder than needed for #1 (took common log instead of log base 2, then had to substitute k=log4 and factor a multivariable quadratic expression before back-substituting) and for #3 (actually solved for x and found the 8th power--you don't get more and more terms, but you do get larger and larger coefficients--then compared that to the quadratic equation result for the 16th-degree polynomial).
Cool way to do the last one.
Thanks.
In second question 1 can be taken in both a² or √a form ...so the question should have mentioned least number of numbers less than 40
The question is fine...
Make a video on how to get the value of d/dx[erf(x)]
A very nice sol !
I didnt know Saitama took over this channel
Busy man. He's been doing videos for CalisthenicMovement for a long time now.
3rd question lai can\t just ake the two roots of the first expression, alpha and beta. so you get
alpha = (3 +sqrt5) / 2 and beta = (3-sqrt5)/2 use substitution y = x^4 so the sum of roots of y = -k or
((3+sqrt5) / 2 )^4 + ((3-sqrt5)/2)^4 = 2207 or k =-2207
that was fun
To do 47^2 in your head quickly, try this: (50-3)^2 = 2500 - 300 + 9 =2209
2:47 Shouldn't it be 5(log2(n)) instead of -5(log2(n)) because you're shifting to the other side?
nah the -5log2(n) wasnt the one he was shifting, he shifted -(log2(n))² and 6
Who is the target participant for the SMT? I can see some good high school (grade 11 or so) students being able to do 1 and 3, but I don't think 2 would be suitable - maybe if "bijection" was replaced with "1-1 and onto".
You are amazing
Hi Gill. If you want to see similar videos in math competitions, please consider
th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
I like #3
Me too
You look like a math master just came down from the mountain or something XD.
Sorry but 6*^ isn't it 36? I did not fully understand the step from logn2 to finding n. Thank you for your answer
If log_2(n) = 6, then n = 2^6, not n = 6^2.
Can you integrate (1/dx) ?
that's not an integral. dx isn't a term to be wielded arbitrarily
Tutorial on partial differentiation please🥺
Nice one
The first two problems are somewhat doable, but I would absolutely have no chance tackling the last problem.
But for the second question, can we rule out that f(1) =/= 1^2? Since it can go either way, isn't it a "maybe"? I would have written 4, and 5 if you count N = 1. But maybe I'm just stupid.
Huh? For a = 1, we obviously have f(a) = a². So this is not a number we have to count when looking for all numbers for which f(a) != a².
Try solving ln(i)
For question 1, that was a 5? I thought that was an S, and figured they were asking about arithmetic progressions or something
Thank you. guide me about: Integral (e^-x)/x Thank you
can we use the quadratic formula on the 3rd question?
yes but what a nightmare that would be
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Dang, I was lost as to what that second question had been asking.
Is there an even quicker way to solve the 2nd one and if the 40 was replaced by 400 or something ........just wanted to ask ........pls reply
For the last one, x² - 3x + 1 = 0 => x¹⁶ - kx² + 1 = 0, I just said x = y⁸, which means
(y⁸)² - 3(y⁸) + 1 = 0
y¹⁶ - 3y⁸ + 1 = 0
since the symbols arbitrary, we can just replace y with x
x¹⁶ - 3x⁸ + 1 = 0
k = 3
The symbols are not arbitrary though, since we have specific values of x that solve the quadratic. You can check that the roots of the quadratic don't satisfy the new equation you wrote.
i dont understand the bijection problem. U assigned 2 to 4, so the 16 couldnt have the 4 again. But why didnt u assign the 4 to the 16 (in blue) and then the 16 to the 4? You u will end up with 5 numbers, not 4
I feel so dumb for not being able to solve 1🤦🏻♂️... I even though about change of bases but I was just like "nah I'll just give up and watch him solve it"
6:17 why isn’t zero in there? 😢
4:48
For the second question, can we omit 0 as a natural number?
No 0 is not a natural number
@@nurssa5991 if you consider the peano axioms, it is.
Hello, I was wondering, when is it okay to divide by x? Because my teacher always tells us, never divide by x because it could be 0, and I’ve mentioned to her that you could just plug in 0 for x to find out whether it is before you divide, but she just told me not to do it.
"she just told me not to do it" - that is obnoxious..... that's EXACTLY what you do.....say x²=4x is the equation.....x=0 is obviously a solution.....instead of dividing by x, you do:
x²-4x=0
x(x-4)=0
x=0 or x-4=0 --> x=4
Hence, x=0 or 4
.....in equations like x²-5x+6=0 there is no way x=0 is a solution ....so if dividing by x helps you get a special useful form of the equation (like in this video), there is no reason in the world not to do it......
BTW what grade are you in rn that your teacher is so restrictive?
@@aayushdhungana360 Thank you! I am in 9th grade right now, however I am taking algebra 2 honors, so my classmates are almost all 10th graders who are slightly advanced, and the actual class is an 11th grade class. I hope that made sense and thank you for the explanation!
In the third problem, you can divide by x because 0 doesn't satisfy the quadratic.
I think most teachers simply get desperate after teaching for years and decades and always having to correct errors like: x² = x is the same as x = 1. So they simply tell _all_ pupils that one should _never_ divide by x. And don't mention that there are cases where this _is_ allowed and even helpful...
Could anyone suggest a method to prepare for this year's SMT? Mainly I struggle in combinatorics so would be nice if someone could suggest a good source.
Hi Hesh. We have a lot of videos from SMT and other countries in our channel. please consider
th-cam.com/video/rkzxdMFEEtw/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
why is it allowed to divide by x in the last question?