Swedish Mathematics Olympiad | 2002 Question 4
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- Опубліковано 27 чер 2024
- We look at a solution to a number theory problem from the 2002 Sweden Mathematics olympiad.
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8:35
underlining inequality signs is usually not the best way to emphasize them ;)
I think that's why he's using a different colour to do this:)
I will draw a line through this equals sign to emphasize it ≠
The best way to emphasize a capital "I" is to draw a horizontal line on top of it. At least, T think so.
@@NuisanceMan / think extreme italics work well.
Can't believe I'm watching this for entertainment
lool me too
I'm watching this to fall asleep
ikr, it's super fun
Lol me too
That's because he is (somewhat) entertaining
I have noticed that if there is an easy way to prove something, then Michael will point it out and then prove it a harder, but more entertaining and often a more general way :-).
That's true. In mathematics important theorems are often proved in different ways using different techniques. Examples are e.g. theorem of pythagoras or the law of quadratic reciprocity. By this you can understand relations between different areas of mathematics.
All he did in this proof was use Calculus instead of induction. You could even say he did it the easy way.
@@wospy1091
Induction is easier.
@@davidbrisbane7206 not for everyone but it's clearly more elementary
@@gabriel7233 Additionally, if you know both techniques, the work for induction in this case is easier. 2^(n-7) > 128n > 128 implies 2^((n+1)-7)=2*2^(n-7)>128n+128=128(n+1).
I like the method that was chosen to prove the inequality - it proves it not only for natural numbers, but for all real numbers as well.
Same!
This is a high school math problem, so derivatives are an overkill. There is a simpler way, and for a kid trying these problems that solution is more useful.
May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.
@@vedhase2370 The first thing you need to know are the statement gaving, by the problem.
@@vedhase2370 cuz let's say 1=n^(1/n-7)
Then taking log on bs we get
0=(1/n-7)xlog(n)
Now (1/n-7) is non zero as n>=8
=> logn =0
=> n=1
Which is a contradiction. QED
What I've learnt from this channel is to start every math problem by considering case n=1.
He explained so many stuffs in a single video. This shows how different math problems are related to each other!!
I think that this is one of the beauty of maths. There's so much ways to interpret something that you can always found a new one.
For example, one of the most relevant theorems of math is The Fundamental Theorem of Algebra and one of the most important consequences is the fact that links algebra with the study of functions (calculus).
@@ckeimel True to say
May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.
@@vedhase2370 wich part of the solution troubles you?
N^(1/n-7)>1
The proof of this is obvious and left as an exercise to the reader
It is obvious. a^x is a strictly increasing function for a>1 and equals 1 at only x=0.
*n^(1/(n-7))>1
I don't quite get the fact he is doing too much to prove that sentence. There is no number greater than 8 multiplied by itself n-7 times going to give a number lesser or equal than 1, enough proof.
@@josem138 Not multiplied by itself (n-7) times, but 1/(n-7) times. But yes the claim is still trivial.
@@AnindyaMahajan I don't see why. a^y means a is a real constant and y is the input. Here a = x (so a is not a constant) and y = 1/(x-7). After x>=8, the function (x^(1/(x-7))) is strictly decreasing. Apart from looking at the graph what other proof is there?
Edit: Actually after giving it some thought I see it is obvious since
x^(1/(x-7)) = 1
when
1/(x-7) = 0
which leads to:
1 = 0
So
x^(1/(x-7)) is never 1 and since the function is strictly decreasing it will never be less than 1 either.
Fun fact, we know that 0
Ha ha! How?
@@35571113 You take 5th root of pi and 17 by hand, everybody is so impressed they ignore the fact that you didn't prove anything yet.
@@volodyanarchist lmao
May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.
@@vedhase2370 I was wondering that too.
Showing the "obvious" parts and "the long way" is so so good. It's how we all learn. Very good job!
great as always. love when you just underline something and mark it as "clear." that's how you know you're watching a real mathematician
Although it was excluded right at the beginning, n=1 would be a solution as 1^x=1\in N for any x.
Without induction, in my opinion the easiest way is to use binomial coefficients. n < 2^(n-7) for n >=11 is equivalent to n+7 < 2^n for n>=4. We check n=4 and n=5 by hand and for n>5 we have
2^n = (1+1)^n > (n choose 0) + (n choose 1) + (n choose n-1) + (n choose n) = 1 + n + n + 1 = 2 + 2n > n + 7.
beautiful idea
I love this channel, awesome the way you spent a several seconds and a couple of sentences in proving that 10 is not a perfect cube.
This was a really lovely problem. Well presented Michael, loving your stuff
Man who I wish I would have had such videos in high-school almost 20 years ago.
I felt this much easier than any other question
For example question 6 of IMO 2020
Yes
Bcz it is
@@roboto12345 I can only solve Q1 of imo2020
Finding the solutions is easy; the hard part is proving that there are no other solutions. It may be easy enough to grasp intuitively, but math requires more proof than just intuition.
7:10 is not equivalent, but you only need the reverse implication here. f(x) >0 everywhere => f(n) > 0 for natural numbers
This channel is such a gem, more people should be watching this stuff.
No, a person.can watch what they want on youtube.
Abacus, should and must are different
Abacus , ahh I see. Thanks for enlightening me.
@@shurik3nz346 I wasn't advocating strapping people to chairs and prying their eyes open Clock of Orange style; I meant that people would benefit but most people are ignorant of this channel's existence.
@@stevenwilson5556 haha alright
I probably would have turned back after taking the derivative. Thanks for an unusually clear and direct work-through!
Me: Sees 1/n-7
*Ezee*
Me: Sees n just below it
*crap*
Same
1/n-7 is integer only for n=6,8
Dorulo ili kvo
May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.
@@vedhase2370 very late,but since 1/(n-7) is positive, n^(1/(n-7)) > n^(0) = 1
when i first looked at the problem, i found it pretty clear that either the exponents needs to be 1 (which gives 8 as a solution) or n needs to be a perfect power (because otherwise we would never get an integer by taking an (n-7)th root). the rest would be quite similar to what you did, just showing that f(x)=x^(1/(x-7)) is decreasing for x>=8.
The work is pleasingly clear and labeled
Love your videos, participating in my first olympiad in a month or two
Good luck!
What olympiad are you attending?
@@James_Moton Norwegian one
All the best✌️👍👍
Good luck!
I tried to solve this from the thumbnail, so I missed that we were solving over the natural numbers. I got n = - W(-1, -ln(z) / z^7 ) / ln(z) where z is an integer and W is the Lambert W function aka the ProductLog. Apparently, we need the -1 branch.
n. n^(1/n-7)
10.375 2
9 3
8.54777 4
8.31611 5
8.17246 6
8.07331 7
8 8
7.94315 9
7.89749 10
7.85982 11
7.82809 12
7.80089 13
7.77725 14
7.75646 15
7.73799 16...
Me after seeing thumbnail: "Oh that's easy, it's 8"
Me after clicking on video: "Ah, the old bait and switch"
Finally, something I managed to solve before watching the video. Beautiful solution.
Amazing explanation
IMO is happening now, it would be fun to feature some of the questions from this year
That thinking of searching for n>11 is really brilliant.
A backflip would have been great.
I love these videos. It makes me kinda sad I didn’t go for pure instead of applied math.
He can't do backflips in videos any more, due to discontinuities outside the domain of the video in the neighborhood of his head. (That is, hanging microphones.)
@@iabervon Excellent !
First impression: Holy moly, that's gonna be hard
Second impression: But wait, that's gonna be impossible very soon
Solution: It's actually just the first 2 allowed numbers
I'd say the perceived difficulty of this exercise is a strictly decreasing function of time.
It has a spike like x^-2, a discontinuity that occurs when he mentions that we're not using induction
Well if feel like doing a challenging prob
You can try this
ua-cam.com/video/igdy05LZj90/v-deo.html
I had the opposite experience. My immediate impression (even before I noticed the ≥ 8 condition) was 8 works, 9 works, and then that's it. The more he explained why higher numbers don't work, the more I started to think that there's going to be some other number, probably a multiple of 7, where it equals 1, but that it would be really hard to prove.
I did another induction:
n < 2^(n-7) is true for n=11, then we can say:
2^(n+1-7) = 2*2^(n-7) > 2n = n+n > n+1 (for n>=11)
So therefore: n+1
The problem with underlining an inequality to emphasize its strictness is then you make it nonstrict 🤣
May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.
@@vedhase2370 Maybe he considered the fact that 1^n = 1 for all real values of n not worth commenting on.
@@vedhase2370 well, any number to the zero ith power is equal to one, ex 3^0 = 1 and 7^0 = 1. The problem states that n is greater than or equal to eight. 1/(n-7) is always going to be greater than 0 if n is greater than eight. That means that the equation basically boils down to n^(a number greater than 0).
In order for the whole equation to be less then or equal to one, n must be raised to the power of 0, or less, but since 1/(n-7) is always greater than zero due to the parameters of the question, we know that n^(1/(n-7)) is greater than one.
Hope this clears things up
I thought of that too, he almost managed to confuse me!
Excellent! You did your number eight like me until I were 8 years old. I remember my teacher scolded me for it. So far, I do my eights "normally". It was just to annoy you but you're doing a great job!
An arguably quicker way of proving 2^n>128n could have been by plugging n=11 and checking that 2^n is indeed larger and then seeing that the slope of 2^n at n=11 is 1024*11 (chain rule) and that the slope of 128n is 128. Plus the slope of 2^n only keeps on growing whereas 128n has a constant (and lower) slope. Combining f(11)>g(11) and f'(x)>g'(x) for x in [11,∞), you know that f(x)>g(x) ,if f(x)=2^x and g(x)=128x
n = 1 is a soln
trivial
good point, missed that.
The problem states that n>=8.
@@AnkhArcRod But he says the reason the problem says n>=8 is that it doesn't work for n
AnkhArcRod yea, but he said that n=1 wasn't a solution when he was talking about why the bounds of the problem were n>=8
One of the best: creative and not too hard!
this guy has neil patrick harris energy, i love it
I just directly differentiated the original function extended to the reals, showed that was negative for n>=8 and then by exploration found that 8&9 were the only solutions.
Nicely explained and question was good
8:20 Another nice way to see 2^11 - 128(11) > 0 is to factor a 2^7 out of each term and observe it's true iff (2^4 - 11) > 0. (Even if it's not memorized, 128=2^7 was used earlier in the problem.)
I believe you could also use binomial expansion for 2^m to show the inequality as well. Use 2^m = SUM( m choose k) from k=0 to m and use however many terms you want to show that 2^m > 128m
My solution substitute x = n - 7 and n = a^x for some integer a to obtain a^x = x + 7. Then obviously for x >= 4 we need a < 2 and we only need to evaluate x = 1,2,3. The only satisfactory pairs (a,x) are (8,1) and (3,2)
Yeah Sir please do IMO questions also like IMO 2003 P2 number theory and algebra questions
Fun problem
Yeah that would be great
From e < 4, we get 1 < 2 ln 2 by taking the natural log on both sides, and then dividing both sides by 2 yields the inequality.
the equivalence at 7:30 is false.
But the implication b=>a is enough to prove the result (b result with x in R).
Why the equivalence is false? I had seen other comments claiming the same thing, one of them said that f(x) is negative for some real numbers, but how can this be true for x>= 11?
@@Nicolas-zf3pv It's not clear that if f(n)>0 for n>=11, n integer then f(x)>0 for x>11, x real. In this case the equivalency holds because f is continuous and monotonous. But it doesn't hold for any f. He proves later that it's monotonous, so with that bit from the proof (and the fact that the function is clearly continuous) he could prove that the equivalency holds as well. He doesn't need to, because the other direction is sufficient. But the way he writes it, it's not sufficient.
after watching Numberphile: *I NOW KNOW AND UNDERSTAND MATH! I CAN SOLVE ANY MATH PROBLEMS NOW, I CAN DO IT!*
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Right? This is nuts
In order for it to be an element of N then n should be written as n=m^(n-7), with m>=3>e cuz n>8>e (the case of n=8) is trivial) also m=n^m when we play around with the inequality and by replacing values we get n
Wonderful video, always a pleasure to watch. I was wondering if you Could make a video on the “points in the space” type of problem (I believe it’s called misc but I’m not sure). An example to this type of problem is “There are a 1000 points in the space, prove you could always pass a line such that half the points would be on one side and the other half on the other side”.
Hey, Michael. Can you please make a video on problem #6 from the 2009 IMO.
Love number theory
One of the few number theory problems I silved on my own
Super awesome !
I like very much your olimpiad problems
Could you please do IMO 2003 P2 its a cool question
Nice suggestion
There’s a lot to learn from that problem
Would be grateful if Michael could solve it for us
Yeah great , I hope sir does it
Fun problem
Thank you
Thank you very much!
What is that called on your hoodie? That is something I found myself but I don't know what it is called or why it works.
Well, generally in questions these inequalities are hard to prove but this was not the case in this problem. This was really a nice problem.
I proved the inequality by swapping n-7=k and proving that 2^k>k+7 for k>=4 - which is easy by expanding (1+1)^k using the binomial theorem and keeping only the kC0, kC1, kC2, kC(n-1)and kCn terms , giving the inequality 2^k>=2+2k+(k)(k-1)/2 > 8+2k for k>=4 >k+7
What about n=1? It's also a solution since 1^(-1/6)=1...
I used a slightly different approach. I took the derivative of x^(1/(x-7)), and showed it was decreasing for all x greater than or equal to 8. Since we know 1 is our lower bound, and the n=11 case is less than 2, then we know we only need to check 8, 9, and 10
By thinking of it as an n-7th root you can also show that n>=2^(n-7) as n!=1 and hence 8
I really like the ending you make in each video. That's a good place to stop. It feels like maths has an infinite domain one can explore. It's like the opposite of claustrophobia.
Hi Septagram. If you are interested in math competitions, please consider our channel ua-cam.com/users/SQRTime. Hope to see you 😊
This is a nice solution.
set f(x) =x^(1/x-7)
arguing the monotony of the function on [8,+∞) let a,b≥8 be real numbers s.t a1/b-7 ⇒ f(a)>f(b)
⇒the function is strictly decreasing on [8,+∞). therefore max(f(x))=f(8)=8.
We have 0
Please correctly state the question by adding: n ∈ N (natural number). Otherwise, there are more real number solutions for n, since the real function f(x) = x ^ (1/x-7) is a continuous function, and we have f(8)=8 and f(11)
11:19 : After reading Richard Feynman's brilliant "Surely You're Joking, Mr. Feynman!" I can't forget that ln 2 = 0.69... which is clearly bigger than 0.5
it's viable also for n=1 as 1^any number=1 which is an integer.
thank you; brilliant
Unless n is in natural, there has to be a solution for f(n) = n^(1/(n-7)) = 2. Note that f is monotonic, continous in (7,+inf> and has an asymptote at f = 1.
Unusual use of the word "clear"
Wow
open my mind
If n^(1/(n-7)) is an integer, then n = m^(n-7) for some integer m greater than or equal to 2. Take a prime divisor p of n and let e be its exponent in the factorization of n. Since n = m^(n-7), the exponent of p in the factorization of m^(n-7) is greater than or equal to n-7 and hence so is a. It thus follows that 2^a - 7 ≦ n - 7 ≦ a. Therefore 2^a - a ≦ 7．Because 2^x - x is increasing on x ≧ 1 and 2^4 - 4 > 7, we have the inequality a ≦ 3; Hence n ≦ 10.
6:55 I don't think it is equivalent because you could take such a function that it would be positive on integers but not always positive for real numbers but anyway that doesn't change the solution
well if you include limit of n as it approaches infinity, then it would equal 1
Let n=b^x. Then we need to observe when b^[x/(b^x-7)] is an integer. That means b^x-7 divides x. For that to be possible, b^x-7 =< x. If x=1, b^x=8. If x>1, then 2^x-7 =< x so x
The function f(x)=x^{1/(x-7)}=e^{ln x/(x-7)} is greater than 1 for x>7 and monotone decreasing, since
f'(x)=f(x)/[x(x-7)^2]*[-7-x(ln x -1)]10 we have 1
Thanks for the great videos!
Simpler final proof (edit: oops, it's the same):
ln(2)>1/2
2>e^(1/2)
4>e
Lol, it's the same proof he used, nothing different.
@@MarcoMate87 Oh, you're right. I just saw that he started differently and wrote this down quickly.
Set k = n-7
(k+7)^(1/k) in N, k > 0
(k+7) = m^k, m,k in N
k = 1 and 2 are trivial solutions, with m = 8 and 3.
Notice that (k+7)^(1/k) = m is strictly decreasing with k, so any other solution would have m = 2.
2^k = k+7 has a solution between 3 and 4, so not in integers.
Hence, n in {8,9} are the only solutions.
Also the fact that the minimum prime factorization can be log2(n) indicates that the forth root is only available at the integer 16 at minimum is sort of a proof.
isn't it easier to take both sides to the power n-7 so we're looking for x^(n-7)=n . x positive integer. since x=1^(n-7) is always 1 and the smallest x is then 2 and 2^(n-7)>n for all n>11 is pretty obvious with no heavy duty math. The induction is just that for n+1, 2*(previous>n)>n+1 .
Do you guys know where can I find olympiad problems like that , I mean I website or something like that ???
Art of problem solving, go to forums and you’ll find problems ranging from national olympiads from each country to imo problems
Wow.. You are so mathematically facile .Very impressive .
I think we can also say that n is between 8 and 10 because the first integer to accept a 4th root is 16 and n=11 gives us 4th root of 11 which is impossible.
You can also take the derivative of the function itself. It takes a bit of implicit differentiation but it's not too complex and you get y((x-7)/x-ln(x))/(x-7)^2=y'. Then since you know y is always positive and so is (x-7)^2 (at least whenever x>8, which is the relevant range anyways) you can show (x-7)/x-ln(x) is always negative since (x-7)/x approaches 1 and ln(x) approaches infinity (a bit trickier than just that but that's the idea). Positive * positive * negative = negative. Since the derivative is always negative, and you know for x=10, y8 or whatever the function is always greater than 1, you know the value will always be between 1 and 2 and therefore can't be an integer.
For ln(2)>1/2 can't we just operate on both sides with e. Get 2>√e then square both for 4>e. That's true, so the starting inequality was true.
So this all makes sense but there's one thing that's messed up and it's that the length of the drawstrings on your sweatshirt are of significantly unequal lengths and I think that really needs to be addressed
at wich age students participate in that exam? (srry 4 my english)
0:47 Actually, n = 1 will work (if not for the restriction on n)
Can't we just say for the last part:
ln(2) > 1/2 e ^ (ln(2)) > e^(1/2) [e to the power of both sides] 2 > e^(1/2) [by definition/properties of logarithm] 4 > e [squaring both sides of the logarithm]
Or should x^(log_x(y)) = y (where log_x is log base x) be proven in olympiad setting as well?
Hey do you have a video explaining whats on your shirt
A suggestion for a quick and easy proof (it feels almost too easy - are there any flaws in my reasoning?) : f(n)=n^(1/(n-7) >= 8^(1/(n-7)) = 2^(3/(n-7) > 1. When does f(n) get >=2? 2^(3/(n-7))>=2 3/(n-7) >=1 n
Was it in the Question that "n" had to be a nautral number? or just that n^(1/(n-7)) had to be a nautural
For the final proof, can't we say that because ln2>1/2, so by applying natural exponential function, we have 2>e^(1/2) which is equivalent to 2>1/(e^2) (by applying power rules), that is true right?
Oh, this was cleverer than taking the derivative of n^(1/(n-7)) and showing that that's always negative. Doing those inequality-preserving algebraic transformations make the derivative rather nicer to write out.
The title card asks when the expression is an INTEGER rather than a NATURAL, and doesn't bound it to n >= 8. Solving that, we can include both 1 and -1.
We can include 1 (at n=1) and 0 (at n=0), but I don't think (-1)^(1/(-8)) is an integer. EDIT: n=0 is undefined because 0^(1/-7)=(1/0)^7, and that's dividing by 0.
ALSO: The title card doesn't specify that n should be an integer, so if we assume that n can be irrational, we can express all integers greater than 1 as n^(1/(n-7)). f(n)=n^(1/(n-7)) is a continuous function for n>7. All integers greater than 8 will lie between 7
0:24 0:48 For n=1, n^(1/(n-7))=1^(-1/6)=1
Super! Next!
Amazing Sbera. For math competitions, please consider ua-cam.com/video/LSypt8uSSW4/v-deo.html and other videos in the Olympiad playlist. Hope you enjoy 😊
take x= n*power(1/n-7) take log both side
logx= 1/n-7 * log n , t= logx
n-7* t = logn