Swedish Mathematics Olympiad | 2002 Question 4

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We look at a solution to a number theory problem from the 2002 Sweden Mathematics olympiad.
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Комментарии : 628

@patrickwienhoft7987 3 года назад

8:35 underlining inequality signs is usually not the best way to emphasize them ;)

@elefesel 3 года назад

I think that's why he's using a different colour to do this:)

@kurt.dresner 3 года назад

I will draw a line through this equals sign to emphasize it ≠

@NuisanceMan Год назад

The best way to emphasize a capital "I" is to draw a horizontal line on top of it. At least, T think so.

@landsgevaer 9 месяцев назад

@@NuisanceMan / think extreme italics work well.

@arpansingh2116 3 года назад

Can't believe I'm watching this for entertainment

@thangphamngoc3897 3 года назад

lool me too

@dananajj 3 года назад

I'm watching this to fall asleep

@NickiRusin 3 года назад

ikr, it's super fun

@RosidinAli 3 года назад

Lol me too

@wassollderscheiss33 3 года назад

That's because he is (somewhat) entertaining

@davidbrisbane7206 3 года назад

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 :-).

@falknfurter 3 года назад

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.

@wospy1091 3 года назад

All he did in this proof was use Calculus instead of induction. You could even say he did it the easy way.

@davidbrisbane7206 3 года назад

@@wospy1091 Induction is easier.

@gabriel7233 3 года назад

@@davidbrisbane7206 not for everyone but it's clearly more elementary

@iabervon 3 года назад

@@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).

@Miju001 3 года назад

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.

@x_gosie 3 года назад

Same!

@SREproducciones 3 года назад

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.

@vedhase2370 3 года назад

May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.

@x_gosie 3 года назад

@@vedhase2370 The first thing you need to know are the statement gaving, by the problem.

@randomdude9135 2 года назад

@@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

@pratikmaity4315 3 года назад

He explained so many stuffs in a single video. This shows how different math problems are related to each other!!

@ckeimel 3 года назад

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).

@pratikmaity4315 3 года назад

@@ckeimel True to say

@vedhase2370 3 года назад

May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.

@ckeimel 3 года назад

@@vedhase2370 wich part of the solution troubles you?

@ireallyhatemakingupnamesfo1758 3 года назад

N^(1/n-7)>1 The proof of this is obvious and left as an exercise to the reader

@AnindyaMahajan 3 года назад

It is obvious. a^x is a strictly increasing function for a>1 and equals 1 at only x=0.

@richardconrardy9673 3 года назад

*n^(1/(n-7))>1

@josem138 3 года назад

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.

@jamesjohnson4695 3 года назад

@@josem138 Not multiplied by itself (n-7) times, but 1/(n-7) times. But yes the claim is still trivial.

@rhaq426 3 года назад

@@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.

@takyc7883 3 года назад

Fun fact, we know that 0

@35571113 3 года назад

Ha ha! How?

@volodyanarchist 3 года назад

@@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.

@84y87 3 года назад

@@volodyanarchist lmao

@vedhase2370 3 года назад

May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.

@ASD128London 3 года назад

@@vedhase2370 I was wondering that too.

@derhamcohomology 3 года назад

What I've learnt from this channel is to start every math problem by considering case n=1.

@xenofurmi Месяц назад

Showing the "obvious" parts and "the long way" is so so good. It's how we all learn. Very good job!

@TheOneSevenNine 3 года назад

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

@hrishikeshkulkarni2320 3 года назад

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.

@tomaszgaazka6777 3 года назад

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.

@robwsur 3 года назад

beautiful idea

@tomatrix7525 3 года назад

This was a really lovely problem. Well presented Michael, loving your stuff

@dariensisko0 3 года назад

Although it was excluded right at the beginning, n=1 would be a solution as 1^x=1\in N for any x.

@lawandeconomics1 3 года назад

I probably would have turned back after taking the derivative. Thanks for an unusually clear and direct work-through!

@stevenwilson5556 3 года назад

This channel is such a gem, more people should be watching this stuff.

@shurik3nz346 3 года назад

No, a person.can watch what they want on youtube.

@shurik3nz346 3 года назад

Abacus, should and must are different

@shurik3nz346 3 года назад

Abacus , ahh I see. Thanks for enlightening me.

@stevenwilson5556 3 года назад

@@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.

@shurik3nz346 3 года назад

@@stevenwilson5556 haha alright

@demenion3521 3 года назад

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.

@red0guy 3 года назад

Man who I wish I would have had such videos in high-school almost 20 years ago.

@harish6787 3 года назад

I felt this much easier than any other question

@roboto12345 3 года назад

For example question 6 of IMO 2020

@pedroafonso8384 3 года назад

Yes

@xz1891 3 года назад

Bcz it is

@xz1891 3 года назад

@@roboto12345 I can only solve Q1 of imo2020

@paulchapman8023 3 года назад

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.

@MK-13337 3 года назад

7:10 is not equivalent, but you only need the reverse implication here. f(x) >0 everywhere => f(n) > 0 for natural numbers

@calebbirnbaum4605 3 года назад

The work is pleasingly clear and labeled

@jole0 3 года назад

Love your videos, participating in my first olympiad in a month or two

@rishabhsinha4765 3 года назад

Good luck!

@James_Moton 3 года назад

What olympiad are you attending?

@jole0 3 года назад

@@James_Moton Norwegian one

@asdfghjkl6506 3 года назад

All the best✌️👍👍

@35571113 3 года назад

Good luck!

@markfischer5044 3 года назад

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...

@utsav8981 3 года назад

Me: Sees 1/n-7 *Ezee* Me: Sees n just below it *crap*

@0xDEAD_Inside 3 года назад

Same

@84y87 3 года назад

1/n-7 is integer only for n=6,8

@alex-cm9fd 3 года назад

Dorulo ili kvo

@vedhase2370 3 года назад

May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.

@ethanyap8680 3 года назад

@@vedhase2370 very late,but since 1/(n-7) is positive, n^(1/(n-7)) > n^(0) = 1

@chilling00000 3 года назад

IMO is happening now, it would be fun to feature some of the questions from this year

@islamgaziev1717 3 года назад

Finally, something I managed to solve before watching the video. Beautiful solution.

@iooooooo1 3 года назад

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.)

@manucitomx 3 года назад

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.

@iabervon 3 года назад

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.)

@speedsterh 3 года назад

@@iabervon Excellent !

@basmamjouel2893 Год назад

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

@islandman9748 3 года назад

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!

@GhostyOcean 3 года назад

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

@lucas0m0james 3 года назад

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.

@riskassure Год назад

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.

@frederickm9823 3 года назад

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

@gmchess7367 3 года назад

Amazing explanation

@PI227 3 года назад

Hey, Michael. Can you please make a video on problem #6 from the 2009 IMO.

@scottrichmond3548 23 дня назад

Me after seeing thumbnail: "Oh that's easy, it's 8" Me after clicking on video: "Ah, the old bait and switch"

@Dusk-MTG 3 года назад

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.

@KingstonCzajkowski 3 года назад

It has a spike like x^-2, a discontinuity that occurs when he mentions that we're not using induction

@RZMATHS 3 года назад

Well if feel like doing a challenging prob You can try this ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-igdy05LZj90.html

@Jivvi Год назад

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.

@kinshuksingh90 Год назад

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

@asa4766 3 года назад

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”.

@patrickgambill9326 2 года назад

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

@ChristianRosenhagen 3 года назад

One of the best: creative and not too hard!

@ananyashukla2749 3 года назад

Nicely explained and question was good

@knvcsg1839 3 года назад

That thinking of searching for n>11 is really brilliant.

@gaufqwi 3 года назад

The problem with underlining an inequality to emphasize its strictness is then you make it nonstrict 🤣

@vedhase2370 3 года назад

May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.

@paulchapman8023 3 года назад

@@vedhase2370 Maybe he considered the fact that 1^n = 1 for all real values of n not worth commenting on.

@kadenxu5142 3 года назад

@@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

@jongu71 3 года назад

I thought of that too, he almost managed to confuse me!

@Omar-of4tz 3 года назад

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

@rezhaadriantanuharja3389 3 года назад

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)

@judecarter6095 10 месяцев назад

By thinking of it as an n-7th root you can also show that n>=2^(n-7) as n!=1 and hence 8

@shreyathebest100 3 года назад

One of the few number theory problems I silved on my own

@mangai3599 3 года назад

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.

@hogehoge1030 Год назад

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.

@jakubdutkiewicz5816 3 года назад

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.

@ameerunbegum7525 3 года назад

Super awesome !

@dneary 3 года назад

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

@hans-juergenbrasch3683 3 года назад

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

@handanyldzhan9232 3 года назад

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

@AntonBourbon 2 года назад

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

@abc13deagosto 3 года назад

Thank you very much!

@rafael7696 3 года назад

I like very much your olimpiad problems

@dominikstepien2000 3 года назад

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

@miketakeahike4741 3 года назад

Yeah Sir please do IMO questions also like IMO 2003 P2 number theory and algebra questions

@parmilakumari3146 3 года назад

Fun problem

@miketakeahike4741 3 года назад

Yeah that would be great

@mohamednejighnimi4860 3 года назад

it's viable also for n=1 as 1^any number=1 which is an integer.

@ramk4004 3 года назад

Thank you

@bcwbcw3741 3 года назад

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 .

@stefanocorno4649 3 года назад

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.

@canaDavid1 4 месяца назад

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.

@martinschulz6832 3 года назад

What about n=1? It's also a solution since 1^(-1/6)=1...

@ruslanadayev589 3 года назад

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?

@AntonioSaracino7 3 года назад

thank you; brilliant

@potatochipbirdkite659 3 года назад

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.

@Marre2795 3 года назад

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

@redolentofmark 3 года назад

well if you include limit of n as it approaches infinity, then it would equal 1

@ussenterncc1701e 3 года назад

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.

@dumbledort9755 3 года назад

the equivalence at 7:30 is false. But the implication b=>a is enough to prove the result (b result with x in R).

@Nicolas-zf3pv 3 года назад

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?

@dyidyirr 3 года назад

@@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.

@oida10000 3 года назад

Just from think about it: n must be greater than 7 if it is less than that we have 1/(n^(7-n)) which isn't an integer, 8 and 9 both work 8^(1/(8-7))=8^(1/1)=8^1=8 and 9^(1/(9-7))=9^(1/2)=3 but are there more?

@ysqure3 3 года назад

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.

@shqotequila 3 года назад

Love number theory

@septagram9491 2 года назад

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.

@SQRTime 2 года назад

Hi Septagram. If you are interested in math competitions, please consider our channel ru-vid.com. Hope to see you 😊

@nasrullahhusnan2289 Год назад

I think it is easier to grasp by using Taylor series to show that ln(2)

@nasrullahhusnan2289 Год назад

Sorry the last sentence must be ln(2) < ½ instead.

@noedeverchere2833 3 года назад

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?

@abhishekkumar-os5zk 3 года назад

take x= n*power(1/n-7) take log both side logx= 1/n-7 * log n , t= logx n-7* t = logn

@lord_hamza3550 Год назад

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.

@antman7673 3 года назад

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.

@stumbling 3 года назад

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.

@sujitsivadanam Год назад

0:47 Actually, n = 1 will work (if not for the restriction on n)

@dorijancirkveni 3 года назад

0:24 0:48 For n=1, n^(1/(n-7))=1^(-1/6)=1

@TheCloudyoshi 3 года назад

Is it mathematically sound to just take the derivative of f(x)=x^(1/(x-7)), and prove that the derivative is negative for all x>=8? (Then by spot checks you can figure out for all x>10, f(x) is between 1 and 2.) The function would be differentiable for all x>7 I believe...

@user-wt1ul7ki6p 3 года назад

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)

@SupremeChickenx 3 года назад

this guy has neil patrick harris energy, i love it

@danieldomert1022 3 года назад

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

@pradipchaterjee9576 3 года назад

Do you want a single value of n or other values satisfying the equation

@jamirimaj6880 3 года назад

after watching Numberphile: *I NOW KNOW AND UNDERSTAND MATH! I CAN SOLVE ANY MATH PROBLEMS NOW, I CAN DO IT!* after watching serious Math like this channel: 😭😭😭😭😭😭😭

@mathadventuress 3 года назад

Right? This is nuts

@LeoDaSoR 3 года назад

at wich age students participate in that exam? (srry 4 my english)

@zouhairees-sqally3830 3 года назад

Do you guys know where can I find olympiad problems like that , I mean I website or something like that ???

@ruggbi 3 года назад

Art of problem solving, go to forums and you’ll find problems ranging from national olympiads from each country to imo problems

@mathunt1130 19 дней назад

This is a nice solution.

@manusiabiasa7305 3 года назад

Wow open my mind

@parmilakumari3146 3 года назад

Could you please do IMO 2003 P2 its a cool question

@mikkijhuria287 3 года назад

Nice suggestion

@mikkijhuria287 3 года назад

There’s a lot to learn from that problem

@mikkijhuria287 3 года назад

Would be grateful if Michael could solve it for us

@miketakeahike4741 3 года назад

Yeah great , I hope sir does it

@parmilakumari3146 3 года назад

Fun problem

@michaelblankenau6598 3 года назад

Wow.. You are so mathematically facile .Very impressive .

@rajatmishra6628 3 года назад

Easy one.. Though 2nd part was pretty obvious and you made it look tough

@TIMS3O 3 года назад

Define f(x)=x^(1/(x-7)). Another way to prove it is to notice that f(11)1 when x>7. Then we are done if we can show that f is strictly decreasing when x>7 which by differentation amounts to showing that x-7-ln(x)x7. But this is clearly true since ln(x)>ln(7)>1.