Greek Mathematics Olympiad

Greek Mathematics Olympiad | 2008 Q2

Рет қаралды 45,477

Күн бұрын

We solve a nice number theory problem from the 2008 Greek math olympiad.
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Пікірлер: 106

@leickrobinson5186 3 жыл бұрын

“That’s going to be, like, 8 or something”? 😂🤣

@wise_math 3 жыл бұрын

😅 yeah that sort of thing can be confusing for students. But overall good presentation.

@tomatrix7525 3 жыл бұрын

@@wise_math Haha, I love his style, personally.. Classic!

@tonyhaddad1394 3 жыл бұрын

What I love about your channel is that you are making many video in 1 day 😍😍😍

@HagenvonEitzen 3 жыл бұрын

08:00 Without reference to irrationality: Do x=-1 "by hand" and then: If x

@andreivila7607 3 жыл бұрын

Wow! I loved the problem! Nice use of the Sophie Germain identity. I would like to see more number theory problems on this channel! :)

@professorpoke 3 жыл бұрын

If you want good number theory problems, then this channel is a good to stop.

@TechToppers 3 жыл бұрын

Do you know? I saw thumbnail for 2 minutes and I solved the problem by intuition.(I got that Ms Sophie would help I mean)

@gilangNoDrop 3 жыл бұрын

Yeah I solved it too by Sophie germain, we can prove that (x^4+2(2^(2^(x-2)))^2 - 2(x^2)(2^2^(x-2)) > 1 for x≥2

@TechToppers 3 жыл бұрын

@@gilangNoDrop Yup!

@acentasecond3721 3 жыл бұрын

great video, you gave me some inspiration for a future video!

@goodplacetostop2973 3 жыл бұрын

12:55 Αυτό είναι ένα καλό μέρος για να σταματήσουμε More than 80k subs, let’s go! Here’s the daily... Considering the function f : ℝ → ℝ with these properties : - f(a + b) = f(a) + f(b), a and b being any real numbers - f(1) = 1 - f(1/x) = (1/x²) • f(x), x being a non-zero real number Determine f and calculate f(1/√2020)

@KuroboshiHadar 3 жыл бұрын

f: R → R f(a+b) = f(a) + f(b) f(1) = 1 f(1/x) = (1/x²)f(x) for all x ≠ 0 Firstly, notice that f(a+0) = f(a) + f(0) = f(a) so f(0) = 0 If p and q are non negative integers, we have → f(p*a) = f(a + a + ... + a) = p*f(a) → q*f(a/q) = f(a/q + a/q + ... + a/q) = f(a) → f(a/q) = f(a)/q Also, f(a-a) = f(a) + f(-a) = f(0) = 0 so f(-a) = -f(a) Now we can take r = p/q to be any rational number, and it will satisfy → f(r*a) = r * f(a) If a = 1, f(r) = r * f(1) = r Assuming f is continuous, this extends to all x in R since the rationals are dense in R, so f(x*1) = x*f(1) = x for all x in R Also, it satisfies f(1/x) = 1/x = 1/x² * f(x) = 1/x² * x = 1/x So f(x) = x f(1/√2020) = 1/√2020 Did I do it right? I don't know if there are other functions that satisfy the conditions though...

@IanXMiller 3 жыл бұрын

Or more trivially: f(x) = f(1+1+1+1+...+1)=f(1)+f(1)+f(1)+f(1)+..+f(1)=1+1+1+1+..+1=x so the only function is the identity function.

@mikelolis3750 3 жыл бұрын

change it to "Αυτό είναι ένα καλό μέρος για να σταματήσουμε"

@mathteachervictor 3 жыл бұрын

@@KuroboshiHadar you are right. There is only one continuous function. I guess the last equality follows that f is continuous.

@light9744 3 жыл бұрын

First, rewrite the function with f(x+c)=f(x)+f(c) where c be a constant and x be a variable. Then differentiate both side becoming f'(x+c)=f'(x) , which shows f'(x)=d where d is a constant. With this situation, we can conclude f(x)=ex+f where e, f are constants. As f(a+b)=f(a)+f(b) which shows e(a+b)+f=e(a+b)+2f , which leads to the result f=0. Then using f(1)=1, we can conclude f(x)=x. Therefore, f(1/√2020) is itself. (However, i don't know how to proof the function f(x) is differentiable or not, so the solution is with flaw. I hope someone else can do the proofing for me!)

@parmilakumari3146 3 жыл бұрын

Wow you find such cool NT contest problems

@soranuareane 3 жыл бұрын

That orange chalk looks quite nice this time of year. You should use it more often! I find it easier to see than the white chalk for some reason, too.

@user-qg6do2xn9t 3 жыл бұрын

Nice problems and solutions

@MrRyanroberson1 3 жыл бұрын

interesting little homework at the end. turns out when you factor to get y^2-z^2 the way you did, you do so exactly such that y+z and y-z are perfect squares themselves

@davidepierrat9072 3 жыл бұрын

if p is prime, p and phi(phi(p)) are coprime so you can't get anything by crt so no need to even try looking at it mod stuff

@tomyato 3 жыл бұрын

That’s powerful 🔥

@wesleydeng71 3 жыл бұрын

Made the same first attempt and then switched to factoring.😀

@egillandersson1780 3 жыл бұрын

Damned ! I got the correct factorization but not the conclusion ☹️. Nice probleme and nice solution ! Thank you

@jayvaghela9888 3 жыл бұрын

Great work you deserve more likes, love from India 🇮🇳

@antoniopalacios8160 3 жыл бұрын

8:00 I think, actually there is a mistake with the parentheses. We have that for x=-1, 2^2^(-1)=0.25. To get sqrt(2) we need to put the parentheses 2^(2^(-1)). And so on for x=-2,-3,etc. Thank you.

@reeeeeplease1178 2 жыл бұрын

No, if you write x^y^z that notation means x^(y^z) instead of (x^y)^z. So something like 2^2^2^2 is evaluated from the "top" down: 2^2^2^2 = 2^2^4 = 2^16 ~ 60,000 If you want to do from the bottom up, you put parantheses: 256 = 16^2 = (4^2)^2 = ((2^2)^2)^2

@TheQEDRoom 3 жыл бұрын

can we simplify this by noting that p must be a sum of two squares so p must be of 1 mod 4? then it follows that x^8 is 1 mod 4 also.

@holomurphy22 3 жыл бұрын

When x is odd we have x^2 = 1 mod 8. And here x is obviously odd. So its kind of trivial that x^8 is 1 mod 4

@demenion3521 3 жыл бұрын

for proving that y-z>1, we can almost factor it: y-z = x^8+2*2^(2^(x-1))-2*x^2*2^(2^(x-2)) = 1/2 {[x^4-2*2^(2^(x-2))]²+x^4}. as x obviously needs to be odd, the squared term is strictly greater than 0 and moreover an odd number. therefore y-z>=1/2 (1+x^4)>1 for all odd x>1.

@devrajdas6544 3 жыл бұрын

You could have used sophie germain i dentity too ig?

@sayansircar360 3 жыл бұрын

Make a video on madhava- Leibniz series

@jkid1134 3 жыл бұрын

Good excuse for the colored chalk, I agree

@hateyou218 3 жыл бұрын

Superb

@aldinofahreza 3 жыл бұрын

I trust that's only 1.. and you prove it 😂😂😂

@raqmet 3 жыл бұрын

Good video

@xchomphk.9788 Жыл бұрын

x = 1 holds, for x>=2, 2^x + 2 = 2 mod 4. Meaning that 2^(2^x+2) = 4 mod 5. Its easy to prove any x^8 is 1 mod 5 through trial and error, therefore for x>=2, the expression is divisible by 5 , therefore the only solution is x = 1. Honestly a really easy olympiad Q2 problem even for my standards

@fivestar5855 2 жыл бұрын

My brain just burned out

@jotaro6390 3 жыл бұрын

I solved it!

@giuseppebassi7406 3 жыл бұрын

Wait wait. Since i have used many times the Sophie-germain identity, i have already thought about that when i saw the title. I did the same method, but i have been stuck to prove that y-z is bigger than 1 for all x>2. In fact, i tried that with induction but without results. How can this be so simple that he gave that to homework problem? 😟

@michaelcampbell6922 3 жыл бұрын

Set f(x)=y-z. Find f(3). Use f’(3) to show f(x) is increasing for x>=3.

@jonathanjacobson7012 3 жыл бұрын

Can anyone recommend me a concise book that goes through the topics mentioned in this video? e.g. Fermat's little theorem, factorization, etc.

@stephenbeck7222 3 жыл бұрын

For Fermat’s Little Theorem, I would look at an abstract algebra book. Plenty of good ones out there and you can google reviews. The factoring stuff is high school level computation plus a lot of intuition/cleverness which is mostly just Michael’s familiarity with contest problems.

@Miguel-xd7xp 3 жыл бұрын

Maybe The Justin steven's website could help you

@jonathanjacobson7012 3 жыл бұрын

@@Miguel-xd7xp not much material there, but thanks.

@Miguel-xd7xp 3 жыл бұрын

@@jonathanjacobson7012 Also the AoPS (art of problem solving) can help, there are more people who can recommend a book

@yashvardhan2093 3 жыл бұрын

What I did was a little complicated I first showed that if x is negative then exponent of 2 is fractional and hence it is not an integer only. Then I got the solution for x=1. Then I proved that there exists no solution for x being even . Thus I proved x is odd But clearly if it’s odd it’s unit digit is 1,3,5,7,9 . By unit digit cyclicity if I showed that in case of 1,3,7,9 unit digit of x^8 is 1 and that of the other part is 1 always thus the unit digit is 5 and hence divisible by 5. But when unit digit is 5 This didn’t work but then Sophie Germain struck me when I saw that 4 and remembered about the factorization

@yashvardhan2093 3 жыл бұрын

There is mistype I showed that the unit digit of other part is 4

@SB-wy2wx 3 жыл бұрын

Could someone please expand on his reasoning for choosing mod 5, his "8 = 2×4, 4 = 5-1" is not making sense. Why cant u do 8 = 8×1, and 8 = 9-1, so choose mod 9? Sorry if question is stupid, just new to Number theory

@AlephThree 3 жыл бұрын

8 is not coprime to 2 so you can’t apply FLT

@T6e6r6o 3 жыл бұрын

Math and flag nerd here, love all your videos. Just a little nitpick. Just like the flag of the United States, the union (ie. the stars part) of which, when displayed vertically, must be on the upper-left corner from the observer's perspective, the Greek flag's canton (ie. the cross part), when displayed vertically, must also be on the upper-left corner from the observer's perspective.

@justforfun2238 3 жыл бұрын

nerd

@caesar_cipher 3 жыл бұрын

Sheldon Cooper presents Fun with Flags

@mondolee 3 жыл бұрын

that’s interesting... does it mean that these flags when vertically viewed have to be mirrored?

@user-ht1vg5we2p 3 жыл бұрын

Also the stripes are horizontal

@user-ht1vg5we2p 3 жыл бұрын

The flag is turned around 90°

@user-A168 3 жыл бұрын

Good

@logicalproofs7276 3 жыл бұрын

Plz do this question Cube root of (5-x)=5-x³.

@IanXMiller 3 жыл бұрын

You get three roots of a degree six polynomial, only one of which is real.

@mehmeterciyas6844 3 жыл бұрын

@@IanXMiller no shit.

@hybmnzz2658 3 жыл бұрын

Notice that if f(x) = 5-x^3 then the inverse of f is cuberoot(5-x). Therefore the function equals its inverse. Notice that f is strictly decreasing. This implies f(x) = x and so 5-x^3 = x. This is a simple third degree polynomial to solve. I'll leave it to you guys to prove that f=f^-1 implies f(x)=x if f is strictly decreasing.

@erenozilgili4634 3 жыл бұрын

@@hybmnzz2658 I got to x^3 +x =5 part but how to solve it can you explain?

@blazedinfernape886 3 жыл бұрын

@@hybmnzz2658 It can be approached intutively because if f and f^-1 are continuous in an interval then f and f^-1 are images of each other with respect to y=x and an image and object can equal only at that line itself.

@tilek4417 3 жыл бұрын

Sophie Germain

@marceliusmartirosianas6104 3 жыл бұрын

8^x+2^x+2= [[[8x+4x]]= [[ 6x+6]]=[[ x =1 ]] marcelius mrtirosianas

@antoniussugianto7973 3 жыл бұрын

Clearly Never prime for even x

@wise_math 3 жыл бұрын

Right. But also have to consider the odds, and this is included in x >= 2.

@yla3727 3 жыл бұрын

9:55 it looks like you used the formula of a^2 + b^2 = ( a + b )^2, which isn't correct as I learnt at school. I had checked your term and get bad result

@AnkhArcRod 3 жыл бұрын

:) He used a^2 + b^2 = (a+b)^2 -2ab. In this case, the 2ab turns out to be a square as well.

@yla3727 3 жыл бұрын

@@AnkhArcRod thanks and excuse me