[Very first IMO problem in history] 1959 IMO Problem #1

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We present three different solutions to the very first - and widely considered to be the easiest - problem in the International Mathematical Olympiad (IMO). Join us to apply the Euclidean algorithm, proof by contradiction, and properties of consecutive integers in this week's math memo.

We thank AoPS for the backbone of these solutions, found here:

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I didn't think the IMO would have such easy problems. I think nowadays they have become harder, at least from what I've seen.

tlqy

Wow IMO questions really have gone a long way from this

jamirimaj

Thanks for the video! Useful for beginners of all ages. I am curious to know what software/ hardware you use to record and share these videos. Thanks again!

pk

Method zero: make the division and obtain 3/2 - 1/(28n+6) which is obviously irreducible. No strange theorems required.

JoseFernandes-jsep

3*(14n + 3) - 2*(21n + 4) = 1.
So by Bezout's lemma,
the gcd(14n +1, 21n + 1) = 1,
so 14n + 3 & 21n + 4 are relatively prime,
so (21n + 4)/(14n + 3) is irreducible.

davidbrisbane

I kinda did a proof by contradiction and using very simple algebra. Not sure if this is correct or if already mentioned.
My method:
Assume that (21n+4)/(14n+3)=a where "a" is a whole number.
(21n+4)=(14an+3a)
(21n-14an)=(3a-4)
n(21-14a)=(3a-4)
n=(3a-4)/(21-14a)
So this means that for (21n+4)/(14n+3) to be reducible. "n" has to be a fraction (3a-4)/(21-14a) where "a" is a whole number. Which means "n" cannot be a natural number for the expression to be reducible.

hkm

Isn't it enough to observe that there's actually no possibility for the expression 21n+4 to be a multiple of 14n+3?
I mean 21n+4 is always larger than 14n+3 for any natural number n and (14n+3)*2 is already larger than 21n+4. Therefore, the gcd can only be 1.

AAA-mvdv

Esse problema foi fácil comparado com os de hoje em dia.

ProjetoEquilatero

When doing these problems how do people know what to do before solving the problem

taopinairlinesmathindustry

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