Tabăra MathTime - Problema 4, Ziua I - JUNIORI
- Laurențiu Ploscaru
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Tabăra MathTime - Problema 4, Ziua I - JUNIORI
Determinați $a,b,c\in \Bbb{N}$ cu $1<a<b<c$ a.î. $(a-1)(b-1)(c-1)\mid (abc-1)$.
People are strange when you're a stranger,
Faces look ugly when you're alone.
Women seem wicked when you're unwanted,
Streets are uneven when you're down.
Faces look ugly when you're alone.
Women seem wicked when you're unwanted,
Streets are uneven when you're down.
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Re: Tabăra MathTime - Problema 4, Ziua I - JUNIORI
Notam : $a-1=x;b-1=y;c-1=z$. Obtinem :
$xyz|(x+1)(y+1)(z+1)-1<=>xyz|xy+yz+zx+x+y+z$.
Evident $xyz\le xy+yz+zx+x+y+z<=>\sum \dfrac{1}{x}+\sum \dfrac{1}{xy}\ge 1$.Avem 0<x<y<z . Daca $x\ge 3=>y\ge 4;z\ge 5$ si vom avea :
$\sum \dfrac{1}{x}+\sum \dfrac{1}{xy}\le \dfrac{1}{3}+\dfrac{1}{4}$$+\dfrac{1}{5}+\dfrac{1}{12}+\dfrac{1}{15}+\dfrac{1}{20}<1$, contradictie.
Deci x=1 sau x=2. In primul caz , $yz|yz+2(y+z)+1$, deci $yz|2(y+z)+1=>y|2z+1$ si $z|2y+1$.Notam $2z+1=my$ si $2y+1=nz$. Adunand si tinanad cont ca 1<y<z, deducem ca 2y+1=z. Atunci $y|4y+3$, adica y|3, deci y=3 si z=7. Analog cazul celalalt.
$xyz|(x+1)(y+1)(z+1)-1<=>xyz|xy+yz+zx+x+y+z$.
Evident $xyz\le xy+yz+zx+x+y+z<=>\sum \dfrac{1}{x}+\sum \dfrac{1}{xy}\ge 1$.Avem 0<x<y<z . Daca $x\ge 3=>y\ge 4;z\ge 5$ si vom avea :
$\sum \dfrac{1}{x}+\sum \dfrac{1}{xy}\le \dfrac{1}{3}+\dfrac{1}{4}$$+\dfrac{1}{5}+\dfrac{1}{12}+\dfrac{1}{15}+\dfrac{1}{20}<1$, contradictie.
Deci x=1 sau x=2. In primul caz , $yz|yz+2(y+z)+1$, deci $yz|2(y+z)+1=>y|2z+1$ si $z|2y+1$.Notam $2z+1=my$ si $2y+1=nz$. Adunand si tinanad cont ca 1<y<z, deducem ca 2y+1=z. Atunci $y|4y+3$, adica y|3, deci y=3 si z=7. Analog cazul celalalt.
Nimic nu-i niciodata asa de simplu cum pare.
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Re: Tabăra MathTime - Problema 4, Ziua I - JUNIORI
Daca inlocuim in enuntul problemei produsul (a-1)(b-1)(c-1) cu cel mai mic multiplu comun al numerelor (a-1);(b-1);(c-1) ce se intampla ?Laurentiu Ploscaru scrie:Determinați $a,b,c\in \Bbb{N}$ cu $1<a<b<c$ a.î. $(a-1)(b-1)(c-1)\mid (abc-1)$.