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Problema numere
Posted: Mon Oct 01, 2007 7:27 pm
by Claudiu Mindrila
Determinati cifrele a,b,c si numerele naturale n si p stiind ca \( \overline{a b} \ =n^3 \), \( \overline{b a c}\ =p^3 \), iar \( \overline{a b}\ \) si \( \overline{b a c}\ \) sunt simultan multiplii ai lui9.
Posted: Sat Mar 01, 2008 1:06 am
by gigel
a=2, b=7, c=9, p=9, n=3. Exista si alte solutii?
Posted: Tue Mar 04, 2008 9:19 am
by mihai++
\( \overline{ab}=9s\rightarrow a+b\vdots 9 \rightarrow a+b=9 \) sau \( a+b=18 \)
Analog \( a+b+c\in\left{9,18,27\right} \).
i) \( a+b=9 \), \( c=0 \) imposibil deoarece \( \overline{bac}=p^3 \) si deci ar trebui sa se termine in 3 zerouri
ii) \( a+b=9 \), \( c=9 \) \( \overline{ab}=n^3\rightarrow n\vdots 3 \rightarrow n^3= \overline{ab}=27 \), care verifica \( 27=3^3, 729=9^3 \).
iii) \( a+b=18\rightarrow a=b=9 \), dar \( 99\neq n^3 \).
Singura solutie \( \left(a,b,c,n,p\right)\in\left{\left(2,7,9,3,9\right)\right} \).