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Cifre consecutive
Posted: Tue Nov 11, 2008 9:17 pm
by alex2008
Stiind ca a, b, c sunt cifre consecutive (in aceasta ordine) si ca \( \frac{\overline{a,(b)}}{51}=\frac{\overline{b,(c)}}{61} \) . Sa se calculeze suma \( S=\overline{a,(b)}+\overline{b,(c)}+\overline{c,(a)} \) .
Posted: Thu Apr 16, 2009 6:09 pm
by salazar
\( \frac{a+\frac{a+1}{9}}{51}=\frac{a+1+\frac{a+2}{9}}{61} \)
\( \frac{10a+1}{9\cdot 51}=\frac{10a+11}{9\cdot 61} \)
\( 51(10a+11)=61(10a+1) \)
\( 510a+661=610a+61 \)
\( 600=100a \)
\( a=6,b=7,c=8 \)
\( 6+\frac{7}{9}+7+\frac{8}{9}+8+\frac{6}{9}=21+\frac{21}{9}=21+\frac{7}{3} \)