Question

aflați numerele naturale a, b, c astfel încât
$\frac{a + 2000}{2000} = \frac{b + 2001}{2001} = \frac{c + 2002}{2002}$
si a + b = 2c - 9. dau si coroana.

Answer 1

   
[tex]\displaystyle\\ \frac{a+2000}{2000}=\frac{b+2001}{2001}=\frac{c+2002}{2002}=k+1\\\\ a=k\times2000\\ b=k\times2001\\ c=k\times2002\\\\ a+b=2c-9\\\\ k\times2000+k\times2001=2\times k\times2002-9\\ k(2000+2001)=k\times2\times2002-9\\ 4001k=4004k-9\\ 4004k-4001k=9\\ 3k=9\\\\ k=\frac{9}{3}\\\\ \boxed{k=3}\\\\ a=k\times2000=3\times2000=\boxed{\bf6000}\\ b= k\times2001=3\times2001=\boxed{\bf6003}\\ c=k\times2002=3\times2002=\boxed{\bf6006} [/tex]