1003+1003+a+1003+b+c+2x1003=100.004
3009+2x1003+a+b+c=100.004
3009+2006+a+b+c=100.004
5015+a+b+c=100.004
notam a+b+c=x
5015+x=100.004
=> x=100.004-5015
=> x=94989 ceea ce reprezinta suma celor 3 numere consecutive impare
94989:3=31663 reprezinta numarul b
deci:
a=31661
b=31663
c=31665
facem proba:
31661+31663+31665=94989
inlocuim a,b,c in relatie si facem verificarea
5015+31661+31663+31665=100.004
100.004=100.004, deci este adevarat, astea sunt valorile lui a,b si c