Răspunsuri

2014-09-28T08:59:52+03:00
1000a+100b+10c+2+100a+10b+c=2004
1100a+110b+11c+2=2004
11(100a+10b+c)+2=2004
a=1
b=8
c=2
abc=182
verificare 1100+880+22+2=2004
2014-09-28T09:01:42+03:00
Abc2+abc=2004 1000a+100b+10c+2+100a+10b+c = 2004 1100a+110b+11c=2002 11(100a+10b+c)=2002 Abc(cu bara deasupra) =2002/11