Aflati un numar de trei cifre care impartit, pe rand, la 40,50,60 da de fiecare data restul = 11

n=40a+11
n=50b+11
n=60c+11

n-11=40a
n-11=50b
n-11=60c
n-11| [40;50;60]
n-11=[40;50;60]*k, k∈N
n-11=[4;5;6]*10*k
n-11=60*10*k
n-11=600k
n=600k+11
n=611, 1211,...
 3 cifre are doar 611
n=611
 

Verificare
 611:30=20 rest 11
600:40 =15 rest 11
 611:60=10 rest11
problema e bine rezolvata

Rezolvarea este in atasare. :)