Determinati a,b, c stiind ca:
abc = 5ab + 4 bc, a, b, c sunt nenule
a, b, c nr nenule
a,b,c apartin {0,1,2,3,4,5,6,7,8,9} si a,b sunt nenuli.
abc=5ab+4bc <=> 100a+10b+c=50a+5b+40b+4c <=> 50a-35b-3c=0 <=> 5(10a-7b)=3c => 5|c. => c apartine {0,5}.
Daca c=0 => 50a-35b=0 <=> 10a=7b => 10|b => b=0 => a=0, ceea ce nu convine.
Daca c=5 => 50a-35b-15=0 <=> 10a=7b+3. => 10|7b+3 => 7b+3 apartine {10,20,30,40,50,60} <=> 7b apartine {7,17,27,37,47,57}. => b=1. => a=1.
In concluzie, a=1, b=1 si c=5.
abc=5ab+4bc <=> 100a+10b+c=50a+5b+40b+4c <=> 50a-35b-3c=0 <=> 5(10a-7b)=3c => 5|c. => c apartine {0,5}.
Daca c=0 => 50a-35b=0 <=> 10a=7b => 10|b => b=0 => a=0, ceea ce nu convine.
Daca c=5 => 50a-35b-15=0 <=> 10a=7b+3. => 10|7b+3 => 7b+3 apartine {10,20,30,40,50,60} <=> 7b apartine {7,17,27,37,47,57}. => b=1. => a=1.
In concluzie, a=1, b=1 si c=5.