Question

determinati a b c d astfel incit abcd+abc +ab +a=1568

Answer 1

abcd+abc+ab+a=1568

a,b,c,d<10;  (abcd=1000a+100b+10c+d)

=> (1000a+100b+10c+d)+(100a+10b+c)+(10a+b)+a=1568

1111a + 111b + 11c +d =1568

a=1; nu poate fi 2 pt ca 1111•2=2222

=> 111b+11c+d=1568-1111

111b+11c+d=457

=> b=4

11c+d=457-444

11c+d=13

=>c=1 si d=13-11

d=2

a=1, b=4, c=1, d=2

abcd=1412

verificare: 1412+141+14+1=1568