Question

Restul impartirii numarului a=1+2+2²+2³+....+2^99 la 7 este:
a)6
b)4
c)0
d)2.

Answer 1

a = (1+2+2²) + 2³·(1+2+2²) +........+2^97·(1+2+2²) = 7(1+2³+....+2^97) = M7 ⇒ r=0