Question

Demonstrații ca nr a=2^0+2^1+2^2+2^3+2^4+2^5 este divizibil (:) cu 7

URGENT VA ROG​

Answer 1

$a = {2}^{0} + {2}^{1} + {2}^{2} + ... + {2}^{5}$

$2a = {2}^{1} + {2}^{2} + {2}^{3} + ... + {2}^{6}$

$2a - a = {2}^{1} + {2}^{2} + {2}^{3} + ... {2}^{6} - ( {2}^{0} + {2}^{1} + {2}^{2} + ... + {2}^{6} )$

$a = {2}^{1} + {2}^{2} + {2}^{3} + {2}^{4} + {2}^{5} + {2}^{6} - {2}^{0} - {2}^{1} - {2}^{2} - {2}^{3} - {2}^{4} - {2}^{5}$

$a = {2}^{6} - 1 = 64 - 1 = 63$

63:7 => a:7