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Disciplina: Arquitetura e Organização de Computadores Docente: Ricardo Ferreira Aluno: Renato Paula de Oliveira Matricula: 201601312873 Exercício Pontuado Respostas: 1) 35/2 1 17/2 1 8/2 0 4/2 0 2/2 0 1/2 1 0 R= 100011 Letra (A) 2) 30/2 0 15/2 1 7/2 1 3/2 1 1/2 1 0 R= 11110 Letra (A) 3) 210/2 0 105/2 1 52/2 0 26/2 0 13/2 1 6/2 0 3/2 1 1/2 1 0 R= 11010010 310/2 0 155/2 1 77/2 1 38/2 0 19/2 1 9/2 1 4/2 0 2/2 0 1/2 1 0 R= 100110110 410/2 0 205/2 1 102/2 0 51/2 1 25/2 1 12/2 0 6/2 0 3/2 1 1/2 1 0 R= 110011010 810/2 0 405/2 1 202/2 0 101/2 1 50/2 0 25/2 1 12/2 0 6/2 0 3/2 1 1/2 1 0 R=1100101010 1610/2 0 805/2 1 402/2 0 201/2 1 100/2 0 50/2 0 25/2 1 12/2 0 6/2 0 3/2 1 1/2 1 0 R= 11001001010 6410/2 0 3205/2 1 1602/2 0 801/2 1 400/2 0 200/2 0 100/2 0 50/2 0 25/2 1 12/2 0 6/2 0 3/2 1 1/2 1 0 R= 1100100001010 12710/2 0 6355/2 1 3177/2 1 1588/2 0 794/2 0 397/2 1 198/2 0 99/2 1 49/2 1 24/2 0 12/2 0 6/2 0 3/2 1 1/2 1 0 R= 11000110100110 12810/2 0 6405/2 1 3202/2 0 1601/2 1 800/2 0 400/2 0 200/2 0 100/2 0 50/2 0 25/2 1 12/2 0 6/2 0 3/2 1 1/2 1 0 R= 11001000001010 25510/2 0 12755/2 1 6377/2 1 3188/2 0 1594/2 0 797/2 1 398/2 0 199/2 1 99/2 1 49/2 1 24/2 0 12/2 0 6/2 0 3/2 1 1/2 1 0 R= 110001110100110 25610/2 0 12805/2 1 6402/2 0 3201/2 1 1600/2 0 800/2 0 400/2 0 200/2 0 100/2 0 50/2 0 25/2 1 12/2 0 6/2 0 3/2 1 1/2 1 0 R= 110010000001010 4) 11 = 1*2^1 + 1*2^0 = 3 110 = 1*2^2 + 1*2^1 + 0*2^0 = 6 101 = 1*2^2 + 0*2^1 + 1*2^0 = 5 1010 = 1*2^3 + 0*2^2 + 1*2^1 + 0*2^0 = 10 1011 = 1*2^3 + 0*2^2 + 1*2^1 + 1*2^0 = 11 10 = 1*2^1 + 0*2^0 = 2 111 = 1*2^2 + 1*2^1 + 1*2^0 = 7 1000 = 1*2^3 + 0*2^2 + 0*2^1 + 0*2^0 = 8 10000 = 1*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 0*2^0 = 16 11111 = 1*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0 = 31
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