IFT2880 ORGANISATION DES ORDINATEURS ET SYSTEMES
Solutions des exercices de la semaine du 10 janvier
2.1b The decimal equivalent for 245316 is 2 · 1296 + 4 · 216 + 5 · 36 + 3 · 6 + 1 = 3655.
2.5 The easiest way to solve this problem is to use the approximation 210 » 1000, and the fact that 2A · 2B = 2A+B. From this, 1,000,000 is approximately 210 · 210, or 20 bits, and 4,000,000 is approximately 220 · 4, or 22 bits. Therefore, the representation of 3,175,000 will require 22 bits, or 3 bytes.
2.7b 2AB3 + 35DC --- 608F
2.7c 1FF9 + F7 --- 20F0
2.8c 11010011 + 10001010 --- 101011101
2.26a 0.10010012 = 1/2 + 1/16 + 1/128 = (64 + 8 + 1) / 128 = .0.570310 2.26c 0.2A112 = 2/12 + 10/144 + 1/1728 = (288 + 120 + 1) /1728 = 0.236610
2.27a 27 = (16 + 8 + 2 + 1) = 11011
for 0.625, either recognize that .625 = 5/8 or multiply: 0.625 * 2 = 1.250
0.250 * 2 = 0.5
0.5 * 2 = 1 The final result is thus 11011.101
The decimal representation can be found by performing group conversion:
0001 1011 . 1010
1 B A
The hexadecimal representation is thus 1B.A
