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assignment 1 solution.docx

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Department
Information Technology
Course
ITEC 1000
Professor
Peter Khaiter
Semester
Winter

Description
AP/ITEC1000 Section N "Introduction to Information Technologies" Winter 2013 Assignment #1 Answers and Solutions 1.a 4607 = 1000111111111 because 10 2 4607 / 2 = 2303 remainder 1 2303 / 2 = 1151 remainder 1 1151 / 2 = 575 remainder 1 575 / 2 = 287 remainder 1 287 / 2 = 143 remainder 1 143 / 2 = 71 remainder 1 71 / 2 = 35 remainder 1 35 / 2 = 17 remainder 1 17 / 2 = 8 remainder 1 8 / 2 = 4 remainder 0 4 / 2 = 2 remainder 0 2 / 2 = 1 remainder 0 1 / 2 = 0 remainder 1 1.b 371.7510 101110011.112because 371 / 2 = 185 remainder 1 And 0.75 185 / 2 = 92 remainder 1 x 2 92 / 2 = 46 remainder 0 1.50 46 / 2 = 23 remainder 0 x 2 23 / 2 = 11 remainder 1 -------- 11 / 2 = 5 remainder 1 1.00 5/ 2 = 2 remainder 1 2 / 2 = 1 remainder 0 1 / 2 = 0 remainder 1 1.c 0.63827 =10.10100.. bec2use 0.63827 x 2 1.27654 x 2 0.55308 x 2 1.10616 x 2 0.21232 x 2 0.42464 6 5 2 1 2.a 1100110 =2 2 + 2 + 2 + 2 = 64 + 32 + 4 + 2 = 102 10 2.b 10111.011 = 23.375 because 2 10 4 2 1 0 -2 -3 10111 =22 + 2 + 2 +2 = 23 10 And 0.011 2 2 + 2 = 0.375 10 -1 -3 -5 2.c 0.10101 =2= 2 + 2 + 2 = 0.5 + 0.125 + 0.03125 = 0.65625 10 3.a D6C8 = 1101011011001000 because 16 2 D 16 = 1101 2 6 = 0110 16 2 C 16= 1100 2 8 161000 2 3.b 9A.DC = 16011010.11011100 because 2 9 16= 1001 2 A 16= 1010 2 D = 1101 16 2 C 161100 2 3.c 0.C3F =16.110000111111 becau2e C16 = 1100 2 316 = 0011 2 F16 = 1111 2 4.a 11010011101 = 69D because 2 16 1101 2 D 16 1001 = 9 2 16 0110 2 6 16 4
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