ITEC 1000 Chapter Notes - Chapter 1-3: Octal

• School: York University
• Department: Information Technology
• Course: ITEC 1000
• Professor: Peter A Khaiter

Document Summary

351 / 2 = 175 remainder 1 and 0. 25. 175 / 2 = 87 remainder 1 x 2. 87 / 2 = 43 remainder 1 0. 50. 43 / 2 = 21 remainder 1 x 2. 21 / 2 = 10 remainder 1 -------- 6. a 67678 = 1102 1112 1102 1112 = 11012 11112 01112 = df716. 6. b cf9a16 = 11002 11112 10012 10102 = 0012 1002 1112 1102 0112 0102 = 7. a 110201. 012 this cannot be a valid number in base 2, since 2 is not a proper digit in base 2. 7. b 763. 688 this cannot be a valid octal number, since 8 is not a proper digit in base. 7. c ae. g516 this cannot be a valid number in base 16, since g is not a proper digit in base. 11. a 397310 = 1111100001012 therefore, it will take 12 bits.