Solutions

1. 302 =1 two-fifty-sixes, zero one-twenty-eights, zero sixty-fours, 1 thirty-two's, zero sixteen's, 1 eight's, 1 four's, 1 two's and zer one's or units; in other words,

1*256 + 0*128 + 0*64 + 1*32 + 0*16 + 1*8 + 1*4 + 1*2 + 0*1 = 100101110

2. 8+1=9

3. 128+64+32+16+8+4+2+1=255

4. 128+64+32+8+4+2=238

5. 64+16+4+1=85

6. 00001000

7. 10000110

8. 11100010

9. 00000100

10. 10111001

11. 1101

12. 1 0101

13. 1100 1111

14. 111 1100 1010

15. 10 1101

16. 100 1001

17. 14

18. 208

19. 42

20. 276

21. 15

22. 255

23. 65535

24. b. Accurate Note: correctness is not guanteed as per the GIGO principle.

25. 256*1024

26. 524288

27. 655360

28. 737280

29. 1 mega byte = 1,048,576 bytes

30. 4 x 1024 x 1024 = 4,194,304 -- roughly 4 million bytes

31. 6 x 1024 = 6,144 megabytes -- roughly 6 thousand MB

32. 10 x 1024 = 10,240 -- roughly 10 thousand PB

33. T = 0101 0100 O = 0100 1111 D = 0100 0100 A = 0100 0001 Y = 0101 1001

34. H = 0100 1000 E = 0100 0101 L = 0100 1100 L = 0100 1100 O = 0100 1111

35. f = 0110 0110

36. l = 0110 1100

37. 111013221

38. 250751

39. 151E9

40. The number "10" read as a binary number represents decimal 2 so the quote distinguishes 2 kinds of people as stated.

41. The MSB place value = 32 so the 'excess' value is 32 which is to be subtracted from the total sum which is 25 so 25 - 32 = -7

42. Since the MSB, the sign bit is a 0 (+), sum up all the 1s and add the sign so the value is +25

43. The MSB place value = 16 so the 'excess' value is 16 which is to be subtracted from the total sum which is 29 so 29 - 16 = +13

44. Since the MSB, the sign bit is a 1 (-), convert to its signed complement which is 00011 or +3 and apply the original sign; so -3

45. Perform subtraction by preserving addition

46. First represent +7 as 0111;

  • since this is the largest number that needs to be represented,
  • the minimum bit-length is 4 in this case;
    the value -6 needs to be represented but we first represent +6 and then convert to its negative signed complement ;
  • so 0110 converts to 1010 for -6
  • Now we add the two numbers representing +7 and -6;
  • 0111 and 1010;
  • The result is initially 10001 but the 5th extra bit is not allowed so truncate it;
  • the final result is 0001 or +1.

47. The integer value left of the radix is 8+2+1 or 11;

  • In the fraction side, the last column with a 1 is the 1/256 column
  • so that is the unit to count;
  • Starting from the unit column, there is 1 unit of 1/256 and
  • moving left we find 2 more, and 16 more and 32 more;
  • So, there are 32+16+2+1 or 51 units of 1/256 or 51/256.
  • Therefore, the value represented is 11 and 51/256th.

48. The MSB of 1 represents the negative (-) sign in floating-point;

  • the exponent is 001 in excess notation which is 1-4 or -3;
  • the mantissa begins as .1101 but,
  • after applying the exponent, the radix moves 3 positions to the left becoming .0001101;
  • the resulting real number represents total units of 1/128th
  • of which there are 8 + 4 + 1 or 13;
  • therefore the value represented is -13/128th.

49. The MSB of 0 represents the negative (+) sign in floating-point;

  • the exponent is 101 in excess notation which is 5-4 or +1;
  • the mantissa begins as .1001 but,
  • after applying the exponent, the radix moves 1 position to the right becoming 1.001;
  • the integer represented is 1 and the fraction represented is 1/8;
  • the resulting real number therefore is +1 and 1/8th.

50. The fraction 3/8 is .01100

  • so the radix must move 1 postion to the left;
  • so the exponent should represent -1.

Note: in normalized floating-point we are working with does not allow the mantissa to begin with a 0.

51. The value 4 and 3/4 in binary is 100.11;

4 ones to the left of the radix and to the right, 1/2 and a 1/4 which is 2/4 and 1/4 or 3/4

  • so the mantissa needs to be 10011;
  • and the exponent should indicate that the radix should move 3 positions to the right of the mantissa (or +3) from its assumed original position at the left of the mantissa;
  • representing +3 in 4-bit excess notation or excess(8) is 1011 [8+3];
  • therefore the final number, including the entire value's positive sign-bit, is 0 1011 10011