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This is the lsd of the tens−complement. For the remainder part of the tens−complement, we subtract 6 , 5 , 3 , and 2 from 9 and we obtain 3 , 4 , 6 , and 7 respectively. Therefore, the tens−complement of 23567 is 76433 . 8642 Solution: We first subtract 2 (lsd) from 10 and all other digits from 9 . 1358 . 562 Solution: We first subtract 2 (lsd) from 10 and all other digits from 9 . 438 . 2 Nines−Complement The nines−complement of a number can be found by subtracting every digit (lsd) of that number from 9 .
4 Exercises 1. Convert ( +74329 ) 10 and ( – 74329 ) 10 into sign magnitude binary representation form. 2. Convert the decimal numbers ( +47387 ) 10 and ( – 47387 ) 10 into sign magnitude binary representation form. 3. Give the IEEE single precision floating point representation for the number ( 1253 ) 10 4. Give the IEEE single precision floating point representation for the number ( – 1253 ) 10 5. 485 ) 10 6. 5 Solutions to End−of−Chapter Exercises 1. For ( +74329 ) 10 , dec2hex(74329) returns ( 12259 ) 16 = ( 0001 0010 0010 0101 1001 ) 2 where the spaces were added to indicate the hexadecimal groupings.
20. Here, the first non−zero least significant digit of the subtrahend is 2 and it is subtracted from 10 . Thus, Minuend = 43561 stays unchanged → 43561 ⎫ + Subtrahend = 13820 take tens – c omplement → 86180 ⎬⎭ Discard end carry → 129741 Therefore, ( 43561 – 13820 ) 10 = ( 29741 ) 10 21. Minuend = 13820 stays unchanged → 13820 ⎫ + Subtrahend = 43561 take tens – c omplement → 56439 ⎬⎭ No end carry → 70259 Since there is no end carry, we take the tens−complement of the sum 70259 and we place a minus (−) sign in front of it resulting in – 29741 .