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The values of P,Q,R, if 47Q80 is the 10's c complement of RPRP0 are

a)4, 3, 2  b) 5,4,4 c)3,4,5  d)2,4,5
in Digital Logic by Boss (49.3k points)
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2 Answers

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Ans is option D.

10's compliment of base 10 number X with n digits  = 10n - X

here, n = 5, X = RPRP0

hence

   100000 - RPRP0 = 47Q80

from unit place : 0-0 = 0

then, 10 - P = 8. Hence P =2.

then, 9 - R = Q ( compute this later)  ...(1)

then 9 - P = 7. Confirmation of P =2.

then, 9 - R = 4. hence R =5.

again from eq (1),   9-R = Q => 9-5 = 4, hence Q =4.

P = 2, Q=4, R =5

by Boss (18.8k points)
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+1 vote
100000-47Q80=RPRP0

now 100000=47Q80+RPRP0

P+8 SHOULD HAVE 0 IN THE UNIT PLACE

ONLY 3 IS MATCHING SO P=2;

NOW P+8 WILL GIVE A CARRY SO R+Q+1 SHOULD HAVE 0 IN UNIT PLACE =4+5+1=10

NOW 100000=47480+52520
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