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Add 648 and 487 in BCD code.Ans. In BCD Code, addition of 648 and 487:
6 4 8 = 0 1 1 0 0 1 0 0 1 0 0 04 8 7 = 0 1 0 0 1 0 0 0 0 1 1 1-------------------------------------------------- 1 0 1 0 1 1 0 0 1 1 1 1 10 12 15--------------------------------------------------In the given problem every three groups are invalid, since the four bit sum is more than 9. In these cases, add +6(that is 0110) to the 4 bit sum to skip the 6 invalid states. Whether a carry is generated while adding 6, so add the carry to the next four bit group that is,
6 4 8 = 0 1 1 0 0 1 0 0 1 0 0 04 8 7 = 0 1 0 0 1 0 0 0 0 1 1 1-------------------------------------------------- 1 0 1 0 1 1 0 0 1 1 1 1 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 1 1 1 1 1------------------------------------------------------0001 0 0 0 1 0 0 1 1 0 1 0 1
1 1 3 5-------------------------------------------------------
Hence addition of 648 and 487 is 1135 in BCD Code.
1) Short both I/Ps of NAND gate & use the gate as an inverter. 2) Connect any one of the two I/Ps to VCC & use the remaining I/P & use the gate as an inverter.
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