1100 or 0110 or both are correct..

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**BCD conversation** :- Convert each digit into groups of four binary digits equivalent.

6248 = 6 - 0110 2-0010 4 - 0100 8-1000

`(6248)10 = (0110 0010 0100 1000)`

_{EXCESS THREE CODE :- }

_{Add +3 in each digit then convert into groups of four binary digit equivalent.}

(9)_{10 = }(1001)_{excess3 , }(5)_{10 = }(0101)_{excess3, }(7)_{10 = }(0111)_{excess3, }(11)_{10 = }(1011)_{excess3}

Now,combine together (6248)_{10}=(1001 0101 0111 1011)_{excess3}

**2421 CODE :- _{ }** Represent each decimal digit in binary with respect to weight 2 4 2 1.

6 = 0x2^{1} + 1x2^{2} + 1x2^{1}^{ }+ 0x2^{0}. 2 = 0x2^{1}+ 0x2^{2} + 1x2^{1}^{ }+ 0x2^{0}.

4 = 0x2^{1} + 1x2^{2} + 0x2^{1}^{ }+ 0x2^{0}. 8 = 1x2^{1} + 1x2^{2} + 1x2^{1}^{ }+ 0x2^{0}.

combine all those. ANS :- (0110 0010 0100 1110)_{2421}

**6311 CODE :- _{ }** Represent each decimal digit in binary with respect to weight 6 3 1 1.

6 = 1x6^{1} + 0x3^{1} + 0x1^{1}^{ }+ 0x1^{1}. 2 = 0x6^{1}+ 0x3^{1} + 1x1^{1}^{ }+ 1x1^{1}.

4 = 0x6^{1} + 1x3^{1}+ 0x1^{1}^{ }+ 1x1^{1}. 8 = 1x6^{1} + 0x3^{1} + 1x1^{1}^{ }+ 1x1^{1}.

combine all those. ANS :- (1000 0011 0101 1011)6311

Thank you....

+1 vote

Assuming the number 6248 in decimal .

BCD : 0110 0010 0100 1000

Excess 3: 1001 0101 0111 1011

2-4-2-1 0110 0010 0100 1110 (This is not unique since the representation in 2-4-2-1 varies) Each digit is represented as the summation like 6 is obtained 4+2 , so bit position 4,2 is made 1 in the first digit )

6-3-1-1 1000 0011 0110 1011 (This representation is also not unique and is one of the representations.

BCD : 0110 0010 0100 1000

Excess 3: 1001 0101 0111 1011

2-4-2-1 0110 0010 0100 1110 (This is not unique since the representation in 2-4-2-1 varies) Each digit is represented as the summation like 6 is obtained 4+2 , so bit position 4,2 is made 1 in the first digit )

6-3-1-1 1000 0011 0110 1011 (This representation is also not unique and is one of the representations.

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