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Design a logic circuit to convert a single digit BCD number to the number modulo six as follows (Do not detect illegal input):

1. Write the truth table for all bits. Label the input bits $I_1, I_2, \ldots$ with $I_1$ as the least significant bit. Label the output bits $R_1, R_2\ldots$ with $R_1$ as the least significant bit. Use $1$ to signify truth.
2. Draw one circuit for each output bit using, altogether, two two-input AND gates, one two-input  OR gate and two NOT gates.

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$${\begin{array}{|cccc|c|ccc|}\hline \bf{I_4}& \bf{I_3}& \bf{I_2}&\bf{ I_1}& &\bf{R_3}& \bf{R_2} & \bf{R_1}\\\hline 0&0&0&0&\bf{0} &0&0&0\\\hline 0&0&0&1&\bf{1}& 0&0&1 \\ \hline 0&0&1&0&\bf{2}& 0&1&0 \\ \hline 0&0&1&1&\bf{3}& 0&1&1 \\ \hline 0&1&0&0&\bf{4}& 1&0&0 \\ \hline 0&1&0&1&\bf{5} &1&0&1 \\ \hline 0&1&1&0&\bf{6}& 0&0&0 \\ \hline0&1&1&1&\bf{7}& 0&0&1\\ \hline 1&0&0&0&\bf{8}& 0&1&0 \\ \hline 1&0&0&1&\bf{9}& 0&1&1 \\ \hline \end{array}}$$

• $R_1 = I_1$
• $R_2 = I_2\overline{ I_3} + I_4$
• $R_3 = I_3\overline{I_2}$

This requires $2$ NOT gates, $2$ two-input AND gates and $1$ two-input OR gate.

by

Simply Y0=D
why other inputs not consider here that is after 9. i think input should go from 0 to 15.
edited
No. of Logic Gates given in question would be insufficient, so how you are going to implement your expression in circuit diagram???

and for Expression why you did not use don't care for (10,11,12,13,14,15) ?
why you did not use don't care for (10,11,12,13,14,15) ?
corrected.
R2=(i2)'(i4) + (i2)(i3')
The question says that a single digit BCD is converted. BCD values range from 0 to 9 only. So we can consider all the output values for (10,11,12,13,14,15) as don't cares. The answer is right. :)

After using Don't care (10,11,12,13,14,15) and after K-Map simplification you will get

R1=I1

R2=I2.I3' + I4

R3= I3.I2'

R4 = 0

Here, 2 input AND Gate used=2

2 input OR Gate used=1

NOT Gate used=2

Don't cares are not shown in the table but considered while constructing K-map for $R_1,R_2,R_3.$