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So here's the question:

We want to design a circuit that determines if a four-bit number is less than 10 and is also even.

a. (10 points) Write an expression of M such that M=1 if the four-bit number X (x3 x2 x1 x0) is less than 10 and is also even.

Is this asking for the answer in POS form since it is using "M" instead of "m"?

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  • \$\begingroup\$ Why do you think they care? Besides, comparator should in fact perform an arithmetical operation, thrn you can use its output together with the bit that depends on oddness. \$\endgroup\$
    – user76844
    Feb 9 '19 at 6:59
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They probably want the simplest form.

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Yes. The conventional representation for the POS form is M(.). Now, since the last bit for the even numbers is 0, it is enough to have 3 bits will all numbers from 0 to 4. So, we just need a mod 5 counter to generate all these numbers and the last bit of the four bits (LSB) being grounded always.

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  • \$\begingroup\$ Can you give some citation or reference for your statement that "POS form is M(.)"? I've been doing this for a long time and never saw this as a "conventional representation". In my experience we add words to the question to make it clear what is expected. \$\endgroup\$ Feb 9 '19 at 21:14

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