Discussion
The most suitable gates to check whether the number of ones in a digital word is even or odd is :
Answer: A
The most suitable gates are EX-OR gates.
The output of an exclusive-OR is 1 when an odd number of its inputs are 1. Chaining EX-OR gates across all the bits of a word (\(b_{0} \oplus b_{1} \oplus \dots \oplus b_{n}\)) therefore gives 1 if the number of ones is odd and 0 if it is even. This is how parity generators and parity checkers are built, for example in the 74180 and 74280 chips.
NAND and NOR are universal gates, so a parity circuit could in principle be built from them, but only by first making EX-OR functions from several gates each. AND and OR alone cannot detect parity, since they cannot tell two ones from one.
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