What is the value of xyz? (1) y! = 6 and x! > 720 (2) z is the least

Statement 1 does not give us enough information on x and z
Not sufficient.
Statement 2 tells us that z is 0. 0 is the least even integer greater than -1. So the product of any x and y with z would be 0.
So sufficient.
IMO B

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hi , eswarchethu135
can 0 be considered as even integer?

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

hi Archit3110

Let's try to evaluate whether 0 is even or odd.

We know that, even \(\pm\) odd = odd ______________(1)
even * odd = even _____________ (2)

Using these 2 rules, we will evaluate whether 0 is even or odd.

Let's assume 0 is odd and substitute in eq (2)

Then, even * odd = even

0 * even = even

0 multiplied by any number gives 0 as the answer. Now in the above eq, we are getting 0 as the answer but we assumed 0 as an odd integer. So this is contradicting.

Now substitute in eq (1),

0 \(\pm\) even = odd

0, when added or subtracted by an even integer gives an even integer. So this is also contradicting.

Now let's assume 0 is even and substitute in eq (1),

0 \(\pm\) odd = odd

This statement now holds perfectly.

Now substituting the same in eq (2),

even * odd = even

0 * odd = even.

As for the above equation, the answer should be an even integer and we considered 0 as an even integer this statement also holds perfectly.

In this way, we can prove in many ways that 0 is definitely an even integer and can never be an odd integer. It is just that it is neither positive nor negative.

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