HOT Competition: A function g is defined, for all positive integers x

We can't say if p is even or odd from this statement.
If p is even, g(p) = odd, and g(g(p)) will be even. even + odd can give 195
similarly, if p is odd, g(p) = even and g(g(p)) will be odd. 195 can be the result still.

Now, can we say directly at this step that statement 1 is insufficient? In this case, it turns out the answer is yes. But, I am not so sure we can say that directly without solving for p.
Because we may run into one of the following cases:
1. What if the final equation comes out to be the same irrespective of us considering p EVEN or ODD
2. What if we consider p as EVEN, then solve for p, and it turns out to be odd (or vice versa).
In either of the above cases, A alone will suffice. In this particular case, however, it turns out p = 11 (if we consider it to be odd), or p = 10 (if we consider p even). So, statement 1 is not sufficient


All that this statement tells us is p is even. It doesn't tell us anything about the value of p. So clearly, multiple possibilities exist. In fact, for any even p, the statement 2 will hold true.

But combining statement 2 with 1, we get that p can only be 10 (and not 11). So, they are together sufficient to answer the above question.
Answer: C

First , notice that g of g will be even if x is even and odd if x is odd

Statement one --- Lets say x is even that means g of g of of p will be even and g of p (which is 4p-5) will be odd… so we have a even and a odd number adding up to 195 which is odd as well. Cool so p can be any even value . We can that the statement is insufficient at this point, but analyze further.

Lets say x is odd that means g of g of of p will be odd and g of p (which is 5p-15) will be even… so again, we have a even and a odd number adding up to 195 which is odd as well.

So p can be any particular (stress on this) even integer or a particular(stress on this) odd integer. Hence Insufficient


Again lets say, p is even … then g of g of p will be even … again g of g of (anything even) will be even .

lets say, p is odd … then g of g of p will be odd … again g of g of (anything odd) will be odd . But, per the statement it has to be even so we'll disregard this case.
Statement 2 simply means that p is even. Insufficient by itself.

On combining , lets solve with p as even
we get 20p-40 + 4p-5 = 195 …. 24p = 240 p = 10.

C it is.

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