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

Given: A function g is defined, for all positive integers x > 3, as g(x)= 4x-5 if x is even and g(x)=5x-15 if x is odd, for all positive integers x > 3. What is the value of the positive integer, p?


(1) g(g(p))+g(p)=195
Case1:p is even
g(4p-5)+4p-5=195
4p-5 is odd
5(4p-5)- 15+4p-5=195
24p=240
p = 10
Case 2:p is odd
g(5p-15)+5p-15=195
5p-15 is even
4(5p-15)-5+5p-15=195
25p=275
p=11
NOT SUFFICIENT

(2) g(g(g(g(p)))) is even.
p is even
The value of p can not be found
NOT SUFFICIENT

(1) + (2)
(1) g(g(p))+g(p)=195
(2) g(g(g(g(p)))) is even.
p is even
p = 10
SUFFICIENT

IMO C

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A function g is defined, for all positive integers x > 3, as g(x)=4x−5 if x is even and g(x)=5x−15 if x is odd, for all positive integers x > 3. What is the value of the positive integer, p?

lets understand the given first:
g(x) = 4x-5 if x=even, so if we put even value we get g(x=even) = odd value and if we apply given function to the output of g(x=even) than we need to use g(g(x)) = 5(g(x) -15

same happens to if x= odd, it starts with g(x) = 5x-15 and than results even value and we need to use g(x) = 4x-5 function for the output

Now lets see the statement

(1) g(g(p))+g(p)=195g(g(p))+g(p)=195
St. 1 assume p = even above equation become:
5(4p-5)-15 + 4p-5 = 195
solve we get p = 10
if p =odd equation becomes
4(5p-15)-5 + 5p-15 = 195
solve for p we get p =11

so Not sufficient

(2) g(g(g(g(p)))) is even.
st. 2 tell about the convoluted function results in even value, so if we consider p = even than only it is possible for an even resluts
if p = even
g(p) = odd
g(g(p))=even
g(g(g(p)))=odd
g(g(g(g(p))))= even

so p can take any even value >3 Not Sufficient

Combining 1 &2 we get p = 10 sufficient

IMO C

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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