shanmugamgsn wrote:
Vips0000 wrote:
MOKSH wrote:
The function g(x) is defined for integers x such that if x is even, g(x) = x/2 and if x is odd, g(x) = x + 5. Given that g(g(g(g(g(x))))) = 19, how many possible values for x would satisfy this equation?
a)1 b)5 ,c)7 ,d)8 ,e)11
Wow, more like a mathmatical puzzle than a gmat question. I love it!
Let me define terms:
in g(x) = R
x is argument, R is result, g() is function,
in g(g(g(g(g(x))))), g1 is inner most, g5 is outermost for identification.
From definition of function g, we can deduce that:
If Result is even then two possibilities for argument = 1 Even 1 Odd
If Result is odd then one possibility for argument = 1 Even
Since final result = 19 = Odd
Possibilities:
g1:
1 Eveng2: 1*(Even,Odd ) =
1 Even 1 Oddg3: 1*(Even,Odd) + 1 Even =
2 Even 1 Oddg4: 2*(Even, Odd) + 1 Even =
3 Even 2 Oddg5: 3*(Even, Odd) + 2 Even =
5 Even 3 Odd = Total 8Ans D it is!
Vips im totally lost in this... can u explain!!!
how u started g1 with even? based on answer choices?
if so how come u calculated g2?
ha ha.. the explanation was this:
If Result is even then two possibilities for argument = 1 Even 1 Odd
If Result is odd then one possibility for argument = 1 Even
Anyway, to start from scratch:
how u started g1 with even? based on answer choices? question says,
g(x) = x/2 , if x is even=> Observation: if x is even, result is even/2 which could be odd or even.
g(x) = x+5, if x is odd => Observation: if x is odd, result is always even. (odd number+5= even number)
Another way to get there :
We know final result is 19. that is:
g(something) =19
Now what is this something? it could be 38 giving 19 when divided by 2. Or it could be 14 when 5 is added.
However, it can not be 14 because 14 is even and g(14) will be 7 not 19 by the definition of g(x). So there is only possiblity 38.
So if result is odd, then argument must have been even.
Therefore for argument of g1, you start with Even since the result is odd (19).
if so how come u calculated g2Lets again see, we found out that argument of g1 was even. Now this even could have been result of another even number or an odd number. Let see the example:
taking forward previous values. We found above that argument for g1 is 38.
now, argument for g2? we know that g2(something) =38
What is this something? it could be 76, which gives 38 when divided by 2. Or it could be 33 which gives 38 when 5 is added. Both of these values are possible as per g(x) definition.
It can not be a gmat question. but its good fun.
to summarize, try to understand these lines:
If Result is even then two possibilities for argument = 1 Even 1 Odd
If Result is odd then one possibility for argument = 1 Even