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by [email protected] » Sun Jul 21, 2013 8:47 am
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?
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by Brent@GMATPrepNow » Sun Jul 21, 2013 9:05 am
[email protected] 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?
One of the great things about almost all GMAT math questions is that they can be solved using at least 2 different approaches. Typically, one approach is much faster than the other(s). In my opinion, this question is too far too time-consuming to be a legitimate GMAT question.

Also, this question basically has one approach, and that approach involves "brute force," which requires far too much time. To see what I mean, check out these solutions:
- https://www.beatthegmat.com/functions-t102958.html
- https://www.beatthegmat.com/killer-probl ... 75629.html

Personally, I think this question is akin to asking test-takers to find the sum of all prime numbers from 2 to 401. Sure, we can find the sum using brute force, but there's no nice (fast) way to do it.

Cheers,
Brent
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by [email protected] » Sun Jul 21, 2013 11:20 am
Hi shibsriz,

While I agree with Brent that this question doesn't have the "feel" of a GMAT question (Where are the answer choices? Those answers would likely give you a hint as to the possibilities.), there are some Number Properties that would help you to answer it (and answer it quicker, if you spot the larger pattern).

The crux of this question is in how the functions work and what you "end up with" (and how this info allows you to work "backwards")

You have to run the function 5 times and at each level...
If x = even, then divide by 2
If x= odd, then add 5
After the 5th run, total = 19

Ending with a total of 19 limits your options.
You can only get to 19 if the prior number was 38 (since 38 is even, you'd divide by 2 to get 19).
You CAN'T get to 19 if the prior number is 14 (14 is even, so you'd divide by 2; you WOULDN'T add 5).

This means that when you're looking at an odd number, there's ONLY 1 way to get to it (by dividing by 2 from the previous number).

If, however, you're looking at an even number, there are 2 ways:
If you're trying to get to 38, then you can get there with....
76 (even, so divide by 2) = 38
33 (odd, so add 5) = 38

By working backwards, you can quickly map out each step. To help you continue on, think about....
how to get to 76
how to get to 33

Then continue on from there.

In the end, you should have 8 answers. And for the super-nerds out there, if you look closely, you'll see the beginning of a Fibonacci sequence, but that's not necessary to solve this question.

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by Matt@VeritasPrep » Sun Jul 21, 2013 10:00 pm
I think the fast approach would be this:

If your final value is odd, then your preceding value must have been even, as you can't have added 5 to an even number to get there, given the definition of the function.

If you preceding value was even, the value preceding that could have been even or odd, as you could've cut a larger even number in half or you could've added 5 to an add number.

Continue like this: every even possibility doubles, every odd possibility simply 'replaces' itself. So every time you have an even number, you have two ways of getting there. Following our tree:

Final time: Odd
Time before: Even
Time before: Odd or Even
Time before: Even, Odd or Even
Time before: Odd or Even, Even, Odd or Even
Time before: Even, Odd or Even, Odd or Even, Even, Odd or Even

Not that hard to generate the possibilities, though it's time-consuming if you bother to compute the actual numbers (and you don't really have to). This seems like a totally reasonable 75th+ percentile question on the GMAT -- it's certainly not as hard as a few I saw last time -- though I haven't seen a question much like this.
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