f(x) is defined as the largest integer n such that x is divi

@honchos @A.Haung the biggest thing that ties me up on problems like this is instead of using a 4-step "Understand, Strategize, Solve, Analyze" approach, I jump into the math before truly understanding the problem--classic GMAT mistake and always a recipe for disaster either 1) time destroyer or 2) not moving to educated guessing after the first minute because you're frantically doing arithmetic with no end in sight, then have to randomly guess.

Functions problems are inherently difficult because you get confused with numbers going in and numbers coming out. That's where I began my rephrase: 1) what are the inputs, 2) what are the outputs, 3) what do I want? 4) what do I need to get there? Bunuel's explanation is great but for a non-natural math mind (aka my mind) it's helpful to talk through the full complexity, tricks and traps here. This explanation may be wordy but the hope is it really gets into some of the interstitial information Bunuel touches on.

Understand:
First read-through (general understanding): This is a functions problem. Input is x, function is something complicated in between, output (what the function is "defined" as) is something called <n>. ACs all say f(something), but what I want is the <n> that results when I put that <x> into the function. The test writers deliberately write the vague description, "the number" in the second sentence to get you confused--if you don't take your time, you don't know if "the number" is x, <n>, f(x), or whatever other gobblygook. Sooo, I'm looking for <n>--that's "the number." First off, have to understand how this function works!
2nd read-through (math): Function works by outputting <n> after x is divided by 2^n. x is divisible by 2^n--which means 2^n goes into x evenly. This means the right AC should have a factor which is either 2 (2^1), 4 (2^2), 8 (2^3), 16 (2^4)... that's what I'm looking for. I want the biggest <n> possible--that means, I want the biggest 2^n that is also a factor of x

Strategize: I am given 5 possible Xs in the 5 ACs. Excellent! I can input these to find which <n> is biggest. In picking which one to try first, natural tendency may be toward E--biggest x value, therefore biggest n value, right? This probably gets a lot of people trying to guess with time running out, but it's a TRAP! just because the x value is large, does not mean that that number has a very high 2^n that divides into it. Other way I thought about it: none of the ACs are equal to a power of 2 (2, 4, 8, 16, 32, 64, 128), so I'm going to have to dig deeper into these x values to discover which has the biggest <n> output.
Luckily, 2 is a prime number, therefore doing a prime factorization should show me the "number of 2s" in each x value. Best way I know to examine factors--dig deeper into the factors behind each AC--is by a factor tree. What are we looking for? (always helpful to re-ask that question as often as you can) We are looking for the x value that contains the most 2s in its prime factorization. Why? Because the total # of 2s in the prime factorization = n value, and we want the biggest <n> value to answer the question.

Solve:
A) 24 prime factorization: 2*2*2*3 --> three 2s, n=3
B) 42 p.f.: 2*7*3 --> one 2, n=1
C) 62 p.f.: 2*31 (can't make 31 any smaller, it's prime.. how do I know? just had to memorize it.. or try it out) --> one 2, n=1
D) 76 p.f.: 2*2*19 --> two 2s, n=2
E) 84 p.f.: 2*2*3*7 --> two 2s, n=2

Analyze:
Looks like A is the winner! Turns out that the biggest <n> is actually a small number, and it looks nothing like the x that was inputted into the function (24). One more check at the wording, and I think I have good grasp on what the question is asking and I answered the right question. Can I get any other powers of 2 bigger than 2^3 that go into any other ACs? Doesn't look like it.

Takeaway: Careful with over-obvious answers! As mentioned, if we were to have picked the biggest x value we saw in the ACs (84). we would have gotten the answer wrong.
Other Big Takeaway: Slow down! Although 2 minutes seems like a flash, it really isn't--although this stuff took a while to write out here, really this was stuff that was racing through my head over the course of only 10-20 seconds. If you're doing math with no daylight in sight, you may as well just guess randomly and move on, 'cause you're just wasting time otherwise. Sometimes "Understand" part at the beginning takes the most time, as everyone says the arithmetic isn't (usually) rocket science.
Final Big Takeaway: Super-solid fundamentals and practice, practice, practice! I find Khan Academy is great for filling in holes from the 6th-8th grade math that I am still continuing to fill today--and it's free. Even if you understand, make sure you walk yourself through your reasoning. Just because you read through the OA doesn't mean you really, truly "get it." If you get the ? wrong initially, save it for later and come back to see if you can solve it after a week--if you're still struggling, commit it to a flash card to take with you for breaks at work, on the subway, in cabs etc

Bunuel,

I have difficulty in understanding the question Can you shower some Insight what the question is saying?

I guess its f(84). Since 2^84 gives the largets divisor. Hence the number divisible by 2^84 will be the greatest. Moreover as all are powers of 2 all others 1-4 options also divide the number N = 2^84.
Any other answers. Pls correct me

Darshan

Oh I got the q wrong thx

bkk145,

i only understand what the question was asking after reading your answer. Can you tell me how you attack the question?

Andy Huang

My 1000 posts!!! yeah!

Actually, I assumed that "N" is the same as "n". I don't know if it is a typo or what, but without that assumption, I wouldn't be able to connect the dots. I hope my answer is right.

I don't think I have any particular way to attack the question. I just wrote out the logic. One thing to remember is that function itself has no relation with its output. Often, a function problem will use the input value (in this case x) and try to confuse you into thinking that it has something to do with the output (in this case n). In this problem, if the x and 2^n condition is not given, just remember that x and n has no logical relation with each other. Maybe that will help you see clearer. Hope this helps.

N is the same as n. Sorry, should've been more meticulous

OA is A

f(x) is defined as the largest integer N such that X is divisible by 2^n. Which of the following numbers is the biggest?

f(24)
f(42)
f(62)
f(76)
f(84))
In my opinion A is the answer. A gives us that f(24). Thus, we have f(24)=24/2^n. 24 is 2^3 *3, therefore n could be 3. in B and C n would be 1, while in D n would be 2.

here A is correct as its value is 8
for rest of all the value is 4
hence A

the wording is kind of confusing..and that's why it took me ~2 minutes to answer the question...
1. 24 = 2*2*2*3 - we have 3 factors of 2.
2. 2*3*7 - we have 1 factor of 2.
C. 2*31 - we have 1 factor of 2.
D. 2*2*19 - we have 2 factors of 2.
E. 2*2*3*7 - we have 2 factors of 2.

A has the greatest number of factors of 2.
A is the answer.

We have to find the value of x such that the value of n is largest.
24 is divisible by 8. n =3
42 is divisible by 2. n = 1
62 is divisible by 2. n = 1
62 is divisible by 4. n = 2
62 is divisible by 4. n = 2
Option A

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