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by wizardofwashington » Tue Oct 07, 2008 8:33 am
"For all positive integers n, F(n) is defined as the product of all even integers 2 through n, inclusive. If P is the smallest prime factor of F(120)+1, then P is"
Between 2 and 15
Between 15 and 30
Between 30 and 40
Between 40 and 60
Greater than 60

I'd like to someone to tell me what is the easiest way to solve this, in under 2 mins?
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by PussInBoots » Tue Oct 07, 2008 9:00 am
All I could figure out in 5 min:
1. Number F(120) has a lot of zeros at the end. At least 20, so F(120) = ....00000000000000000000
2. All of prime numbers below 60 are its factors
3. Say X is a prime number smaller than 60, then F(120) is divisible by X, and next such number would be only F(120) + X.
4. F(120) + 1 = ...000000000000001, smallest prime factor has to be bigger than 60
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by wizardofwashington » Tue Oct 07, 2008 9:06 am
That is the approach I took and ended up with the same answer. Unfortunately I have no way of confirming if this is the right answer. There in lies the problem -- not knowing what the OA is!
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by pre-gmat » Tue Oct 07, 2008 9:06 am
Is the answer E <greater than 60>?
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by wizardofwashington » Tue Oct 07, 2008 9:09 am
pre-gmat wrote:Is the answer E <greater than 60>?
Unfortunately, I do not know what the OA is..Sorry!
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by pre-gmat » Tue Oct 07, 2008 9:15 am
This is what I think is the correct answer.

People feel free to correct me.

h(120) is basically 2^60* 60!

the greatest prime number here is 59.

h(120) +1 will have the prime number greater than 60.


I remember solving a similar problem from GPREP test 2.
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by Ian Stewart » Tue Oct 07, 2008 10:20 am
Yes, exactly; f(120) = 2^60 * 60!, so is divisible by every prime less than 60 (since 60! is divisible by every prime less than 60). When you add 1, the remainder will be 1 when you divide f(120) + 1 by any prime less than 60. So f(120) + 1 is not divisible by any prime less than 60. If you want a more detailed explanation, search the forum for 'h(n)', which is the notation used in the original version of this problem.

Incidentally, where on earth is the question from? It's an exact copy of a GMAT prep question with one number changed. If it's a test prep company question, that seems a bit dicey to me!
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by pre-gmat » Tue Oct 07, 2008 10:25 am
Yes, I was surprised too.

The original GPrep question is h(100)+1.

But never saw repeat of this question anywhere.

WizardofWashington, where did you get this from?
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by vivek.kapoor83 » Tue Oct 07, 2008 10:40 am
IMO D
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by vivek.kapoor83 » Tue Oct 07, 2008 10:44 am
Ian Stewart wrote:Yes, exactly; f(120) = 2^60 * 60!,
Can you please explain how u got this eqution.Pls elaborate it.
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by pre-gmat » Tue Oct 07, 2008 10:51 am
F(120) is product of all even integers from 2 to 120.

so 2*4*6*.......120

2^60(1*2*3*......60)

2^60 * 60!
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by gabriel » Tue Oct 07, 2008 2:46 pm
Ian Stewart wrote:Yes, exactly; f(120) = 2^60 * 60!, so is divisible by every prime less than 60 (since 60! is divisible by every prime less than 60). When you add 1, the remainder will be 1 when you divide f(120) + 1 by any prime less than 60. So f(120) + 1 is not divisible by any prime less than 60. If you want a more detailed explanation, search the forum for 'h(n)', which is the notation used in the original version of this problem.

Incidentally, where on earth is the question from? It's an exact copy of a GMAT prep question with one number changed. If it's a test prep company question, that seems a bit dicey to me!
Ian, how many times have you answered this question now :D .

This question (and some of its variants) has been answered several times now. People please use the search function, you would find several threads on similar questions, it would also save you the time and effort of typing the question.
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by wizardofwashington » Tue Oct 07, 2008 8:13 pm
pre-gmat wrote:Yes, I was surprised too.

The original GPrep question is h(100)+1.

But never saw repeat of this question anywhere.

WizardofWashington, where did you get this from?
I ended up finding this in my RSS feed, but then went searching for the real source, which is this Test Prep company's (MLIC, aka ETS) sample test. This question along with a few other tough ones are offered as "samples" for those visiting their website:

https://www.mlicets.org/gmat/gmat-practice-test.html
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by Ian Stewart » Wed Oct 08, 2008 10:13 am
wizardofwashington wrote:
pre-gmat wrote:Yes, I was surprised too.

The original GPrep question is h(100)+1.

But never saw repeat of this question anywhere.

WizardofWashington, where did you get this from?
I ended up finding this in my RSS feed, but then went searching for the real source, which is this Test Prep company's (MLIC, aka ETS) sample test. This question along with a few other tough ones are offered as "samples" for those visiting their website:

https://www.mlicets.org/gmat/gmat-practice-test.html
Thanks for the info. Yeah, for a few of those questions, they've just changed a detail or two in questions that are freely available in GMATPrep. I just completed that test, and in case anyone else bothers with it, I'd point out that there are at least four fundamental mathematical errors in their twenty questions (question 5 needs Pi in the correct answer choice, 6 can't possibly be answered with the information given, 10 doesn't have the right answer among the choices and 18 needs to specify that m and n are integers), so proceed with caution.
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by forgetneoiamtheone » Wed Oct 08, 2008 11:16 am
Ian Stewart wrote:
wizardofwashington wrote:
pre-gmat wrote:Yes, I was surprised too.

The original GPrep question is h(100)+1.

But never saw repeat of this question anywhere.

WizardofWashington, where did you get this from?
I ended up finding this in my RSS feed, but then went searching for the real source, which is this Test Prep company's (MLIC, aka ETS) sample test. This question along with a few other tough ones are offered as "samples" for those visiting their website:

https://www.mlicets.org/gmat/gmat-practice-test.html
Thanks for the info. Yeah, for a few of those questions, they've just changed a detail or two in questions that are freely available in GMATPrep. I just completed that test, and in case anyone else bothers with it, I'd point out that there are at least four fundamental mathematical errors in their twenty questions (question 5 needs Pi in the correct answer choice, 6 can't possibly be answered with the information given, 10 doesn't have the right answer among the choices and 18 needs to specify that m and n are integers), so proceed with caution.
Q5: yes, Pi is needed for this question.

Q6: I choose (6,6), the choice that changes the mean least. SD is a function of "spread of samples from mean". Numerator captures the absolute spread and the denominator normalises it. Denominator would be same for every choice, so what matters is what choice gives you the least "numerator", and it should be the choice that changes it by least amount(in this case 6,6 wouldn't change mean at all). In all other choices, mean will change and atleast one of the added numbers will be "far" from the resulting mean. Agreed, this is not very mathematically rigorous approach, and just an intuitive one.

Q10: I got 9/55 with a crucial assumption.
From the info, number of blue balls compute to 3.
So third blue ball: (6/11)*(3/10), of course the assumption here is that third ball is being drawn after probablistically drawing first two red balls.
The answer would be simply 3/10, if the first event of drawing two red balls is given to have occured.

Q 12? What is the cleanest and fastest way to the answer for this one?
And what is the part played by this piece of info: "equally space points"

Thanks
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