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Sadowski
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PostSun Jul 22, 2007 4:22 pm
I've seen a post about this question, but no one ever actually answered the question.

For every positive even integer n, the function h(n) equals the product of all even integers from 2 to n, inclusive. If P is the smallest prime factor of h(100)+1, then P is...

1) Between 2 and 10
2) Between 10 and 20
3) Between 20 and 30
4) Between 30 and 40
5) >40

3rd question on my 1st GMAT Prep Shocked I would know how to do this if it were the summation of all even integers from 2 to n, but I'm not sure how to deal with the product of all even integers. Anyone?
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gabriel
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PostMon Jul 23, 2007 10:17 am
Sadowski wrote:
I've seen a post about this question, but no one ever actually answered the question.

For every positive even integer n, the function h(n) equals the product of all even integers from 2 to n, inclusive. If P is the smallest prime factor of h(100)+1, then P is...

1) Between 2 and 10
2) Between 10 and 20
3) Between 20 and 30
4) Between 30 and 40
5) >40

3rd question on my 1st GMAT Prep Shocked I would know how to do this if it were the summation of all even integers from 2 to n, but I'm not sure how to deal with the product of all even integers. Anyone?
hmmm .. are u sure bcoz i remeber to have answered a similar question on the forum ...

nywayz this is one of those GMAT questions where math alone wont help u solve the problem ..

first thing first .. i dont think there would be any GMAt question that actually needs you to multiply 50 numbers together to find the answer ..

.. now to the question .. the question says that h(n) = 2*4*6*8* ... *n ...

.. so for example .. h(4) = 2*4 = 2^2(1*2)

h(12) = 2*4*6*8*10*12 = 2^6( 1*2*3*4*5*6)

So h(100)= 2*4*6*8*....*100 = 2^50(1*2*3*4*5*6* ... *50) ...

Now, what are the prime factors for h(100) .. that is what are the prime numbers that divide the expression h(100)=2^50(1*2*3*4*....*50) .. the prime factors would be 2,3,5,7 and so on .. that is all the prime numbers between 1 and 50 would be a factor of the expression h(100) = 2^50(1*2*3*4 ....*50) ..

.. so now consider the expression h(100)+1 = 2^50(1*2*3*4*....*50) + 1 .. since 1 is not exactly divisible by any of the prime numbers, none of the prime factors between 1 and 50 will be able to exactly divide h(100) + 1 .. so the answer is > 40 .. that is E ..

... i hope the explanation made some sense ..
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PostMon Aug 13, 2007 9:28 pm
can someone clarify this solution? I'm stumped at this question as well and this solution doesn't make sense to me.
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PostTue Aug 14, 2007 12:43 am
Well i did my best .. if some one else has a better way of explaining the solution plz do post it .. i have seen this question being asked many times before too ..

... btw i am sure about the answer .. but am not able to explain as to how i got it Sad ..
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PostTue Aug 14, 2007 3:56 am
You may check this earlier post.

http://www.beatthegmat.com/viewtopic.php?t=222&highlight=
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PostWed Oct 17, 2007 8:37 pm
The previous explanation is correct. I will try to elaborate in more detail, and hopefully that will help.

Let's break it down piece by piece.....

h(100) = multiply every even number from 2 to 100

h(100) =2x4x6x8....etc

Can we simplify this? Yes.

h(100) = 2 * (2*2) * (2*3) * (2*4)....all the way up to (2*50)

I think the first "trick" of this question is that they are trying to get you to misapply the distributive property of addition.

If the question asked 2+4+6+8, you could solve by figuring out (2*1)+(2*2)+.... and you could pull out all your 2's, so you have (2*(1+2+3+4)....) - and then find the sum of all the numbers from 1 to 50).

However, this is not that kind of question. It is asking you to MULTIPLY every even number from 2 to 100.

So we can't distribute. It is literally (2*1)*(2*2)*(2*3)......so we are multiplying the number 2 times every single number between 1 and 50 in a composite way. Why is it not between 1 and 100, you ask........because its only the even numbers between 1 and 100, so there are only 50 numbers.

This is the same as saying multiply 2, 50 times, and then multiply all the numbers from 1 to 50. Translated into math, this is 2^50 * 50!.

So, h(100) = 2^50 * 50!

Ok - now we have a probem. We have simplified our equation, but we still are working with an incalculable number for GMAT purposes. So we can't use our standard divisibility rules in any predicable way.

This is where the solution using prime numbers comes into play.

Let's play around with a smaller version of h(100) for practice. Let's try h(6).

h(6) = 2x4x6 = 48

h(6) + 1 = 49

What is the SMALLEST PRIME NUMBER THAT DIVIDES h(6)?

Well, 49 = 7x7 - so 7 is the smallest prime number that divides h(6)

Let's try h(4).

h(4) = 2x4 = 8

8+1 = 9

What's the smallest prime number that divides 9?

Well, it's 3.

Hopefully you are noticing a pattern. The smallest prime number that divides h(n)+1 is always larger than the largest prime number in h(n).

So, before we even get to the formal explanation, we see that rule. If we extend that rule to h(100) +1 - then the largest prime number that divides h(100)+1 is larger than the largest prime in the set H(100). We can easily see from above that 47 is the largest prime number in h(100). So our answer has to be greater than 47.....so we know that it must be E.

Now, a more formal "proof" for this question is along the lines of what someone explained before.

We know h(100) = 2^50 * 50!, as explained earlier.

We know from that if a number is divisible by a prime number, and then you add 1 to the number in question, it CANNOT be divisible by that same prime. For example. if 10 is divisible by 5, can 11 be divisible by 5? Absolutely not.

Take this logic, and apply it to the problem. We are looking for the largest PRIME divisor. So by the above logic, if h(100) is divisible by a certain prime divisor, h(100) + 1 CANNOT be divisible by that same prime divisor.

We know that h(100) = 50! * 2^50, and 50! contains every prime from 1 to 50. Therefore, h(100) must be divisible by every prime from 1 to 50. (If you don't get the part about being divisible by primes, then you need to study this for the GMAT. While this type of question in particular will not come up often, you need to know prime factorization and its applications to do well on GMAT Math).

Since h(100) is divisible by every prime from to 1 to 50, H(100) + 1 CANNOT be divisible by ANY of those primes. Therefore, our prime in question must be greater than 50.
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PostSun Apr 13, 2008 4:41 pm
Great solution.
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PostThu Nov 27, 2008 9:52 pm
In fact, no consecutive numbers share any of the same factors, prime or not, with the exception of one.
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PostSat Nov 14, 2009 9:53 am
Answer is E
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PostSat Nov 14, 2009 11:16 am
This is probably the most posted question on the forums, so not sure how you didn't find other solutions - just do a search on h(100) or h(n) and you'll get pages of solutions.
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