GMAT Prep Stumper

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GMAT Prep Stumper

by Sadowski » Sun Jul 22, 2007 4:22 pm

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 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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Re: GMAT Prep Stumper

by gabriel » Mon Jul 23, 2007 10:17 am

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 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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by maolivie » Mon 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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by gabriel » Tue 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 ..

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by Prasanna » Tue Aug 14, 2007 3:56 am

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by yalephd2007 » Sun Apr 13, 2008 4:41 pm
Great solution.

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by rackbar » Thu 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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E

by dotnetuncle » Sat Nov 14, 2009 9:53 am

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by Stuart@KaplanGMAT » Sat 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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by mcdesty » Thu Aug 30, 2012 4:45 pm
Let's Do h(10) instead of h(100)

h(10) = 2 * 4 * 6 * 8 * 10
= 2*1 * 2*2 * 2*3 * 2*2*2 * 2*5
Combine all the 2's using exponent rules you get 2^8 * 1*3*5
For the next step you have to see that 6 + 15 is divisible by 3 only because 6 is divisible by 3 and because 15 is divisible by 3 (2(3) + 3(5))
Notice that it isn't divisible by 2 because 15 is not divisible by 2
Now 2(3) + 1 will not be divisible by either 2 or 3 because 1 isn't divisible by 2 or by 3
All the prime factors you are likely to isolate by doing h(100) would not be divisible by h(100) + 1.
Have fun isolating the largest prime number and solving the rest of the problem..(:

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