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# Number properties/primes

Author Message
Intern
Joined: 28 Mar 2012
Posts: 2

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22 Apr 2012, 15:06
Hi, I got this question incorrect on the new gmat prep software and i can't figure out how to do it:

Q: For every positive even integer n, the function h(n) is defined to be the product of all the 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) greater than 40

The right answer is (5), but i don't understand how to go about solving this?

Any help is really appreciated! thanks!
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7440
Location: Pune, India

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23 Apr 2012, 11:50
1
KUDOS
Expert's post
shreya32 wrote:
Hi, I got this question incorrect on the new gmat prep software and i can't figure out how to do it:

Q: For every positive even integer n, the function h(n) is defined to be the product of all the 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) greater than 40

The right answer is (5), but i don't understand how to go about solving this?

Any help is really appreciated! thanks!

Here is some theory first:
Pick any two consecutive numbers. Can they both be even i.e. have 2 as a factor?
e.g. (5,6) or (101, 102) or (999, 1000) etc .. Only one number in these pairs will have 2 as a factor.
Now think: If you pick any two consecutive integers, can they both have 4 as a factor? or 7 as a factor? or 99 as a factor? No!
(For that matter, once you get one multiple of 99, you will not get another one in the next 98 numbers. The next multiple will appear when you add 99 to this multiple. e.g. you pick 99. Can 100, 101, 102... be multiples of 99? The next one is 198 so numbers from 100 to 197 are not multiples of 99.)

We can say that consecutive numbers will not have any common factor other than 1. (1 is a factor of every number.)
e.g. if N = 2*3*5*7*11,

Think consecutive numbers now:
(N - 4), (N - 3), (N - 2), (N - 1), N, (N + 1), (N + 2), (N + 3), (N + 4)...

(N + 1) and (N - 1) will not have any of 2, 3, 5, 7 and 11 as factors. (Since they are consecutive with N)
(N + 2) and (N - 2) will not have 3, 5, 7 and 11 as factors. (2 will be a factor of both these numbers since 2 is a factor of N)
(N + 3), ( N + 4), (N -3 ), (N - 4) will not have 5, 7 and 11 as factors. (But (N + 3) and (N - 3) will have 3 as a factor. (N + 4) and (N - 4) will have 2 as a factor)
and so on...

Let's get back to the original question:
2*50! = 2*1*2*3*4*5*6*7*8*9*10*11.....*50

This number has all numbers till 50 as its factors. So 2*50! + 1 cannot have all these numbers as its factor.
Therefore the smallest prime number that will be the factor of 2*50! + 1 is greater than 50.

A trick question can be "What is the smallest factor of (2*50! + 1)?"
The smallest factor will be 1.

I have also discussed the theory related to this question in detail here: http://www.veritasprep.com/blog/2011/09 ... h-part-ii/
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Re: Number properties/primes   [#permalink] 23 Apr 2012, 11:50
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