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help needed on a problem (properties of integers)

Expert replies
by utjianwang » Wed Jul 31, 2013 6:07 pm
Hi there,
I have a difficulty to solve the following problem:

From 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 n(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 on 40.

You are be appreciated by providing help on solving this problem.
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Source: — Problem Solving |

by Brent@GMATPrepNow » Wed Jul 31, 2013 6:15 pm
For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, the p is

A: Between 2 & 10
B: Between 10 & 20
C: Between 20 & 30
D: Between 30 & 40
E: Greater than 40
Important Concept: If k is a positive integer that’s greater than 1, and if k is a factor (divisor) of N, then k is not a divisor of N+1
For example, since 7 is a factor of 350, we know that 7 is not a factor of (350+1)
Similarly, since 8 is a factor of 312, we know that 8 is not a factor of 313

Now let’s examine h(100)
h(100) = (2)(4)(6)(8)….(96)(98)(100)
= (2x1)(2x2)(2x3)(2x4)....(2x48)(2x49)(2x50)
Factor out all of the 2's to get: h(100) = [2^50][(1)(2)(3)(4)….(48)(49)(50)]

Since 2 is in the product of h(100), we know that 2 is a factor of h(100), which means that 2 is not a factor of h(100)+1 (based on the above rule)

Similarly, since 3 is in the product of h(100), we know that 3 is a factor of h(100), which means that 3 is not a factor of h(100)+1 (based on the above rule)

Similarly, since 5 is in the product of h(100), we know that 5 is a factor of h(100), which means that 5 is not a factor of h(100)+1 (based on the above rule)

.
.
.
.
Similarly, since 47 is in the product of h(100), we know that 47 is a factor of h(100), which means that 47 is not a factor of h(100)+1 (based on the above rule)

So, we can see that none of the primes from 2 to 47 can be factors of h(100)+1, which means the smallest prime factor of h(100)+1 must be greater than 47.

Answer = E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by utjianwang » Thu Aug 01, 2013 7:30 am
Hi Brent, great answer. Thanks a lot. utjianwang
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by mgm » Fri Aug 02, 2013 3:34 pm
Just curious where is this question from ?
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by Brent@GMATPrepNow » Fri Aug 02, 2013 4:06 pm
mgm wrote:Just curious where is this question from ?
This one is often the first questions in the Official GMATPrep practice tests (https://www.mba.com/the-gmat/download-fr ... tware.aspx)

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by ela07mjt » Sat Aug 03, 2013 3:39 am
Hi Brent

Thanks for the explanation.

You mention 47 cannot be the factor. How about 48, 49 and 50?
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by Brent@GMATPrepNow » Sat Aug 03, 2013 5:57 am
ela07mjt wrote:Hi Brent

Thanks for the explanation.

You mention 47 cannot be the factor. How about 48, 49 and 50?
The question asks us to find the smallest prime factor of h(100) + 1. So, 47 is the last prime factor to consider in the product (1)(2)(3).....(49)(50).

Having said that, we could use the same process to show that 48, 49 and 50 are also not factors.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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