Help on a PS question please.

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srinivasapriyan.r: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Sat May 04, 2013 9:49 am

by srinivasapriyan.r » Wed Oct 23, 2013 4:38 am

Help on a PS question please.

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Source: Beat The GMAT — Problem Solving | Wed Oct 23, 2013 4:38 am

theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Wed Oct 23, 2013 4:52 am

h(100) = 2 x 4 x 6 x 8 ..... 100

h(100) = 2(1x2x3..50) = (2^50)*50!

h(100) + 1 = (2^50)*50! + 1

here, 2 to 50 are factors of h(100) so they cannot be factor of h(100)+1

So, the factor must be greater than 50.

Answer [spoiler]{E}[/spoiler]

R A H U L

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srinivasapriyan.r: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Sat May 04, 2013 9:49 am

by srinivasapriyan.r » Wed Oct 23, 2013 5:59 am

Thanks Rahul for the explanation.

Can you please elaborate on the below point?
here, 2 to 50 are factors of h(100) so they cannot be factor of h(100)+1

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Wed Oct 23, 2013 6:45 am

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 Hanneson - Creator of GMATPrepNow.com

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theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Wed Oct 23, 2013 7:51 am

srinivasapriyan.r wrote:Thanks Rahul for the explanation.

Can you please elaborate on the below point?
here, 2 to 50 are factors of h(100) so they cannot be factor of h(100)+1

I will give you examples to explain this.

Consider 15 = 3 x 5 .. Now if we increase 15 by 1 i.e. 16 = 2x2x2x2 --> both "3" & "5" are not factors

Similarly 20 = 2x5x2 .. & 21 = 7x3 --> No common factors

Similarly 50 = 2x5x5 ... & 51 = 3x17 ---> No common factors

So, if a number has some factors . .then number+1 will not have same factors..

Same logic implies here.

I hope it's better now

R A H U L

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by GMATGuruNY » Wed Oct 23, 2013 9:03 am

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

Since the difference between them is 1, h(100) and h(100)+1 are consecutive integers.
Consecutive integers are COPRIMES: they share no factors other than 1.

Let's examine why:

If x is a multiple of 2, the next largest multiple of 2 is x+2.
If x is a multiple of 3, the next largest multiple of 3 is x+3.

Using this logic, if we go from x to x+1, we get only to the next largest multiple of 1.
So 1 is the only factor common both to x and to x+1.
In other words, x and x+1 are COPRIMES.

Thus:
h(100) and h(100)+1 are COPRIMES. They share no factors other than 1.

h(100) = 2 * 4 * 6 *....* 94 * 96 * 98 * 100
Factoring out 2 from every value above, we get:
h(100) = 2â�µâ�°(1 * 2 * 3 *... * 47 * 48 * 49 * 50)

Looking at the set of parentheses on the right, we can see that every prime number between 1 and 50 is a factor of h(100).
Since h(100) and h(100)+1 are coprimes, NONE of the prime numbers between 1 and 50 can be a factor of h(100)+1.

Thus, the smallest prime factor of h(100) + 1 must be greater than 50.

The correct answer is E.

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