It is currently 23 Oct 2017, 14:12

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Gmatprep function

Author Message
Intern
Joined: 19 Jan 2009
Posts: 18

Kudos [?]: 11 [0], given: 0

Schools: Kellogg, Ross, Darden, Kelley, UNC

### Show Tags

20 May 2009, 11:28
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Can someone explain?

The function should result in 100! + 1, which is odd. But I can't move further.
Attachments

PS3.jpg [ 55.86 KiB | Viewed 1204 times ]

Kudos [?]: 11 [0], given: 0

Intern
Joined: 19 May 2009
Posts: 5

Kudos [?]: [0], given: 0

### Show Tags

22 May 2009, 09:57
Quote:
Can someone explain?

The function should result in 100! + 1, which is odd. But I can't move further.

You made a small mistake: h(n) is DIFFERENT than n!

Please note that h(n) is defined as the multiplication of the EVEN integers for 2 to n

so h(6)= 2*4*6 = 2^4 * 3

and h(100)=2*4*6*...*100

obviously h(100)+1 can not be divided for any non trivial factor of h(100) and this includes of course any prime number P from 1 to 50 which appears in h(100) at least as 2*P, for example 17 appears as factor in 34 and 68 in h(n). From that follows the right answer you showed.

Just for fun it would have been the same answer if h(n)=n! .

Kudos [?]: [0], given: 0

Intern
Joined: 19 Jan 2009
Posts: 18

Kudos [?]: 11 [0], given: 0

Schools: Kellogg, Ross, Darden, Kelley, UNC

### Show Tags

24 May 2009, 23:26
Thanx Jorge, excellent explanation. I already spot my error after posting, however it didn't make much difference because I was lacking the concept to solve this problem.

Let me just fix up a small error on your side:

so h(6)= 2*4*6 = 2^3 * 3!
because 2*4*6 = 2*2(2)*2(3) = 2^3*3!

Take care

Kudos [?]: 11 [0], given: 0

Manager
Joined: 14 May 2009
Posts: 191

Kudos [?]: 48 [0], given: 1

### Show Tags

25 May 2009, 23:52
$$h(100) + 1 = 2(4)(6)(8)...(98)(100) + 1 = 2(1*2*3*...*49*50) + 1$$

Stare at it really hard... it's in the quotient remainder form. Anytime a number can be written as $$Q + R$$ where $$0<=R<=Q$$ and $$R & Q are coprime$$ (ie they have no common factors in common) then that number is NOT divisible by Q or any factor of Q.

Hence the largest prime factor has to be bigger than 50, and conversely, 40.

Final Answer, $$E$$.
_________________

Kudos [?]: 48 [0], given: 1

Manager
Joined: 04 Sep 2006
Posts: 113

Kudos [?]: 62 [0], given: 0

### Show Tags

01 Jun 2009, 10:08
smallest prime factor of h(100)+1 is asked

Kudos [?]: 62 [0], given: 0

Senior Manager
Joined: 16 Jan 2009
Posts: 354

Kudos [?]: 233 [0], given: 16

Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)

### Show Tags

01 Jun 2009, 11:27
I think I have seen this preblem before.
let me look around for the link.
_________________

Lahoosaher

Kudos [?]: 233 [0], given: 16

Senior Manager
Joined: 16 Jan 2009
Posts: 354

Kudos [?]: 233 [0], given: 16

Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)

### Show Tags

01 Jun 2009, 11:33
smallest-prime-factor-76565.html
_________________

Lahoosaher

Kudos [?]: 233 [0], given: 16

Re: Gmatprep function   [#permalink] 01 Jun 2009, 11:33
Display posts from previous: Sort by