BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

GMAT Prep - Functions

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 21
Joined: Fri Aug 06, 2010 1:30 am

GMAT Prep - Functions

by anhe123 » Sat Oct 16, 2010 3:30 am
For every positive 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

a) between 2 and 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 40

Answer: e


I find functions quite difficult, so if you know of any good excercises or have any tips, I would be glad to hear it.

Thanks!
André
Top
Source: — Problem Solving |
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 am
Location: Bangalore, India
Thanked: 146 times
Followed by:24 members

by shovan85 » Sat Oct 16, 2010 3:39 am
h(100) = 2*4*6*...*100
take all of the 2 common then
h(100) = (2*2*...50 times) * (1*2*3*...*50) = 2^50*(1*2*3*...*50).

Hence, all integers up to 50 are factors of h(100).

Now if h(100) + 1 is divided by any integers from 1 to 50 will have a remainder 1 as h(100) is divisible by all integers below 50.

So the least prime factor will be definitely > 50

IMO E
Top
Junior | Next Rank: 30 Posts
Posts: 21
Joined: Fri Aug 06, 2010 1:30 am

by anhe123 » Sat Oct 16, 2010 3:52 am
shovan85 wrote:h(100) = 2*4*6*...*100
take all of the 2 common then
h(100) = (2*2*...50 times) * (1*2*3*...*50) = 2^50*(1*2*3*...*50).

Hence, all integers up to 50 are factors of h(100).

Now if h(100) + 1 is divided by any integers from 1 to 50 will have a remainder 1 as h(100) is divisible by all integers below 50.

So the least prime factor will be definitely > 50

IMO E
I don't really understand why 2 or 3 cannot be the smallest prime factor. Ii it because you add 1 at the end? What affect does the 1 have? Is this equation the same as the smallest prime of 100! + 1 ?

Thanks for your answer :)
Top
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 am
Location: Bangalore, India
Thanked: 146 times
Followed by:24 members

by shovan85 » Sat Oct 16, 2010 4:07 am
andre.heggli wrote:
shovan85 wrote:h(100) = 2*4*6*...*100
take all of the 2 common then
h(100) = (2*2*...50 times) * (1*2*3*...*50) = 2^50*(1*2*3*...*50).

Hence, all integers up to 50 are factors of h(100).

Now if h(100) + 1 is divided by any integers from 1 to 50 will have a remainder 1 as h(100) is divisible by all integers below 50.

So the least prime factor will be definitely > 50

IMO E
I don't really understand why 2 or 3 cannot be the smallest prime factor. Ii it because you add 1 at the end? What affect does the 1 have? Is this equation the same as the smallest prime of 100! + 1 ?

Thanks for your answer :)
What is the question asking.... smallest prime number as the factor of h(100)+1.
What I proved (from my prev post).... from 1 to 50 (including all primes) are not a factor of h(100) +1 (as the remainder will be 1).

Now tell me if you get remainder as 1 when u divide this h(100)+1 by 2 or 3 or 7 or 11 or .... 41 are they factors of h(100) +1 .

Obviously no as whatever prime below 50 u divide with h(100)+1 you will get a remainder and so those are not factors of h(100)+1.

The addition of 1 makes the primes below 50 not divisible.

yes it is kind of same to the logic 100! + 1. Here if you see we get 50! which says all the primes below 50 is a factor of h(100) n0t (h100) +1
Top
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

by goyalsau » Sat Oct 16, 2010 4:37 am
shovan85 wrote:h(100) = 2*4*6*...*100
take all of the 2 common then
h(100) = (2*2*...50 times) * (1*2*3*...*50) = 2^50*(1*2*3*...*50).

Hence, all integers up to 50 are factors of h(100).

Now if h(100) + 1 is divided by any integers from 1 to 50 will have a remainder 1 as h(100) is divisible by all integers below 50.

So the least prime factor will be definitely > 50

IMO E
Luckily i got it right,
Even i have taken 2 common from all the terms but that was just one 2 not 2^50.

Thanks for making the point. Clear
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Sun Oct 17, 2010 7:07 am
andre.heggli wrote:For every positive 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

a) between 2 and 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 40

Answer: e


I find functions quite difficult, so if you know of any good excercises or have any tips, I would be glad to hear it.

Thanks!
André
Here is the rule that is being tested with this problem:

If x is a positive integer, the only factor common both to x and to x+1 is 1. They share no other factors.

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. They share no other factors.

Thus, in the problem above, we know that 1 is the only factor common both to h(100) and to h(100) + 1. They share no other factors.

h(100) = 2 * 4 * 6 *....* 94 * 96 * 98 * 100

Factoring out 2, we get:

h(100) = 2^50 (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). This means that NONE of the prime numbers between 1 and 50 is a factor of h(100) + 1, because h(100) and h(100) + 1 share no factors other than 1.

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

The correct answer is E.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Post new topic Post Reply
• Page 1 of 1