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raunekk: Legendary Member; Posts: 1159; Joined: Wed Apr 16, 2008 10:35 pm; Thanked: 56 times

by raunekk » Sat Jun 28, 2008 11:02 pm

1. h(n) is the product of the even numbers from 2 to n, inclusive, and p is the least prime factor of h(100)+1. What is the range of p?

i need a short method for this kinda problems..

thx in advance

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Source: Beat The GMAT — Problem Solving | Sat Jun 28, 2008 11:02 pm

TrizMA: Senior | Next Rank: 100 Posts; Posts: 36; Joined: Sun Jun 22, 2008 5:47 am

by TrizMA » Sun Jun 29, 2008 12:52 pm

what's the source of qn and pls give answer choices.

Qn doesn't seem right. Can't understand range of P. Shouldn't P be one number i.e. least prime factor.

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raunekk: Legendary Member; Posts: 1159; Joined: Wed Apr 16, 2008 10:35 pm; Thanked: 56 times

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by raunekk » Mon Jun 30, 2008 1:47 am

Hi trizma ... its from d gmat prep tests..Have to tried them.??.i took one yesterday.. and i was surprised to c that the level of gmatprep is higher than kaplan and priceton...if u havent taken it yet ..take it first..!!tat wud help u to judge where u stand...
i was quite confident about quant ..coz i used to solve both kaplan workbook and princeton with 90% hit rate.. untill yesterday when i tried one of the 2 tests of gmat prep..

d answer is p>47...i f u get it do explain it to me 2...

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TrizMA: Senior | Next Rank: 100 Posts; Posts: 36; Joined: Sun Jun 22, 2008 5:47 am

by TrizMA » Mon Jun 30, 2008 2:20 am

raunekk, here is the solution

All prime numbers less than 47 are multiples of h(100). this mean h(100) is divisible by the prime number. However when you add 1 to h(100) the prime number won't be able to divide this new number. The idea is whenever a number x divides a certain number y and u add 1 to y then x won't divide y+1. It will always give a remainder of 1.

For expample; take 2: as 2 is part of h(100) it will divide h(100) but when u add 1 to h(100) it becomes even. similarly 3 is part of h(100) since 6 is part of h(100). When u add 1 to h(100) it's no longer divisible by 3 but gives a remainder of 1.

For all number less than or equal to 47 there multiples are part of h(100). For example 94 which is a multiple of 47 is a multiple of h(100). same goes for 43 for which 86 is a multiple of h(100). so all prime number less than 50 have their multiple of 2 included in h(100). Hence when you add 1 to h(100) they won't be able to divide [h(100)+1] and always give a remainder of 1.

Only a prime number greater then 47 is not a multiple of h(100). for example 53 is not a multiple of h(100). Hence there is a chance that 53 can be a least prime factor of [h(100)+1]. So we are sure range of p is above 47

reagrds
Triz

p.s: Pls post answer choices because a question is asked within that framework. though least prime factor of h(100) is certainly one number and not range. Answer choices tells they are looking for.

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TrizMA: Senior | Next Rank: 100 Posts; Posts: 36; Joined: Sun Jun 22, 2008 5:47 am

by TrizMA » Mon Jun 30, 2008 2:23 am

Sorry a typo above:

"For expample; take 2: as 2 is part of h(100) it will divide h(100) but when u add 1 to h(100) it becomes even". it becomes odd

thanks

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raunekk: Legendary Member; Posts: 1159; Joined: Wed Apr 16, 2008 10:35 pm; Thanked: 56 times

by raunekk » Mon Jun 30, 2008 3:09 am

sorry mistake... its p>40..

i didnt understand onethg..how did u come to the conclusion that prime numbers greater than 47 r not factors h(100)+1..

i mean ,if i would had not given u d answer than how would u decide that .. do we have to pick up every prime number from 1 to 100 and check..
thx in advance..

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TrizMA: Senior | Next Rank: 100 Posts; Posts: 36; Joined: Sun Jun 22, 2008 5:47 am

by TrizMA » Mon Jun 30, 2008 3:23 am

lol

even gave explanation for 47 as an answer. pls post all answers choices.

well, actually i said that prime number greater than 47 can be factor of h(100)+ 1. the point was prime numbers less than 47 cannot be factors of h(100)+1.

let's take 43 as an example. 43*2=86 and 86 is multiple of h(100). so h(100) is divisible by 86. this means h(100) is divisible by both 43 and 2. If u add 1 to h(100) then neither 86, 2 or 43 can divide [h(100)+1]. They will all give a remainder of 1

So least common prime factor of h(100)+1 should be greater than 47. Are u sure OA is greater than 40 and answer choice for greater than 47 is not included. After all greater than 47 is also greater than 40

thanks!

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TrizMA: Senior | Next Rank: 100 Posts; Posts: 36; Joined: Sun Jun 22, 2008 5:47 am

by TrizMA » Mon Jun 30, 2008 3:33 am

raunekk, u don't have to pick up every number from 0 to 100.

here is how i came with 47:

i) All prime numbers are odd
ii) all even numbers between 2 & 100 are multiples of 100
iii) Odd number multiplied by 2 gives even. so all odd number less than 50 when multiplied by 2 will be less than 100 and will also be even

The above is detailed explanation. Faster way is to divide 100 by 2 and you get 50. You know that all numbers less than 50 are part of h(100) because when these number are multiplied by 2 they give an even number less than 100.

For example if it was h(110)+1. I would say p>53. simply divide by 2 and find the greatest prime number below that.

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