Problem Solving

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

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.

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...

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.

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

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..

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!

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.