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zaur, as a rule, if n is divisible by a factor, (n+1) will not be divisible by the same factor. You can check this for any number. For example, if 6 is divisible by 2, 3 and 6 then (6+1) will not be divisible by any of 2,3, or 6.

Extending the same logic, h(100) is divisible by all the integers from 1 to 50. Hence, h(100)+1 will not be divisible by any number from 1 to 50 and hence its factor will be greater than 50.

Thanks scthakur! Its much clear now! + 1 kudo for you.

For every positive 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, then p is

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

Thanks - I am sure its easy I just don't know how to attack it.

you can se that :h(100)=2*4*6*...*100=(2*1)(2*2)(2*3)...(2*50) or h(100)=2^50(1*2*3...*50) h(100)+1=2^50(1*2*3*...*50) + 1 if h(100)+1 is divisible by p then p must be greater than 47. Here we go: we can see that h(100) is divisible by 2,3,..47(47 is the greatest prime which is smaller than 50) so if h(100) +1 is is divisible by p which p <=47 then 1 is divisible by p so p=1 ridiculous then if h(100) +1 divisible by p, p must be greater than 47 and answer is E We use the characteristic : a+ b is divisible by c, and a is divisible by c too. So b is divisible by c

So what you are saying is that I should guess and move on? LOL? Can anyone else break that down for the only slightly intelligent people in this forum.

So what you are saying is that I should guess and move on? LOL? Can anyone else break that down for the only slightly intelligent people in this forum.

i am one of those slightly intel.

however this is based on a concept that says

if x is devisible by z then sure x+1 is not devisible by z also, the largest prime that devide x has to be less than the largest prime that devide x+1.

This is probably the most hardest question i ever encountered.

h (2) = 2 h (4) = 2 x 4 = 2! x 2^2 h (6) = 2 x 4 x 6 = 3! x 2^3 h (8) = 2 x 4 x 6 x 8 = 4! x 2^4 h (10) = 2 x 4 x 6 x 8 x 10 = 5! x 2^5 . . . . h (100) = 2 x 4 x 6 x .........x 100 = 50! x 2^50 = (1x2x3x4x5x......x40x....48x49x50) (2^50). The pattren identified is that for every n, h (n) = (n/2)! x 2^(n/2)

Now lets find the p, the smallest prime factor of h (100) + 1. lets try from the begaining:

if p = 2, [h(100)+1] should be divisible by 2. But it is not. [h (100) + 1]/2 = h (100)/2 + 1/2 = (50! x 2^50)/2 + 1/2. it doesnot result in an integer. so 2 cannot be p.

if p = 3, [h (100) + 1]/3 = h (100)/3 + 1/3 = (50! x 2^50)/3 + 1/3. it doesnot result in an integer. so 3 cannot be p. if p = 4, [h (100) + 1]/4 = h (100)/4 + 1/4 = (50! x 2^50)/4 + 1/4. it doesnot result in an integer. so 3 cannot be p. if p = 4, [h (100) + 1]/4 = h (100)/4 + 1/4 = (50! x 2^50)/4 + 1/4. it doesnot result in an integer. so 4 cannot be p. if p = 5, [h (100) + 1]/5 = h (100)/5 + 1/5 = (50! x 2^50)/5 + 1/5. it doesnot result in an integer. so 5 cannot be p. . . . . . if p = 50, [h (100) + 1]/50 = h (100)/50 + 1/50 = (50! x 2^50)/50 + 1/50. it doesnot result in an integer. so 50 cannot be p.

The finding is that: Something [h(n)] divisible by k is not divisible by k if 1 is added to something [h(n)]. Therefore h (n=100) + 1 has the smallest prime i.e. > 50. So it is E.
_________________

For every positive even 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 100 b) between 10 and 20 c) between 20 and 30 d) between 30 and 40 e) greater than 40.

please explain ... this question is from gmatprep1.

For every positive 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,then p is 1. between 2 and 10 2. between 10 and 20 3. between 20 and 30 4. between 30 and 40 5. greater than 40

Schools: HBS(08) - Ding. HBS, Stanford, Kellogg, Tuck, Stern, all dings. Yale - Withdrew App. Emory Executive -- Accepted, Matriculated, Withdrewed (yes, I spelled it wrong on purpose). ROSS -- GO BLUE 2011.

For every positive 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,then p is 1. between 2 and 10 2. between 10 and 20 3. between 20 and 30 4. between 30 and 40 5. greater than 40

For every positive 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,then p is 1. between 2 and 10 2. between 10 and 20 3. between 20 and 30 4. between 30 and 40 5. greater than 40

the smallest prime factor of 2*4*6*8...*100 is 2.

2+1=3

answer is 1.

I gues this is not correct . You did not take the +1 in consideration , which changes the answer. Please see my response below.
_________________

Lahoosaher

gmatclubot

Re: Smallest Prime Factor
[#permalink]
17 Mar 2009, 14:36

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