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Smalles prime factor [#permalink]
08 Jun 2008, 11:21

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, the 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

Re: Smalles prime factor [#permalink]
08 Jun 2008, 11:35

ldpedroso 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, the 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

before posting a question..please search for it on this forum...this question has been solved 1000 times here..

Function h(n) Gmat prep question [#permalink]
07 Jul 2008, 20:04

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 h(100)+1 , then p is

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 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

OA TO FOLLOW

My choice is E. h(100)= 2*4*6....*100 so it contains all the prime number from 2 to 49 so h(100)+1 is not divisible by 2, 3,..., 49 so its smallest prime factor must be greater than 49. Is it right?

There is something I don't understand here. Just because the largest prime factor of h(100) is 47, shouldn't mean that the smallest prime factor of h(100)+1 will be bigger than 47. Let me give you an example:

9 = 3*3, the largest prime number here is 3, so when we look at the next higher number:

10=2*5, so the smallest prime factor here is 2, which is not bigger than 3. So how can we say that the bigger number's smallest prime factor will be bigger than the biggest prime factor of the number just under it??? Can someone please explain this concept to me? thanks

There is something I don't understand here. Just because the largest prime factor of h(100) is 47, shouldn't mean that the smallest prime factor of h(100)+1 will be bigger than 47. Let me give you an example:

9 = 3*3, the largest prime number here is 3, so when we look at the next higher number:

10=2*5, so the smallest prime factor here is 2, which is not bigger than 3. So how can we say that the bigger number's smallest prime factor will be bigger than the biggest prime factor of the number just under it??? Can someone please explain this concept to me? thanks

False analogy h(100) is divisible by all prime numbers from 2, ..., 47 while 9 is not divisible by all prime numbers from 2 to 3. If 9 were, 10 would never divisible by 2. If A is divisible by p and A+1 is divisible by p too, 1 is divisible by p. It's wrong.

GMAT Prep: Function - part 2 [#permalink]
23 Jul 2008, 16:47

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

Re: GMAT Prep: Function - part 2 [#permalink]
23 Jul 2008, 19:24

h (100) would mean 2*3*4*5*6...............................100 = X Now X has the following properties:

1) it's an even number. 2) it ends in more than three zeros thus a multiple of 5,25, 125 etc. 3) its a multiple of all the prime factors till 100.

Now X+1 would not be an even number. would not have zeros at the end would not be a multiple of any of the prime numbers till 100.

Thus there are two possibilities: 1) Either X+1 is a prime number 2) X+1 is a composite number.

Now for 1) there's no option to proove its validity. For 2) the only option one can pick up is e. Coz' from the above we can see that X+1 would not be a multiple of any number from 2 to 100.

GMATPresp: Smallest Prime Factor [#permalink]
13 Aug 2008, 22:19

Any mathematical way to get the answer? I tried the lengthy way of finding pattern...but unsuccessful.

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 and 10 B) between 10 and 20 C) between 20 and 30 D) between 30 and 40 E) greather than 40

Re: GMATPresp: Smallest Prime Factor [#permalink]
14 Aug 2008, 03:15

scthakur wrote:

Any mathematical way to get the answer? I tried the lengthy way of finding pattern...but unsuccessful.

PS take the trouble of typing the question mate..

Anyways..

h(100) = 2*4*6*....*100 = 2^50 * 50!

Now, a general rule. If x>1 is a factor of a number n, then x will not a factor of (n+1). This can be tried out with a few examples. Factors of 15 are 3,5,15. None of these will be factors of 16. Factors of 16 are 2,4,8,16. None of these are factors of 17.. so on and so forth.

So none of the numbers 1-50 will be a factor of h(100)+1. The prime factor has to be greater than 50. So E

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

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.

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