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This question is testing the concept of coprimes. 2 positive integers are coprime when their greatest common factor (their only common factor) is 1. Now note that two different prime numbers are always coprime. For example, 3 and 7 are coprime. So are 13 and 19. But the two integers need not be prime numbers in order to be coprime. For example, 4 and 9 are coprime (1 is their only common factor). Also, Important! Two consecutive integers are always coprime. The question is testing you on this concept. Let's solve it now: h(n) = 2*4*6*....................*n. ----n is even. h(100) = 2*4*6*........ 94*96*98*100. h(100) = (2^50)*(1*2*3*.......47*48*49*50). Note:I have pooled together all the 2s from all the even integers from 2 to 100; that's how I got 2^50. Now, the largest prime number involved in the above factorization is 47. All the prime from 2 to 47 are also involved in the above factorization. Actually, 47 is the greatest prime factor of h(100). Since h(100) and h(100) + 1 are consecutive integers, they are necessarily coprime (see above). h(100) and h(100) + 1 have no common factor except 1, so they have no common prime factor either. The smallest prime factor of h(100) +1 must then be greater than 47. Clearly, this prime factor is greater than 40.

That's all folks! Asan Azu, The GMAT Doctor.

Wow, lot of info. I understood everything except for this part: h(100) = (2^50)*(1*2*3*.......47*48*49*50) Could you explain this part a little more clearly? Thanks. _________________

Factorials were someone's attempt to make math look exciting!!!

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

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..

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.

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

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.

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

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

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