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but they wont be prime factors of h(100) + 1, since they wont divide into a number one larger than that. Therefore, the largest prime factor must be larger than 50, so E.

For Every Positive Integer n, the function h(n) is defined to be product of all even integers from 2 to n.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

I was able to find answer using extrapolation but would like to know what techniques can be used for these types of questions.

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 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. over 40

Solution is E.
H(100)+1=2*4*...100+1=2^50*(1*2*3...*47*48*49*50)+1
From above, it is seen that first part is divisible by all prime numbers ranging from 2 to 47. But 1 is divisible by none of them. Hence, if it is ever divisible by any prime number, then the divisor must be greater than 47.

As h(100) is a multiple of every positive integer from 2 to 50, h(100) + 1 is not a multiple of any positive integer from 2 to 50. Thus, the smallest prime factor of h(100) +1 must be greater than 50.
Note: h(100) + 1, when divided by any positive integer from 2 to 50, will yield a remainder of 1, and thus is not a multiple of that integer.

For every positive integer n, 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

For every positive integer n, 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

would the answer be A? Here is my logic but I am not sure:

for h(100)+1, we know the answer will be odd
since n = 100 and we are multiplying the even integers 2*4*6......*100 and adding 1 to the product.

this is a sequence so we can say that 2 * (difference of the first and last term) * # of terms = 2 * 98 * 50 = 9800 + 1 = 9801

For every positive integer n, 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

The answer is E and this has been discussed many times in the last few weeks. Do search, and you will get the solution.

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, the p is...

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, the p is...

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