enigma123 wrote:
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 10
B. Between 10 and 20
C. Between 20 and 30
D. Between 30 and 40
E. Greater than 40
This is how I am trying to solve this. Please help me if you think I am not right. OA is not provided in the book.
h(100) = 2 * 4 * 6 ****************100
Tn = a1 + (n-1) d-----------------------(1) where Tn is the last term, a1 is the first term and d is the common difference of the evenly spaced set.
100 = 2 + (n-1) 2
n = 50
Product of terms = Average * number of terms
Average = (a1+an)/2
Therefore average = 102/2 = 51
Product of the series = 51*50 = 2550.
H(100) + 1 = 2550+1 = 2551 which is prime. And prime numbers have exactly 2 factors 1 and the number itself. Therefore for me D is the answer i.e. < 10
Dear
avigutmanCould you validate my approach?
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?Thus, h(n) = 2*4*6 .... = (2*n+2)!
h(100)+1 = (2*100+2)!+1 = 202!+1
Because 2 consecutive numbers do not share common factors but 1, the answer will be grate then XXX+1 or grater then 40 ?
Thanks beforehand.