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

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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

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\[\Huge h(n)=2^{n/2} \cdot (n/2)!\] the smallest prime number that is not a repeat is 29.
h(100) = 2*4*6...100 if you observe properly, it is written (2^50)*(1*2..50) or (2^50)*(50 !) we are concerned about (2^50)*(50!) +1 note that h(100) and h(100) +1 are both consecutive nos. , thus they dont share ANY common factor apart from 1 . h(100) clearly has all prime factors from 1 to 50 as its factors, thus h(100)+1 cant have those as factors. hence ans would be greater than 50 according to me, or E
ah, h(100) + 1. i missed that +1

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though for that part u r absolutely correct

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