H(74) = 6*8*10*12*......*74
= (2*3)(2*4)(2*5)(2*6)*.....(2*37)
= 2^35(3*4*5*......*37)
= 2^34 (2*3*4*5....*37)
= 2^34 (37!)

So H(37)+1 = (2^34)*37!+1

(2^34)*37! - Highest prime of this expression would be 37.

(2^34)(37!)+1 will have a prime factor greater than 37

SO, Smallest prime factor would be > 37. E IMO

if i'm not misapplying the distributive rule can be enforced only for the summation
here's the product, hence 2*(3*4*5*6...*37) is not the same as 6*8*10*12 ... but rather is equivalent to 6*4*5*...
agree?

shankar.ashwin wrote:H(74) = 6*8*10*12*......*74
= 2*(3*4*5*6*.....*37)
= 1*2*3*4*5*......*37
= 37!

So H(37)+1 = 37!+1

SO, Smallest prime factor would be > 37. E IMO

Oh my!! Terrible mistake.. Blunder!! Let me correct my post

pemdas wrote:if i'm not misapplying the distributive rule can be enforced only for the summation
here's the product, hence 2*(3*4*5*6...*37) is not the same as 6*8*10*12 ... but rather is equivalent to 6*4*5*...
agree?

@Shankar, another modification of this question could start from 10 as well, and then you would loose factorial advantage (available here for the previous numbers allow 1*2*3=6)

anymore approaches?

I think that would be > than 37 as well, because prime numbers less than 10 are 2,5 and 7 and these would appear in (2,10 and 14) respectively.
Whats the OA btw?

pemdas wrote:@Shankar, another modification of this question could start from 10 as well, and then you would loose factorial advantage (available here for the previous numbers allow 1*2*3=6)

anymore approaches?

OA is E This is my first topic on BTG and it's not a prep. company question; i have made this up to check myself. Thanks for contributing, I have a different solution which factors 2 out of all even numbers (2^35)*(3,4...37). Alternatively for 10 -> (2^33)*(5,6,7,8...37). The idea is the same as that sounded in your post...