Thinking out loud back to ya:

"...So I would go for E on the basis that the lowest prime that it could be a multiple of would have to be more than 80."

I think the lowest prime that it could be a mutliple of would have to be more than 37, no? (74 is the the biggest even number between 2 and 78, which is itself a product of a prime factor (2x37)). so if all prime numbers up to 37 are used up, the next biggest prime number would be 41... which would still leave us with choice E. no?

I hope this is not too confusing.

Here is the explanation from the previous discussion on this board. I then tried to convey the concept through another example with small number.

h(100) = 2* 4* 6* ..... *98*100

=> 2*(2*2)*(2*3)....*(2*49)*(2*50)

=> (2^50)(1*2*3....*49*50)

Means - all integers from 1 to 50 are factors of h(100). So, none of them will be factor of h(100)+1

So, smallest prime factor of h(100)+1 will be greater than 50

To understand this problem, try h(10) instead of h(100)

H(10) = 2 * (2*2) * (2*3) * (2*4) * (2*5) [this value is 3840]
 2^5 (1 * 2 * 3 * 4 *5) [this is also 3840]
So the smalles prime factor of h(10) will be greater than 5.

The answer is 3840 ( 2, 3 and 5 are factors of 3840)