The answer E. this question has been asked before but I don't where the thread is but anyways;

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

= 2(1*2*3*4................*50)

so for a divisor of h(100) + 1, to happen, it would have to be greater than 50 or in this case 40.

prindaroy wrote:The answer E. this question has been asked before but I don't where the thread is but anyways;

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

= 2(1*2*3*4................*50)

so for a divisor of h(100) + 1, to happen, it would have to be greater than 50 or in this case 40.

could u please explain the last line please.

and in the previous step its 2^50...u cant take just 2 out as its not summation but multiplication inside the bracket.

Okay,

so ever number from 1 to 50 will divide h(100) right? But not one of those numbers will divide h(100)+1, because they already divide h(100). So the smallest number that will do so, has to be greater than 50. In the choices given, greater than 40 is the only choice that reflects our solution. So E is the answer

this question has been discussed a zillion times..
Please go here

https://www.beatthegmat.com/gmat-prep-q-t14339.html

prindaroy wrote:Okay,

so ever number from 1 to 50 will divide h(100) right? But not one of those numbers will divide h(100)+1, because they already divide h(100). So the smallest number that will do so, has to be greater than 50. In the choices given, greater than 40 is the only choice that reflects our solution. So E is the answer

great understood