Please help on this one: how can I solve it fast?

This problem has been discussed many times. Please search it on Beat the GMAT.


Answer >50

Hence E

akhp77 wrote:This problem has been discussed many times. Please search it on Beat the GMAT.


Answer >50

Hence E

hahahhha Prasad Bhai.. I know u r a strong patron of BTG..I appreciate that!!

So when r u taking test?

Anton8103 wrote:5) For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n inclusive. If p is the smallest prime factor of h(100) + 1, then p is between

a. 2 and 10
b. 10 and 20
c. 20 and 30
d. 30 and 40
e. > 40

h(n) = 2x3x4................xn = n!/1=n!
so h(100) + 1 = 100! + 1
since 100! is divisible by all the numbers between 1 to 100 and further numbers too so whenever 100! + 1 would be divided by these numbers they will leave a remainder as 1.

so smallest prime factor would be surely [spoiler]greater than 100 and hence >40
ans E[/spoiler]

try

https://www.manhattangmat.com/forums/for ... t1152.html

@ shashank
h(100) = 2*4*6.......100
= 2(1*2*3......50)
h(100) + 1 = 2(1*2....50)+1

p = 53?