MGMAT - Need Help

g(80) = 2*4*6*8............*80

= 2*(2*2)*(2*3)*(2*4)........*(2*40)

= (2^40)*(1*2*3*4.....*40)

= (2^40)(40!)

Means g(80) is divisible by all integers till 40. The prime no less than 40 is 37 that divides g(80). So, 37 will not divide g(80)+1

So, next prime no that would divide g(80)+1 should be greater than 40.

Ans should be E. IMO

The question says, "g(80) + 1 is divisible by the lowest prime number p".
And, 37 is the highest prime number that divides g(80). I think I'm not able to understand the solution.

Any number less than 40 is a factor of g(80) so it is not a factor of g(80)+1. Hence, the smallest factor of g(80)+1 will be greater than 40.

Hope this clears the doubt...

kajcha wrote:Any number less than 40 is a factor of g(80) so it is not a factor of g(80)+1. Hence, the smallest factor of g(80)+1 will be greater than 40.

Hope this clears the doubt...

g(2n) = 2^n (n!)

g( 6 ) = 48; lowest prime number that divides 49 is 7.
g( 8 ) = 384; lowest prime number that divides 385 is 5.

How can you generalize that the smallest (prime) factor of g(80)+1 will be greater than 40. May be I need go back and open the basic mathematics books again.

g(2n) = 2^n (n!)

g( 6 ) = 48; lowest prime number that divides 49 is 7.
g( 8 ) = 384; lowest prime number that divides 385 is 5.


as we see here , g(2n) = g(8). So n = 4
the lowest prime number that divides 385 is 5 which is > 4

g(80) = 2*3*.....79*80

so any prime nos, as a matter of fact any nos from 2..80 will divide g(80) evenly, hence it will give a remainder of 1 when dividing g(80) +1

so any prime factor of g(80) +1 should be greater that 80 hence greater than 40.

mtg/niks

g(6) = 2*3*4*5*6 = 720. How did you calculate it be 48???

similarly g(8) = 40320 :

g( 6 ) = g( 2x3 ) = 2^3 x (3!) = 8 x 6 = 48.

And I got the point. Thanks for taking time to explain.