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MGMAT - Need Help

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MGMAT - Need Help

by wizardofwashington » Mon Sep 24, 2007 4:02 pm
If g(n) represented the product of every even integer from 2 to n, then g(80) + 1 is divisible by the lowest prime number p. P is:

A. Between 1 and 10
B. Between 11 and 20
C. Between 21 and 30
D. Between 31 and 40
E. Greater than 40

Can you help with this. Not sure what the OA is.
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by kajcha » Mon Sep 24, 2007 6:43 pm
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
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by niks_01.27 » Mon Sep 24, 2007 7:13 pm
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.
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by kajcha » Mon Sep 24, 2007 7:20 pm
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...
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by niks_01.27 » Mon Sep 24, 2007 7:55 pm
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.
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by mtg123 » Tue Sep 25, 2007 4:30 am
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
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by samirpandeyit62 » Tue Sep 25, 2007 6:11 am
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.
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by kajcha » Tue Sep 25, 2007 6:53 am
mtg/niks

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

similarly g(8) = 40320 :?:
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by niks_01.27 » Tue Sep 25, 2007 2:17 pm
g( 6 ) = g( 2x3 ) = 2^3 x (3!) = 8 x 6 = 48.

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