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by harsh.champ » Thu Feb 04, 2010 3:16 am
N! is defined for non negative integers as N!=N*(N-1)*(N-2)* ...3*2*1 . The number of positive integers which divide (25 )! are (a) (b) (c) (d)

(1)(2^13).(3^3).(5^2)
(2)(2^8).(3^2).(5^2)
(3)(2^11).(3^2).(5)
(4)(2^8).(3^3).(5^3)
Last edited by harsh.champ on Thu Feb 04, 2010 12:58 pm, edited 1 time in total.
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by ajith » Thu Feb 04, 2010 7:22 am
harsh.champ wrote:N! is defined for non negative integers as N!=N*(N-1)*(N-2)* ...3*2*1 . The number of positive integers which divide (25 )! are (a) (b) (c) (d)

(1)2^13.3^3.5^2
(2)2^8.3^2.5^2
(3)2^11.3^2.5
(4)2^8.3^3.5^3
25! = 2^22*3^10*5^4*7^3*11^2*13^1*17^1*19^1*23^1

Total num of factors = 23*11*5*4*3*2*2*2*2
=2^6*11*23*5
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by Mom4MBA » Thu Feb 04, 2010 8:23 am
hi ajith, in 25!, 5 has power of 6 not 4, I think you missed 25

answer 23*11*7*4*2*2*2*2 = 23*11*7*(2^6)
but this does not match with the given choices.
Stay focused
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by ajith » Thu Feb 04, 2010 8:28 am
Mom4MBA wrote:hi ajith, in 25!, 5 has power of 6 not 4, I think you missed 25

answer 23*11*7*4*2*2*2*2 = 23*11*7*(2^6)
but this does not match with the given choices.
Thanks I agree
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