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(#M40019092) ELITMUS Numerical Ability question Number System Keep an EYE

How many factors of the number 2^5*3^4*5^2 are divisible by 2?

Asked In Elitmus Kishore Kumar (11 years ago)

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pushpa yadav  (10 years ago)

total no. of factors= (5+1)(4+1)(2+1)= 90

no.of odd factors= (4+1)(2+1)= 15

so, no. of even factors= 90-15=75

Is this solution Helpfull? Yes (26) | No   

surabhi sugandha  (10 years ago)

75 is the answer,
(32*81*25)/2 is 2^4*3^4*5^2..so total no.of factors are (4+1)*(4+1)*(2+1)=75

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raghu  (10 years ago)

for the given number the required factors are
with atleast one 2 and remaining any thing from the given expression
2 : 1 combination
2*(_) 3 or 5 : 2 combinations
2*(_*_) 3*3 or 3*5 or 5*5: 3combs
2*(_*_*_) 3*3*3 or 3*3*5 or 3*5*5 :3 combs
2*(_*_*_*_) 3*3*3*3 or 3*3*3*5 or 3*3*5*5 :3combs
2*(_*_*_*_*_) 3*3*3*3*5 or 3*3*3*5*5 : 2 combs
2*(_*_*_*_*_*_) 3*3*3*3*5*5 : 1comb

total 15 combs for 2
repeating same for 2^2 2^3 2^4 2^5 we get
15*5=75 factors

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Ravindra  (11 years ago)

ans) would be 19

12 = (2^2)*3

(2^2)*3 this should be multiple of the no.

remaining is 2^3*3^3*5^2

3*3*2=18

total would be 18+1=19

Is this solution Helpfull? Yes | No (8)  

hemanth chowdary  (10 years ago)

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