Latest Elitmus Aptitude Question SOLUTION: Number of Odd divisor of 9,000,000 a) 21 b) 73 c )74

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(#M40095172) ELITMUS Logical Reasoning question Number System Keep an EYE

Number of Odd divisor of 9,000,000
a) 21 b) 73 c )74

Asked In Elitmus Ritesh Siriya (9 years ago)

Unsolved Read Solution (7)

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Satya B  (9 years ago)

9,000,000=2^6*3^2*5^6
odd divisors=(2+1)*(6+1)=21

Is this solution Helpfull? Yes (33) | No (1)  

Dilraj vyas  (9 years ago)

9,000,000=2^6*3^2*5^6
so No of odd divisors=
for 2= 2^0
for 5= 5^0,5^1,5^2,5^3,5^4,5^5,5^6
for 3=3^0,3^1,3^2

so No of divisors=1*7*3
=21 Ans...

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Nikhil  (9 years ago)

ans is 21 , same was asked in october 28.

Is this solution Helpfull? Yes (4) | No   

aditya sharma  (9 years ago)

9,000,000=9*1,000,000
=3*3*10*10*10*10*10*10
odd number=3*3*5*5*5*5*5*5
so 2+6+6+6+1=21

Is this solution Helpfull? Yes (3) | No (2)  

arun kumar singh  (9 years ago)

9000000=2^6*3^2*5^6
we find odd diviser then evem power is =1
2^0*(2+1)*(6+1)=21

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Pragati Kumari  (7 years ago)

9000000=3^2*10^6=3^2*2^6*5^6
odd divisors=(2+1)(0+1)(6+1) we will get odd divisors when power of 2 should be zero.
so, 3*1*7=21

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ashish chandra kesharwani  (6 years ago)

answer (a) is the correct answer
the factor of the 9,000,000 is =3^2 * 2^6 * 5^6
so total no of the factor is =(2+1)(6+1)(6+1)=147
so total no of even factor is =(2)(6+1)(6+1)=126
so total no of odd factor is =total factor-even factor
=147-126
=21

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