Latest Elitmus Aptitude Question SOLUTION: How many odd factors are there for the value 9,0000000.

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(#M40028857) ELITMUS Numerical Ability question quant Keep an EYE

How many odd factors are there for the value 9,0000000.

Asked In Elitmus saurabh prakash (11 years ago)

Unsolved Read Solution (14)

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manish yadav  (11 years ago)

90000000= 2^7 *3^2 *5^7 so no. of even factors (2+1)*(7+1)=24

Is this solution Helpfull? Yes (38) | No (8)  

RAKESH  (11 years ago)

9,0000000=3^2*10^7=2^7*3^2*5^7
total factors=(7+1)*(2+1)*(7+1)=8*3*8=192
no. of even factor=(7+1)=8 (only 2^7 gives even factor)
no. of odd factors=(2+1)*(7+1)=24 (3^2*5^7 gives odd factor)
ans 24

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Akshay Kumar Singh  (11 years ago)

actual question is how many odd factors are there for the value 9,000,000
so we 3x3
(2x5)pow6 so we get
power of 5 is 6 and power of 3 is 2
(6+1)*(2+1)=21
answer is 21

Is this solution Helpfull? Yes (20) | No (9)  

ankit chauhan  (11 years ago)

in question it is 9,00,00,00 = 1,3,9,5,25,125,625,3125,15625,15,75,15,75,375,1875,9375,46875,45,225,1125,5625,28125,140625= 21 factors

and shorcut for it..2^6*3^2*5^6 we cant take 2 and its any power because 2when 2 is multiply with any no its became even therefore 5^6*3^2=(6+1)(2+1)=21 factors

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Akshay Kumar Singh  (11 years ago)

@ankit yar option me 21 tha..
a.5
b.21
c.73
d.76

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saurabh prakash  (11 years ago)

ohh ya @akshay kumar singh u r rite actually the value was -9,000,000

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sumit singh  (11 years ago)

Concept
10=5*2
100=(5^2)*(2^2)
.
.
.
10000000=(5^7)*(2^7)

9*10000000=(3^2)*(5^7)*(2*7)
No. of odd factors in above term are 3^2 and 5^7 so (2+1)*(7+1)=24

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Chirag Ranjan Lahiri  (10 years ago)

9,0000000=3^2*10^7=2^7*3^2*5^7
total factors=(7+1)*(2+1)*(7+1)=8*3*8=192
no. of even factor=(7+1)=8 (only 2^7 gives even factor)
no. of odd factors=(2+1)*(7+1)=24 (3^2*5^7 gives odd factor)
ans 24

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ankit chauhan  (11 years ago)

@ akshay yr mera bhi answer 21 he aa rha tha but yr 21 is not in option i think so....20 tha shayad in options me..do u remember options???

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ankit chauhan  (11 years ago)

@akshay thanks for option hmm shayad main he todha confuse ho rha tha..fir toh maine 21 he lagya hoga..confm

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saurabh prakash  (11 years ago)

@ankit-option was-
a> 8
b>21
c>28
the last opt i dont remember..and the correct was b>21..

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rajveer singh  (11 years ago)

21 is the right answer we can't take 2 , simply because it's a even number and we are told to tell odd factors only.....

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harish kumar  (11 years ago)

14 will be the number of odd factors.

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harish kumar  (11 years ago)

ans will be 3

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