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If N = 2^2 x 3^3 x 5^5, find the total number of odd and even ...

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If N = 2^2 x 3^3 x 5^5, find the total number of odd and even ... [#permalink]

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New post 30 Oct 2017, 16:15
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Question Stats:

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If \(N = 2^2 * 3^3 * 5^5\), find the total number of odd and even factors of \(N\).

A) 24, 96
B) 15, 30
C) 24, 48
D) 15, 96
E) 15, 48
[Reveal] Spoiler: OA

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Re: If N = 2^2 x 3^3 x 5^5, find the total number of odd and even ... [#permalink]

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New post 30 Oct 2017, 18:28
sasyaharry wrote:
If \(N = 2^2 * 3^3 * 5^5\), find the total number of odd and even factors of \(N\).

A) 24, 96
B) 15, 30
C) 24, 48
D) 15, 96
E) 15, 48


even factor = 2 * 4 * 6 = 48 (we need to exclude 2^0 here)
Odd factor = 6 * 4 = 24 (Exclude all 2's factor here )

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Re: If N = 2^2 x 3^3 x 5^5, find the total number of odd and even ... [#permalink]

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New post 30 Oct 2017, 18:46
total no. of factors for the above set will be = 3 * 4 * 6 = 72
and (total odd factors ) + ( total even factors ) = 72

only option C has the sum of 72, hence answer C.

sobby can please explain your approach that you used to determine odd and even factors independently.

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Re: If N = 2^2 x 3^3 x 5^5, find the total number of odd and even ... [#permalink]

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New post 30 Oct 2017, 19:39
kunalsinghNS wrote:
total no. of factors for the above set will be = 3 * 4 * 6 = 72
and (total odd factors ) + ( total even factors ) = 72

only option C has the sum of 72, hence answer C.

sobby can please explain your approach that you used to determine odd and even factors independently.



hi..

\(N = 2^2 * 3^3 * 5^5\)

ODD factors when there ar eno 2s in it..
so \(3^3*5^5\).... factors = (3+1)(5+1)=4*6=24

even factors = total factor - odd factors...
total = 3*4*6=72
so even factor = 72-24=48

so 24,48

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Re: If N = 2^2 x 3^3 x 5^5, find the total number of odd and even ...   [#permalink] 30 Oct 2017, 19:39
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