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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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Manager
Joined: 22 Nov 2016
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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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30 Oct 2017, 17:15
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35% (medium)

Question Stats:

70% (00:48) correct 30% (00:57) wrong based on 33 sessions

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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

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Director
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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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30 Oct 2017, 19:28
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 )
Manager
Joined: 14 Sep 2016
Posts: 151
Re: If N = 2^2 x 3^3 x 5^5, find the total number of odd and even ... [#permalink]

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30 Oct 2017, 19:46
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.
Math Expert
Joined: 02 Aug 2009
Posts: 5784
Re: If N = 2^2 x 3^3 x 5^5, find the total number of odd and even ... [#permalink]

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30 Oct 2017, 20:39
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

C

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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, 20:39
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