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# How many of the factors of 72 are divisible by 2? 4 5 6

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CEO
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How many of the factors of 72 are divisible by 2? 4 5 6 [#permalink]

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21 Nov 2007, 07:07
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How many of the factors of 72 are divisible by 2?

4
5
6
8
9

Is there a shortcut to this question? It is rather tedious to list out all factors of 72
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 548

Kudos [?]: 3560 [0], given: 360

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21 Nov 2007, 07:32
E.

72=2^3*3^2

N0=(3+1)(2+1)=12 - numbers of all factors

N1=(2+1)=3 - numbers of all factors that are not divisible by 2

Therefore, 12-3=9

9.
CEO
Joined: 21 Jan 2007
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Location: New York City
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Kudos [?]: 855 [0], given: 4

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23 Nov 2007, 12:28
bmwhype2 wrote:
walker wrote:
E.

72=2^3*3^2

N0=(3+1)(2+1)=12 - numbers of all factors

N1=(2+1)=3 - numbers of all factors that are not divisible by 2

Therefore, 12-3=9

9.

Can you explain the logic of that?

lets say we have some random number N and its factors are
2^4 * 5^2 *7^2

The number of factors that are not divisible by 2 are 3*3=9?
Senior Manager
Joined: 09 Oct 2007
Posts: 466
Followers: 1

Kudos [?]: 42 [0], given: 1

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24 Nov 2007, 09:34
bmwhype2 wrote:
bmwhype2 wrote:
walker wrote:
E.

72=2^3*3^2

N0=(3+1)(2+1)=12 - numbers of all factors

N1=(2+1)=3 - numbers of all factors that are not divisible by 2

Therefore, 12-3=9

9.

Can you explain the logic of that?

lets say we have some random number N and its factors are
2^4 * 5^2 *7^2

The number of factors that are not divisible by 2 are 3*3=9?

Any answers on this? Thought I should keep this alive (:
Intern
Joined: 04 Sep 2007
Posts: 12
Followers: 0

Kudos [?]: 0 [0], given: 0

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25 Nov 2007, 02:23
bmwhype2 wrote:
bmwhype2 wrote:
walker wrote:
E.

72=2^3*3^2

N0=(3+1)(2+1)=12 - numbers of all factors

N1=(2+1)=3 - numbers of all factors that are not divisible by 2

Therefore, 12-3=9

9.

Can you explain the logic of that?

lets say we have some random number N and its factors are
2^4 * 5^2 *7^2

The number of factors that are not divisible by 2 are 3*3=9?

I think so.
1
5
7
5*5
7*7
5*7
5*5*7
7*7*5
5*5*7*7

If this is proven to be a general rule, is very very useful.
CEO
Joined: 21 Jan 2007
Posts: 2756
Location: New York City
Followers: 11

Kudos [?]: 855 [0], given: 4

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28 Nov 2007, 04:40
bmwhype2 wrote:
bmwhype2 wrote:
walker wrote:
E.

72=2^3*3^2

N0=(3+1)(2+1)=12 - numbers of all factors

N1=(2+1)=3 - numbers of all factors that are not divisible by 2

Therefore, 12-3=9

9.

Can you explain the logic of that?

lets say we have some random number N and its factors are
2^4 * 5^2 *7^2

The number of factors that are not divisible by 2 are 3*3=9?

its 5*3*3 = 45 total factors
45 - (3 + 3) = 26
28 Nov 2007, 04:40
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