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# What is the number of positive integers that are factors of

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VP
Joined: 06 Jun 2004
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What is the number of positive integers that are factors of [#permalink]

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06 Dec 2005, 00:23
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What is the number of positive integers that are factors of 360?

A. 24
B. 12
C. 8
D. 6
E. 4

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VP
Joined: 22 Aug 2005
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06 Dec 2005, 00:44
360 = 2^3 * 3^2 * 5

total number of factors(including 1 and 360) = 4 * 3 * 2 = 24
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Director
Joined: 14 Sep 2005
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06 Dec 2005, 00:48
360
= 2 x 180
= 2 x 2 x 90
= 2 x 2 x 2 x 45
= 2 x 2 x 2 x 3 x 15
= 2 x 2 x 2 x 3 x 3 x 5
= 2^3 x 3^2 x 5^1

Thus, the numberof factors of 360 is (3+1) x (2+1) x (1+1) = 4 x 3 x 2 = 24.

I pick (A).

* Formular of getting number of factors of (a^p) x (b^r) x (c^s)
(a, b, and c are prime numbers)
= (p+1) x (r+1) x (s+1)
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Director
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06 Dec 2005, 04:39
2,4,5,6,8,9,10,12,15,18,20,24,30,etc.....

There are more than 13. A is the answer.

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Senior Manager
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06 Dec 2005, 12:44
Great explanations guys, but is there a specific rule for this?

Seeing the workings above, I get a little lost if I don't see some kind of theory...

Thx.

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VP
Joined: 06 Jun 2004
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06 Dec 2005, 12:59
OA is A.

gamjatang nailed it with the formula!

One should memorize that formula as it will come in handy

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Senior Manager
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06 Dec 2005, 13:14
Great explanations guys, but is there a specific rule for this?

Seeing the workings above, I get a little lost if I don't see some kind of theory...

Thx.

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VP
Joined: 06 Jun 2004
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06 Dec 2005, 14:09
desiguy wrote:
Great explanations guys, but is there a specific rule for this?

Seeing the workings above, I get a little lost if I don't see some kind of theory...

Thx.

Its a combination problem...so

* Formular of getting number of factors of (a^p) x (b^r) x (c^s)
(a, b, and c are prime numbers)
= (p+1) x (r+1) x (s+1)

Basically you have to take into account (a^0) * (b^0) * (c^0) because for every a^p, there is a a^0, that is why you have to add the 1.

So for example:

how many positive factors are there for 6?

Working: 2^1 * 3^1
Formula: (1+1)*(1+1) = 4 ==> there are a total of 4 positive factors of 6 (1, 2, 3, 6)

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Director
Joined: 27 Jun 2005
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06 Dec 2005, 19:00
Allright guys I am little confused here ....

Quote:
The sum of all factors of 128 is

A. 0
B. 128
C. 1124
D. 4362
E. 5180

the OA to this question is 0
and the explanation given by members is Factor/Divisior can be both positive and negitive.

So if the Given question is asked in little differnt way like

Quote:

What is the number of integers that are factors of 360?

would the answer be 2*24 ( including -ve factors also).....

if thats true then the formula should say the no of +ve factor/divisior = (p+1)*(q+1).......

pls explain [/quote]

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06 Dec 2005, 19:36
360=2^3*3^2*5

S(360)=(3+1)*(2+1)*(1+1)=24

A.
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06 Dec 2005, 19:36
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