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# number prop

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Intern
Joined: 11 Sep 2009
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Location: Tampa ,FL
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27 Oct 2009, 20:49
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how many different positive integers are factors of 441?

4
6
7
9
11

441=3*3*7*7

So

Can you explain me why these ( 3*3, 7*7 , 3*7 , 3, 7 )are also factors of 441.
Manager
Joined: 11 Sep 2009
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27 Oct 2009, 21:05
The practical way to break it down is to think of the number as a product of factor pairs. You can do this by breaking down the integer into its prime factors:

441 = 3 * 3 * 7 * 7

SO,

441 = (3*3*7*7) * 1 [441, 1]
441 = (3*3*7) * 7 [56, 7]
441 = (3*7*7) * 3 [147, 3]
441 = (3*3) * (7*7) [9, 49]
441 = (3*7) * (3*7) [21]

So there are 9 unique factors.

A much easier method to use when presented with a problem like this is to use the following rule: For each unique prime factor, take the exponent (i.e number of occurances) and add 1. Then multiply these numbers together to determine the number of unique factors. For example,

441 = 3^2 * 7^2

Number of factors = (2+1)(2+1) = 9
Manager
Joined: 15 Sep 2009
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27 Oct 2009, 23:07
441 = 21^2.

441 = (3*7) ^2.

So total no.of factors is 3*3 = 9
Senior Manager
Joined: 18 Aug 2009
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Schools: UT at Austin, Indiana State University, UC at Berkeley
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28 Oct 2009, 17:54
I would recommend to use formula for this problem.
1. Factor 441 into prime factors. You will get 3*3, and 7*7
2. We have 3^2 * 7^2
3. So, we multiply (2+1)*(2+1), thus you get 9.
You do not have to calculate each factor of it separately.
_________________

Never give up,,,

Re: number prop   [#permalink] 28 Oct 2009, 17:54
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