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# total no of factors from prime factors?

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total no of factors from prime factors? [#permalink]

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27 Mar 2011, 11:45
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48 = (2)^4 *(3)^1

How would you calculate procedure total number of factors? i know the total number of prime factors is 4+1=5.

Thanks
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Re: total no of factors from prime factors? [#permalink]

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27 Mar 2011, 11:54
yasirilyas wrote:
48 = (2)^4 *(3)^1

How would you calculate procedure total number of factors? i know the total number of prime factors is 4+1=5.

Thanks

$$48=2^4*3^1$$

(4+1)*(1+1) = 5*2=10

http://gmatclub.com/forum/math-number-theory-88376.html
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Re: total no of factors from prime factors? [#permalink]

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14 May 2016, 19:55
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Re: total no of factors from prime factors? [#permalink]

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15 May 2016, 11:10
Fluke is right. And here's why this works how it does:

If there are four twos and one three in the prime factorization of 48, then all of the numbers that divide into 48 will be composed only of twos and threes. (Think of prime factors as the 'building blocks' of a number - they can't be split up, so if a number is composed of a certain set of primes, all of its factors can only be constructed from those primes as well.)

Also, any factor of 48 will have at most four twos, and at most one three. If it had more, then it wouldn't divide evenly into 48 - there'd be a two or a three left over that didn't divide evenly. For example, 9 isn't a factor of 48, because 3*3 doesn't divide evenly into 2*2*2*2*3.

What that means is that any factor of 48 will have either no threes, or exactly one three. And, it will have no twos, one two, two twos, three twos, or four twos. Any combination of these will divide evenly into 48:

no twos and no threes = 1
one two and no threes = 2
two twos and no threes = 4
three twos and no threes = 8
four twos and no threes = 16

no twos and one three = 3
one two and one three = 6
two twos and one three = 12
three twos and one three = 24
four twos and one three = 48

That's actually the list of all of the factors of 48! So, how do you calculate how many factors there are without actually writing them all out like that? It's just combinatorics. A factor has either zero or one threes, and either zero, one, two, three, or four fours. That means there's two possibilities for how many threes a factor can have, and five possibilities for how many twos it can have. Combining those gives you 2*5 = 10 factors.

In general, the formula is (1+a)(1+b)(1+c)(...) where a is the number of twos in the prime factorization, b is the number of threes, c is the number of fives, etc. The reason 1 gets added is because there's an additional possibility of having no twos in that particular factor, or no threes, no fives, etc.
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Re: total no of factors from prime factors?   [#permalink] 15 May 2016, 11:10
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