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Re: The number 10010 has how many positive integer factors? [#permalink]
Top Contributor
TomerSdt wrote:
What's the reasoning for this?



When we break a number into a prime factor form, say N = \(a^p\), then the number of factors n = (p + 1). We add +1 to include 1 as a factor.

For eg if we have N = \(2^2\), then it will have (2 + 1) = 3 factors which are 1, 2 and 4

If N = \(2^2 * 3^1\), then the number of factors = (2 + 1) * (1 + 1) = 3 * 2 = 6

Why do we multiply? We know that the factors of 4 are (1, 2, 4) and the factors of 3 are (1,3)

Therefore (1, 2, 4) * (1, 3) = (1*1, 1*3, 2*1, 2*3, 4*1, 4*3) = 6 factors = (1, 2, 3, 4, 6 and 12) which are nothing but the factors of 12.

So any number N, in the form N = \(a^p * b^q * c^r\) and so on will have the number of factors n = (p + 1) * (q + 1) * (r + 1) and so on.



Arun Kumar
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Re: The number 10010 has how many positive integer factors? [#permalink]
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Re: The number 10010 has how many positive integer factors? [#permalink]
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