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Intern  Joined: 04 Mar 2012
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How many even different factors does the integer P have?  [#permalink]

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How many even different factors does the integer P have?

(1) P = (x^2)(y^2)(z^3)(2^4), where x, y, and z are different odd prime numbers.
(2) The total number of different factors in P are 180 and P is a multiple of 16 but not a multiple of 32. Also only other prime numbers that are factors of P are 3, 5 and 7.
Math Expert V
Joined: 02 Sep 2009
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Re: How many even different factors does the integer P have?  [#permalink]

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How many even different factors does the integer P have?

(1) P = (x^2)(y^2)(z^3)(2^4), where x, y, and z are different odd prime numbers --> the number of factors of P is (2+1)(2+1)(3+1)(4+1)=180 and the number of odd factors of P is (2+1)(2+1)(3+1)=36, so the number of even factors of P is 180-36=144. Sufficient.

(2) The total number of different factors in P are 180 and P is a multiple of 16 but not a multiple of 32. Also only other prime numbers that are factors of P are 3, 5 and 7 --> basically the same here: P=3^p*5^q*7^r*2^4 --> the number of factors of P is 180=(p+1)(q+1)(r+1)(4+1) and the number of odd factors of P is 180/(4+1)=36, so the number of even factors of P is 180-36=144. Sufficient.

Finding the Number of Factors of an Integer

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

So, the # of factors of x=a^2*b^3, where a and b are different prime numbers is (2+1)(3+1)=12.

Hope it helps.
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Manager  Joined: 06 Apr 2010
Posts: 62
Re: How many even different factors does the integer P have?  [#permalink]

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Quote:
Finding the Number of Factors of an Integer

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

So, the # of factors of x=a^2*b^3, where a and b are different prime numbers is (2+1)(3+1)=12.

Hope it helps.

Using the example above, can we conclude that:
1. to find odd factors, we exclude the power of 2
therefore # odd factor= 3^2*5^2
2.to find factors that contain 5,we could rewrite the 450 as
450=5*(2^1*3^2*5^1)
therefore # factors that contain 5=(1+1)(2+1)(1+1)=12
3.2.to find factors that contain 6,we could rewrite the 450 as
450=2*3*(3^1*5^2)
therefore # factors that contain 6=(1+1)(2+1)=6
and so on..
Is that correct?
Director  V
Joined: 27 May 2012
Posts: 932
Re: How many even different factors does the integer P have?  [#permalink]

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Bunuel wrote:
How many even different factors does the integer P have?

(1) P = (x^2)(y^2)(z^3)(2^4), where x, y, and z are different odd prime numbers --> the number of factors of P is (2+1)(2+1)(3+1)(4+1)=180 and the number of odd factors of P is (2+1)(2+1)(3+1)=36, so the number of even factors of P is 180-36=144. Sufficient.

(2) The total number of different factors in P are 180 and P is a multiple of 16 but not a multiple of 32. Also only other prime numbers that are factors of P are 3, 5 and 7 --> basically the same here: P=3^p*5^q*7^r*2^4 --> the number of factors of P is 180=(p+1)(q+1)(r+1)(4+1) and the number of odd factors of P is 180/(4+1)=36, so the number of even factors of P is 180-36=144. Sufficient.

Finding the Number of Factors of an Integer

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

So, the # of factors of x=a^2*b^3, where a and b are different prime numbers is (2+1)(3+1)=12.

Hope it helps.

Is it possible to get a little bit more on this total number of even factors concept?

I realized we are finding total factors then subtracting total number of odd factors , is there no direct way for finding the total number of even factors ?

what if a number has no odd factors such as 16 or 32 etc ? In this case we are looking at the powers of 2's, to find total number of even factors, aren't we ?

But when there are both even and odd factors in a number such as 30 or 18 or 20 etc in this case why are we going other way round , is there no way of looking at the powers of 2 to get total number of even factors ?
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Intern  B
Joined: 30 Jan 2017
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Re: How many even different factors does the integer P have?  [#permalink]

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Bunuel wrote:
How many even different factors does the integer P have?

(1) P = (x^2)(y^2)(z^3)(2^4), where x, y, and z are different odd prime numbers --> the number of factors of P is (2+1)(2+1)(3+1)(4+1)=180 and the number of odd factors of P is (2+1)(2+1)(3+1)=36, so the number of even factors of P is 180-36=144. Sufficient.

(2) The total number of different factors in P are 180 and P is a multiple of 16 but not a multiple of 32. Also only other prime numbers that are factors of P are 3, 5 and 7 --> basically the same here: P=3^p*5^q*7^r*2^4 --> the number of factors of P is 180=(p+1)(q+1)(r+1)(4+1) and the number of odd factors of P is 180/(4+1)=36, so the number of even factors of P is 180-36=144. Sufficient.

Finding the Number of Factors of an Integer

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

So, the # of factors of x=a^2*b^3, where a and b are different prime numbers is (2+1)(3+1)=12.

Hope it helps.

Bunuel, I did not quite understand how the number of odd factors were calculated. Could you please elaborate.
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Re: How many even different factors does the integer P have?  [#permalink]

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_________________ Re: How many even different factors does the integer P have?   [#permalink] 10 Apr 2019, 10:56
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