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19 May 2012, 08:55
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D
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Difficulty:
45%
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Question Stats:
69% (01:59) correct
31%
(02:14)
wrong
based on 468
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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.
(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.
20 May 2012, 11:09
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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.
Answer: D.
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.
(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.
Answer: D.
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.
General Discussion
23 May 2012, 04:09
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Quote:
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?
