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16 Sep 2014, 01:28
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If \(x\) and \(y\) are positive integers, is the total number of positive factors of \(x^3\) a multiple of the total number of positive factors of \(y^2\)?
(1) \(x=4\)
(2) \(y=6\)
(1) \(x=4\)
(2) \(y=6\)
16 Sep 2014, 01:28
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Official Solution:
If \(x\) and \(y\) are positive integers, is the total number of positive factors of \(x^3\) a multiple of the total number of positive factors of \(y^2\)?
Finding the Number of Factors of an Integer
First, make the 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 respective 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: \(450 = 2^1 * 3^2 * 5^2\)
The total number of factors of 450, including 1 and 450 itself, is \((1+1)(2+1)(2+1) = 2*3*3 = 18\) factors.
(1) \(x=4\).
From this statement we deduce that \(x^3=64=2^6\), hence the number of factors of 64 is \(6+1=7\).
Now, is it possible for \(y^2\) to have a number of factors that is a factor of 7, in other words 1 or 7 factors? It could be, but it's not a certainty. The number of factors of a perfect square is always odd. So \(y^2\) should have either 1 factor (for example if \(y^2=1^2\)) or 7 (for example if \(y^2=27^2=3^6\) or \(y^2=8^2=2^6\)), both scenarios are possible. However, \(y^2\) can also have an odd number of factors that is not a factor of 7, like 3 (for example if \(y=5^2\)). Therefore, this statement is not sufficient.
(2) \(y=6\).
From this statement we deduce that \(y^2=36=2^2*3^2\), hence the number of factors of 36 is \((2+1)*(2+1)=9\).
Is it possible for \(x^3\) to have a number of factors that is a multiple of 9 (9, 18, 27, ...)? Let's represent \(x\) as the product of its prime factors: \(x^3=(a^p*b^q*c^r)^3...=a^{3p}*b^{3q}*c^{3r}...\). The number of factors would be \((3p+1)(3q+1)(3r+1)...\) and this should be a multiple of 9. However, \((3p+1)(3q+1)(3r+1)...\) is not divisible by 3, therefore it cannot be a multiple of 9. This information is sufficient to answer the question: No, the number of factors of \(x^3\) is not a multiple of 9.
Just to elaborate, \(x^3\) has a total of \(3k+1\) distinct factors (1, 4, 7, 10, ... either odd or even). The number of factors of \(x^3\) is one more than a multiple of 3, so it's not divisible by 3, and consequently not by 9 either.
Answer: B
If \(x\) and \(y\) are positive integers, is the total number of positive factors of \(x^3\) a multiple of the total number of positive factors of \(y^2\)?
Finding the Number of Factors of an Integer
First, make the 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 respective 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: \(450 = 2^1 * 3^2 * 5^2\)
The total number of factors of 450, including 1 and 450 itself, is \((1+1)(2+1)(2+1) = 2*3*3 = 18\) factors.
(1) \(x=4\).
From this statement we deduce that \(x^3=64=2^6\), hence the number of factors of 64 is \(6+1=7\).
Now, is it possible for \(y^2\) to have a number of factors that is a factor of 7, in other words 1 or 7 factors? It could be, but it's not a certainty. The number of factors of a perfect square is always odd. So \(y^2\) should have either 1 factor (for example if \(y^2=1^2\)) or 7 (for example if \(y^2=27^2=3^6\) or \(y^2=8^2=2^6\)), both scenarios are possible. However, \(y^2\) can also have an odd number of factors that is not a factor of 7, like 3 (for example if \(y=5^2\)). Therefore, this statement is not sufficient.
(2) \(y=6\).
From this statement we deduce that \(y^2=36=2^2*3^2\), hence the number of factors of 36 is \((2+1)*(2+1)=9\).
Is it possible for \(x^3\) to have a number of factors that is a multiple of 9 (9, 18, 27, ...)? Let's represent \(x\) as the product of its prime factors: \(x^3=(a^p*b^q*c^r)^3...=a^{3p}*b^{3q}*c^{3r}...\). The number of factors would be \((3p+1)(3q+1)(3r+1)...\) and this should be a multiple of 9. However, \((3p+1)(3q+1)(3r+1)...\) is not divisible by 3, therefore it cannot be a multiple of 9. This information is sufficient to answer the question: No, the number of factors of \(x^3\) is not a multiple of 9.
Just to elaborate, \(x^3\) has a total of \(3k+1\) distinct factors (1, 4, 7, 10, ... either odd or even). The number of factors of \(x^3\) is one more than a multiple of 3, so it's not divisible by 3, and consequently not by 9 either.
Answer: B
General Discussion
manlog
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03 Aug 2016, 09:22
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Very nice and tough question. In (2) I guess it's enough to use just that x=a^m, and not a product of three prime factors, since we don't know the actual number of prime factors. In any case, even for a^m, x^3, gives (3m+1) factors which can't be a multiple of 9 factors of 6^3
