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If x and y are positive integers, is the total number of positive divi

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vienbuuchau
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If x and y are positive integers, is the total number of positive divi [#permalink] If x and y are positive integers, is the total number of positive divi   Updated on: 25 Dec 2016, 06:38
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If x and y are positive integers, is the total number of positive divisors of x^3 a multiple of the total number of positive divisors of y^2?

(1) x = 4

(2) y = 6

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Originally posted by vienbuuchau on 25 Dec 2016, 06:23.
Last edited by vienbuuchau on 25 Dec 2016, 06:38, edited 2 times in total.
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Re: If x and y are positive integers, is the total number of positive divi [#permalink] If x and y are positive integers, is the total number of positive divi   25 Dec 2016, 06:34
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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 \(y^2\)?

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.


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Re: If x and y are positive integers, is the total number of positive divi [#permalink] If x and y are positive integers, is the total number of positive divi   19 Sep 2017, 00:36
AWESOME, I saw a very similar question somewhere else. Hats off to Bunuel for such a wonderful explanation. Bunuel: you were right. they are very similar. I made a note. thanks
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Re: If x and y are positive integers, is the total number of positive divi [#permalink] If x and y are positive integers, is the total number of positive divi   10 Oct 2018, 16:04
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vienbuuchau wrote:
If x and y are positive integers, is the total number of positive divisors of x^3 a multiple of the total number of positive divisors of y^2?

(1) x = 4

(2) y = 6

\(x,y\,\, \geqslant 1\,\,\,{\text{ints}}\)

\(\frac{{\,\# \,\,{\text{posit}}\,\,{\text{div}}\,\,{\text{of}}\,\,{x^3}\,}}{{\# \,\,{\text{posit}}\,\,{\text{div}}\,\,{\text{of}}\,\,{y^2}}}\,\,\,\mathop = \limits^? \,\,\,\operatorname{int}\)


\(\left( 1 \right)\,\,\,x = {2^2}\,\,\,\,\, \Rightarrow \,\,\,\,\,\# \,\,{\text{posit}}\,\,{\text{div}}\,\,{\text{of}}\,\,{x^3}\left( { = {2^{\,6}}} \right)\,\,\, = \,\,\,7\)

\(\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,y = 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\,\left( {\frac{7}{1} = \operatorname{int} } \right)\,\, \hfill \\\\
\,{\text{Take}}\,\,y = 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\left( {\frac{7}{3} \ne \operatorname{int} } \right)\,\, \hfill \\ \\
\end{gathered} \right.\)



\(\left( 2 \right)\,\,\,y = 2 \cdot 3\,\,\,\,\, \Rightarrow \,\,\,\,\,\# \,\,{\text{posit}}\,\,{\text{div}}\,\,{\text{of}}\,\,{y^2}\left( { = {2^2} \cdot {3^2}} \right)\,\,\, = \,\,\,9\)


\(x = 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\# \,\,{\text{posit}}\,\,{\text{div}}\,\,{\text{of}}\,\,{1^3}\left( { = 1} \right)\,\,\, = \,\,\,1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\,\)

\(x = {\text{prime}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\# \,\,{\text{posit}}\,\,{\text{div}}\,\,{\text{of}}\,\,{x^3}\,\,\, = \,\,\,3 + 1 = 4\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\,\)

\(x = {\text{prime}}{{\text{s}}^{\,{\text{positive}}\,\,{\text{integers}}}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\# \,\,{\text{posit}}\,\,{\text{div}}\,\,{\text{of}}\,\,{x^3}\,\,\, = \,\,\,\,\left[ {({\text{posit}}\,{\text{mult}}\,\,{\text{of}}\,\,3) + 1} \right]\,\,\,{\text{product}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: If x and y are positive integers, is the total number of positive divi [#permalink] If x and y are positive integers, is the total number of positive divi   12 Feb 2022, 05:54
Brilliant Question!
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Re: If x and y are positive integers, is the total number of positive divi [#permalink] If x and y are positive integers, is the total number of positive divi   15 Aug 2023, 07:45
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Re: If x and y are positive integers, is the total number of positive divi [#permalink]
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