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

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If x and y are positive integers, is the total number of positive divi  [#permalink]

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New post 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

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]

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New post 25 Dec 2016, 06:34
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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


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.




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) \(x=4\). From this statement we have that\(x^3=64=2^6\), thus the number of factors of 64 is 6+1=7.

Now, may \(y^2\) have the number of factors which is factor of 7, so 1 or 7 factors? Well may have or may not. Number of factors of a perfect square is odd. So \(y^2\) should have either 1 factor (for example if y^2=1^2) or 7 (for example if y^2=81^2=3^6 or y^2=8^2=2^6), both are possible, BUT \(y^2\) can have other odd number of factors say 3 (for example if y=5^2) and 3 is not factor of 7. Not sufficient

(2) \(y=6\). From that: \(y^2=36=2^2*3^2\), thus the number of factors of 36 is (2+1)*(2+1)=9.

Can \(x^3\) have the number of factors which is 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 multiple of 9. BUT \((3p+1)(3q+1)(3r+1)\) is not divisible by 3, hence it can not be multiple of 9. The answer is NO. Sufficient.

Not to complicate \(x^3\) has \(3k+1>\) number of distinct factors (1, 4, 7, 10, ... odd or even number), so the number of factors of \(x^3\) is 1 more than a multiple of 3, thus it's not divisible by 3 and hence not by 9.


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Re: If x and y are positive integers, is the total number of positive divi  [#permalink]

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Re: If x and y are positive integers, is the total number of positive divi  [#permalink]

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New post 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]

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New post 10 Oct 2018, 16:04
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] 10 Oct 2018, 16:04
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