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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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Updated on: 25 Dec 2016, 05:38
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30% (02:08) correct 70% (01:41) wrong based on 118 sessions

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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, 05:23.
Last edited by vienbuuchau on 25 Dec 2016, 05:38, edited 2 times in total.
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Joined: 02 Sep 2009
Posts: 53066
Re: If x and y are positive integers, is the total number of positive divi  [#permalink]

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25 Dec 2016, 05:34
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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 $$3k+1&gt;$$ 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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25 Dec 2016, 05:33
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

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Joined: 01 Sep 2016
Posts: 193
GMAT 1: 690 Q49 V35
Re: If x and y are positive integers, is the total number of positive divi  [#permalink]

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18 Sep 2017, 23:36
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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10 Oct 2018, 15:04
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, 15:04
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