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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## If x and y are positive integers ##### This topic has 4 expert replies and 1 member reply ## If x and y are positive integers 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 OA:B _________________ Fiza Gupta ### GMAT/MBA Expert Elite Legendary Member Joined 23 Jun 2013 Posted: 10130 messages Followed by: 494 members Upvotes: 2867 GMAT Score: 800 Hi fiza gupta, This question can be solved by TESTing VALUES. We're told that X and Y are POSITIVE INTEGERS. We're asked if the number of positive divisors of X^3 is a multiple of the total number of positive divisors of Y^2. This is a YES/NO question. 1) X = 4 With this Fact, we know that X^3 = 64 and the divisors are: 1 and 64, 2 and 32, 4 and 16, and 8 --> 7 total divisors IF.... Y = 1, then Y^2 = 1 and the total number of divisors is 1... and the answer to the question is YES. Y = 2, then Y^2 = 4 and the total number of divisors is 3... and the answer to the question is NO. Fact 1 is INSUFFICIENT 2) Y = 6 With this Fact, we know that Y^2 = 36 and the divisors are: 1 and 36, 2 and 18, 3 and 12, 4 and 9 and 6 --> 9 total divisors At first glance, this might appear to be a similar issue to what appeared in Fact 1. However, you're going to find that you can't get to 9 total divisors when cubing a positive integer. IF... X = 1, then X^3 = 1 and the total divisors is 1... and the answer to the question is NO. X = 2, then X^3 = 8 and the total divisors is 4... and the answer to the question is NO. X = 3, then X^3 = 27 and the total divisors is 4... and the answer to the question is NO. X = 4, then X^3 = 64 and the total divisors is 7... and the answer to the question is NO. X = 5, then X^3 = 125 and the total divisors is 5... and the answer to the question is NO. X = 6, then X^3 = 216 and the total divisors is 16... and the answer to the question is NO. Etc. Thus, the answer to the question is ALWAYS NO. Fact 2 is SUFFICIENT Final Answer: B GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at Rich.C@empowergmat.com Senior | Next Rank: 100 Posts Joined 04 Jun 2014 Posted: 53 messages Followed by: 2 members Upvotes: 6 fiza gupta 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 OA:B Can someone please explain a more methodical approach? _________________ If you want to fly,you have to give up the things that weighs you down! PS ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 fiza gupta 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. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br ### GMAT/MBA Expert GMAT Instructor Joined 22 Aug 2016 Posted: 1898 messages Followed by: 30 members Upvotes: 470 prabsahi wrote: fiza gupta 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 OA:B Can someone please explain a more methodical approach? Given: x and y are positive integers. Question: Is the total number of positive divisors of x^3 a multiple of the total number of positive divisors of y^2? Let's take each statement one by one. (1) x = 4 Given x = 4, we have x^3 = 4^3 = 2^6 => there are a total of 6 + 1 = 7 positive divisors. If say y = 4, then y^2 also has 7 numbers of positive divisors. Thus, the total number of positive divisors of x^3 IS a multiple of the total number of positive divisors of y^2. The answer is Yes. However, if say y = 2, then y^2 = 2^2 => there are a total of 2 + 1 = 3 positive divisors. Thus, the total number of positive divisors of x^3 IS NOT a multiple of the total number of positive divisors of y^2. The answer is No. (2) y = 6 Given y = 6, we have y^2 = 2^2 * 3^2 => there are a total of (2 + 1)*(2 + 1) = 9 positive divisors. If x = 1, then x^3 = 1^3 = 1. Thus, the total number of positive divisors of x^3 IS NOT a multiple of the total number of positive divisors of y^2. The answer is No. If x = a, where a is a prime factor. x^3 = a^3 => there are a total of 3 + 1 = 4 positive divisors. Since 4 is even and 9 (number of factors of y^3) is odd, the total number of positive divisors of x^3 CANNOT BE a multiple of the total number of positive divisors of y^2. The answer is No. Even if we take x = product of two or more prime factors, we will have a total number of positive divisors = EVEN, which is not divisible by 9, an odd number. Thus, the total number of positive divisors of x^3 CANNOT BE a multiple of the total number of positive divisors of y^2. The answer is No. Sufficient. Say x = a*b*c, where a, b and c are prime factors. x^3 = a^3*b^3*c^3 => there are a total of (3 + 1)*(3 + 1)*(3 + 1) = 64 positive divisors. We see that 64 is even and 9 (number of factors of y^3) is odd, the total number of positive divisors of x^3 CANNOT BE a multiple of the total number of positive divisors of y^2. The correct answer: B Hope this helps! -Jay _________________ Manhattan Review GMAT Prep Locations: New York | Bangalore | Guangzhou | Buenos Aires | and many more... Schedule your free consultation with an experienced GMAT Prep Advisor! Click here. ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 I finished my solution (posted above) with the following statement: $x = {\text{prime}}{{\text{s}}^{\,{\text{positive}}\,\,{\text{integers}}}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\# \,\,{\text{posit}}\,\,{\text{div}}\,\,{\text{of}}\,\,{x^3}\,\,\, = \,\,\,\,\left[ {({\text{posit}}\,{\text{mult}}\,\,{\text{of}}\,\,3) + 1} \right]\,\,\,{\text{product}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,$ Let me give an example to avoid being misunderstood! $$x = {2^2} \cdot 3\,\,\,\, \Rightarrow \,\,\,\,{x^3} = {2^6} \cdot {3^3}$$ $$\# \,\,{\rm{positive}}\,\,{\rm{divisors}}\,\,{\rm{of}}\,\,x\,\,\,\, = \,\,\,\left( {6 + 1} \right)\left( {3 + 1} \right)$$ $${{\left( {6 + 1} \right)\left( {3 + 1} \right)} \over 9} \ne {\mathop{\rm int}}$$ In other words, although we may have odd factors and also even factors in the numerator, (only odd factors is also a possibility), all these factors are NOT multiples of 3 (they are always one unit more than a multiple of 3), we are sure 9 = 3^2 is not a factor of the numerator. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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