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If x and y are positive integers

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If x and y are positive integers

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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

OA:B

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Fiza Gupta

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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

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Contact Rich at Rich.C@empowergmat.com

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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?

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If you want to fly,you have to give up the things that weighs you down!

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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.

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Fabio Skilnik :: www.GMATH.net (Math for the GMAT)
Course release PROMO : finish our test drive till 31/Oct with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 60% discount!

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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
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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.

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Fabio Skilnik :: www.GMATH.net (Math for the GMAT)
Course release PROMO : finish our test drive till 31/Oct with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 60% discount!

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