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# Is the total number of divisors of x^3 a multiple of the

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Is the total number of divisors of x^3 a multiple of the [#permalink]

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06 Nov 2009, 09:27
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Is the total number of divisors of x^3 a multiple of the total number of divisors of y^2 ?

1. x = 4
2. y = 6

***************************************************************************

my question: doesn't a perfect square always have an odd number of factors?

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06 Nov 2009, 09:46
B - 2 is sufficient

Total number of divisors for a perfect cube $$x^3$$ is one more than a multiple of 3 i.e $$3m+1$$ divisors
Total number of divisors for a perfect square $$y^2$$ will be odd i.e $$2n+1$$ divisors

1. $$x=4$$

$$4^3$$ has 7 divisors if you check. 7 is not a multiple of anything except 7 and 1.
So $$y^2$$ can have 7 divisors if say $$y=64$$, or $$y^2$$ can have 1 divisor, where $$y=1$$.
NOT SUFF.

2. $$y=6$$
$$Y^2 = 36$$ has 9 total divisors.
$$x^3$$ has $$3m+1$$ total divisors, which cannot be divided by 9.
SUFF.

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06 Nov 2009, 10:14
4test1 wrote:
B - 2 is sufficient

Total number of divisors for a perfect cube $$x^3$$ is one more than a multiple of 3 i.e $$3m+1$$ divisors
Total number of divisors for a perfect square $$y^2$$ will be odd i.e $$2n+1$$ divisors

1. $$x=4$$

$$4^3$$ has 7 divisors if you check. 7 is not a multiple of anything except 7 and 1.
So $$y^2$$ can have 7 divisors if say $$y=64$$, or $$y^2$$ can have 1 divisor, where $$y=1$$.
NOT SUFF.

2. $$y=6$$
$$Y^2 = 36$$ has 9 total divisors.
$$x^3$$ has $$3m+1$$ total divisors, which cannot be divided by 9.
SUFF.

The trick here is that we are not told that x and y are integers, so for (2) the logic would be correct ONLY for integers. y^2=36=x^3 then total number of divisors of x^3 is equal to the total number of divisors of y^2. Hence insufficient.

amitgovin wrote:
my question: doesn't a perfect square always have an odd number of factors?

1. The number of distinct factors of a perfect square is ALWAYS ODD.
2. The sum of distinct factors of a perfect square is ALWAYS ODD.
4. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors.
5. Perfect square always has even number of powers of prime factors.
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06 Nov 2009, 11:16
Thanks for pointing out the catch, Bunuel. If we can't assume they are integers, then none of the logic really holds true. How do you define a divisor for 1.53 - is it fair to say that the quotient must be an integer?

So would it then be right to skip straight to C since we haven't been told they are integers?

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06 Nov 2009, 11:34
4test1 wrote:
Thanks for pointing out the catch, Bunuel. If we can't assume they are integers, then none of the logic really holds true. How do you define a divisor for 1.53 - is it fair to say that the quotient must be an integer?

So would it then be right to skip straight to C since we haven't been told they are integers?

Well you are partially right. Remember we are not asked to determine # of divisors of x or y, which may not be the integers, but the divisors of x^3 and y^2. x can be 3^1/3 which is not an integer and it makes no sense to ask about the factors of x. But x^3=3 and 3 has 2 distinct factors. So this is not the problem in the question.

BUT you are right about the factors of non integers and here is the tip from GMAT tutor Ian Stewart:

Every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

So we can say that this question is not real GMAT one.
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06 Nov 2009, 22:45
4test1 wrote:
B - 2 is sufficient
Total number of divisors for a perfect cube $$x^3$$ is one more than a multiple of 3 i.e $$3m+1$$ divisors
Total number of divisors for a perfect square $$y^2$$ will be odd i.e $$2n+1$$ divisors

Hi.

whats m & n in your statements above?

TX
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07 Nov 2009, 02:11
3m+1 is to show that the number is one more than a multiple of 3. And 2n + 1 is to show that the number is odd. We do not know exactly what m, n as it will depend on each value for x^3 and y^2.

In general, m,n are integers from {0,1,2,3,4,5,..}

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Re: # of Divisors   [#permalink] 07 Nov 2009, 02:11
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