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# M28-02

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:28
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95% (hard)

Question Stats:

29% (01:55) correct 71% (01:36) wrong based on 54 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) $$y=6$$

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16 Sep 2014, 01:28
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Official Solution:

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 $$y^2$$?

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.

(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=27^2=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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03 Aug 2016, 09:22
Very nice and tough question. In (2) I guess it's enough to use just that x=a^m, and not a product of three prime factors, since we don't know the actual number of prime factors. In any case, even for a^m, x^3, gives (3m+1) factors which can't be a multiple of 9 factors of 6^3
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18 Aug 2016, 05:13
In option 2, if we take X=1 ,then total no. of positive divisor of x3 is 1, which is multiple of total number of positive divisors of y2.
Then the option should be E.
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18 Aug 2016, 07:53
ravirjn wrote:
In option 2, if we take X=1 ,then total no. of positive divisor of x3 is 1, which is multiple of total number of positive divisors of y2.
Then the option should be E.

1 is not a multiple of 9, it's a divisor of 9.
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18 Jun 2017, 07:56
I didn't understand the solution. Can you help in understanding the solution better.
Math Expert
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18 Jun 2017, 09:56
I didn't understand the solution. Can you help in understanding the solution better.

You should be more specific when asking a question.
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31 Jul 2017, 21:56
I got this question wrong
I sort of knew but i wasn't certain about the following - so this is for everyone like myself -

Divisor = Factor
Divisor is the same as factor. Greatest Common Divisor (GCD) is same as greatest common factor (GCF) or highest common factor (HCF).

Now when i re-read the question - it makes more sense!!!
Hope this helps...
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02 Sep 2017, 17:02
1
On the part of the solution that says: for example if y^2=81^2=3^6 .. didnt you mean f y^2=27^2=3^6 ?

81^2 = 3^8
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03 Sep 2017, 04:52
jjacosta37 wrote:
On the part of the solution that says: for example if y^2=81^2=3^6 .. didnt you mean f y^2=27^2=3^6 ?

81^2 = 3^8

_________________________
Yes. Edited the typo. Thank you.
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04 Sep 2017, 10:43
Bunuel wrote:
Official Solution:

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 $$y^2$$?

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.

(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=27^2=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.

Dear Bunuel

Does the formula of total factor have limitation? it does not work for 1? If I apply the rule for 1 it will yield 2 factors which is wrong.

can you clarify please?
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04 Sep 2017, 10:47
Mo2men wrote:
Bunuel wrote:
Official Solution:

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 $$y^2$$?

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.

(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=27^2=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.

Dear Bunuel

Does the formula of total factor have limitation? it does not work for 1? If I apply the rule for 1 it will yield 2 factors which is wrong.

can you clarify please?

1 can be written as a^0*b^0*c^0... --> # of factors = (0+1)(0+1)(0+1)... = 1.

But it's obvious even without it that 1 has 1 factor...
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04 Sep 2017, 10:55
Bunuel wrote:

1 can be written as a^0*b^0*c^0... --> # of factors = (0+1)(0+1)(0+1)... = 1.

But it's obvious even without it that 1 has 1 factor...

I know it is obvious but I thought the rule failed to consider 1.

Thanks for clarification.
Re: M28-02   [#permalink] 04 Sep 2017, 10:55
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