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M28-02

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M28-02 [#permalink]

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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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Re M28-02 [#permalink]

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New post 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 \(3k+1>\) 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.


Answer: B
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Re: M28-02 [#permalink]

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New post 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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Re: M28-02 [#permalink]

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New post 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.
Please correct me
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Re: M28-02 [#permalink]

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New post 18 Aug 2016, 07:53
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Re: M28-02 [#permalink]

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New post 18 Jun 2017, 07:56
I didn't understand the solution. Can you help in understanding the solution better.
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Re: M28-02 [#permalink]

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Re: M28-02 [#permalink]

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New post 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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Re: M28-02 [#permalink]

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New post 02 Sep 2017, 17:02
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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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Re: M28-02 [#permalink]

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Re: M28-02 [#permalink]

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New post 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 \(3k+1>\) 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.


Answer: B



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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Re: M28-02 [#permalink]

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New post 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 \(3k+1>\) 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.


Answer: B



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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M28-02 [#permalink]

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New post 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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