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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 \(y^2\)?

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.

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

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

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

1 is not a multiple of 9, it's a divisor of 9.
_________________

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...

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.

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.