5
$\begingroup$

The coolness of an integer is equal to the integer divided by the total number of factors that it has. For example, $48$ has $10$ factors therefore, coolness $(48) = \frac { 48 }{ 10 } =\quad 4.8$

1. Provide an explanation for why coolness$(xy)$ cannot be equal to an integer if both $x$ and $y$ are different prime numbers.

My attempt: Just from a couple of trials a found that the number of factors of $xy$ seems to always be $4$. Therefore, we can note that coolness$(xy)$ can only be a whole number if $xy$ is divisible by $4$. I am pretty sure that we can’t make a multiple of $4$ by multiplying any combination of 2 prime numbers.

The above working is not valid without proofs and I am unsure of how to approach them. How can I prove that $xy$ has $4$ factors and that it is not possible to get a multiple of $4$ by multiplying two prime numbers?

2. $x$ and $y$ are different prime numbers. Identify the numbers of the form $x y^4$ which have a coolness that is equal to an integer.

My attempt: I am not sure how to solve this one but I think listing the factors of $xy^4$ might help.

3. Prove that the square of any prime number $x$ is equal to the coolness of some integer.

My attempt: No idea other than just listing a couple of prime numbers and then squaring them to check if the result is equal to the coolness of some integer.

Please help me solve the above problems.

$\endgroup$
  • 1
    $\begingroup$ Hint: If $a=p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_n^{\alpha_n}$ is the prime factorization of $a$, then $\operatorname{coolness}(a)=(\alpha_1+1)(\alpha_2+1)\cdots(\alpha_n+1)$. This follows because a factor of $a$ has the form $p_1^{\beta_1}\cdots p_n^{\beta_n}$ with $0\leq \beta_i\leq \alpha_i$. $\endgroup$ – Luiz Cordeiro May 26 '16 at 21:58
  • 1
    $\begingroup$ Possible duplicate of Solving a Word Problem relating to factorisation $\endgroup$ – TheRandomGuy May 27 '16 at 11:41
3
$\begingroup$

The divisors of $xy$ when $x$, $y$ are distinct primes are $1,\ x,\ y,\ xy$. If $x$, $y$ are distinct prime numbers then either both are odd, in which case their product is odd and therefore not a multiple of $4$, or one of them is $2$ and the other is odd, and the product of $2$ and an odd number is never divisible by $4$.

For your second problem, the divisors of $xy^4$ are $\underbrace{1,\ y,\ y^2,\ y^3,\ y^4}_\text{No $x$ appears here.},\ \ \underbrace{x,\ xy,\ xy^2,\ xy^3,\ xy^4}_\text{One $x$ divides each of these.}$.

$\endgroup$
1
$\begingroup$

This answers Question 3.

If $p\ge5$, then

$${24p^2\over\tau(24p^2)}={24p^2\over\tau(2^3)\tau(3^1)\tau(p^2)}={24p^2\over4\cdot2\cdot3}=p^2$$

The factorization in the denominator doesn't hold for $p=2$ or $3$, so this leaves the problem of finding (small?) integers $m$ and $n$ such that

$${m\over\tau(m)}=4\quad\text{and}\quad{n\over\tau(n)}=9$$

A quick look at the OEIS finds that these are solved by $m=36$ and $n=108$, respectively. (The general inequality $\tau(N)\le2\sqrt N$, obtained by pairing each divisor $d\gt\sqrt N$ with its companion divisor $N/d\lt\sqrt N$, reduces the search for solutions to $N/\tau(N)=k$ to the range $1\le N\le(2k)^2$. Note that $1\le36\le8^2$ and $1\le108\le18^2$.)

$\endgroup$
0
$\begingroup$

1. Provide an explanation for why coolness$(xy)$ cannot be equal to an integer if both $x$ and $y$ are different prime numbers.

You are correct in your thinking for this part:

As $x$ and $y$ are distinct primes, $xy$ has $4$ factors: $1, x, y, xy$. This means that $xy$ must be divisible by $4$. However, as $2$ is the only even prime number and cannot be used twice, this means that at least one of $x$ or $y$ must be odd. The product of a integer with an odd number will never yield a multiple of $4$, therefore coolness$(xy)$ will never equal a integer where $x$ and $y$ are distinct primes.

2. $x$ and $y$ are different prime numbers. Identify the numbers of the form $x y^4$ which have a coolness that is equal to an integer.

$xy^4$ has $10$ factors: $1, x, y, y^2, y^3, y^4, xy, xy^2, xy^3, xy^4$. This means that for coolness$(xy^4)$ to be an integer, $xy^4$ must be divisible by $10$. Then $x$ and $y$ must multiple to a factor of $10$. The only prime numbers which serve this are $5$ and $2$. Therefore $x=2, y=5$, or $x=5, y=2$.

3. Prove that the square of any prime number $x$ is equal to the coolness of some integer.

Given the prime $x\neq 3$, the integer number which has its coolness as the square of any prime number $x$, can be written $n(x) = 3^2x^2$. It has $3^2 = 9$ factors ($1$, $3$, $x$, $3x$, $3^2x$, $3x^2$, $3^2$, $x^2$, $3^2x^2$) and $$ \mbox{Coolness}(n(x)) = \frac{3^2x^2}{3^2} = x^2 $$ If $x=3$ then the number $n(x)=2^2 3^3 = 108$. In this case

$$ \mbox{Coolness}(108) = \frac{2^23^3}{2^2 3} = 3^2 $$

Therefore the square of any prime number $x$ is equal to the coolness of some integer.

I hope this helps you on your problems!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.