Positive Pairs

$gcd(x,y)=5!=120$

$x=120a$ $y=120b$ $gcd(a,b)=1$

$lcm(x,y)=lcm(120a,120b)=120lcm(a,b)=120ab/gcd(a,b)=120ab$

$120ab=50!$ $ab=\frac{50!}{120}$

Theorem: Then number of positive integer solutions to $ab=n$ with $a \le b$ and $gcd(a,b)=1$ is equal to $2^{k-1}$ where $k$ is the number of distinct prime divisors of $n$

Proof: take any prime divisor $p$ of $n$ . Let $p^t$ be the largest power that divides $n$ . Assume that not all of these powers are in single factor ( $a$ or $b$ ), then that means $p$ is a common factor of $a$ and $b$ which contradicts $gcd(a,b)=1$ . Thus for any given prime divisor $p^t$ in the factorization of $n$ , either $p^t|a$ or $p^t|b$ but not both.

This means that for each of the $k$ prime divisors we have two choices of which factor to place it in, giving us $2^k$ possible solutions, however we have $a \le b$ thus we need to divide this by half, giving us $2^{k-1}$ solutions.

QED

So, to find the number of solutions we simply need to know the number of prime divisors of $\frac{50!}{120}$ . the number of prime divisors of $50!$ is equal to the number of primes less than $50$ which is $15$ . Now we need to consider if any of these prime factors are eliminated when dividing by $120=2^3*3*5$ now since all of $16,9,25$ are divisors of $50!$ that prevents the prime factors $2,3,5$ from being eliminated. Thus the number of prime divisors of $\frac{50!}{120}$ is 15

Thus our solution is $2^{14}=16384$

Let $p_1,p_2,...,p_{12}$ denote in, increasing order, the primes from $7$ to $47$

Then $5!=2^3 \cdot 3^1 \cdot 5^1 \cdot p^0_1 \cdot \cdot \cdot p^0_{12}$

$50!=2^{\alpha_1}\cdot 3^{\alpha_2}\cdot 5^{\alpha_3}\cdot p^{b_1}_1 \cdot \cdot p^{b_2}_2 \cdot \cdot \cdot p^{b_{12}}_{12}$

Note that $2^4,3^2,5^2,p_1,p_2,....,p_{12}$ all divide $50!$ so all its prime powers differ from those of $5!$

Since $x,y | 50!$ they are of the form $x=2^{n_1} \cdot 3^{n_2} \cdot \cdot \cdot p^{n_{15}}_{12}$ $y=2^{m_1} \cdot 3^{m_2} \cdot \cdot \cdot p^{m_{15}}_{12}$

Then $max(n_i,m_i)$ is the $ith$ prime power in $50!$ and $min(n_i,m_i)$ is the $ith$ prime powerin $5!$

Since the prime powers for $p_{12}$ and under differ in $5!$ and $50!$ , there are $2^{15}$ choices for $x$ , only half of which will be less than $y$ . (Since for each choice of $x$ , we have that $y$ is forced and either $x<y$ or $y<x$ .) So the number of pairs is $2^{15}/2=\boxed{2^{14}}$ .

The prime factorization of 50! consists of powers of the primes < 50 or 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 = first 15 primes, after we remove the gcd 5! each remaining term (=prime to some power) must be 100% in x or y because any other split will change the gcd since each of 15 primes have 2 possibilities that gives $2^{15}$ combinations with $\frac{2^{15}}{2}=2^{14}=\fbox{16384}$ having $x<y$ and due to the unequal splits of each prime factor $x\neq y$

for example 7^8 can only be split into (x, y) as $(7^{0}, 7^{8})$ or $(7^{8},7^{0})$ or 7 will occur in x and y and ruin gcd=5! while maintaining gcd requires 2 split into $2^{3}, 2^{44}$ and 3 split $3^{1},3^{21}$ and 5 split $5^{1},5^{11}$

[prime factorization not necessary] $\frac{50!}{5!} =2^{47-3}\cdot 3^{22-1}\cdot 5^{12-1} \cdot7^{8}\cdot7^{8}\cdot11^{4}\cdot13^{3}\cdot17^{2}\cdot19^{2}\cdot23^{2}\cdot29^{1}\cdot31^{1}\cdot37^{1}\cdot41^{1}\cdot43^{1}\cdot47^{1}$

I find this solution really elegant but this explanation less so, so please et me know how I can improve on it :)