Estonian Mathematical Olympiad

Solution 1: If $x\le y$, then $4\cdot y!<3\cdot x!+4\cdot y!\le 7\cdot y!$, so $\frac{4}{5}\cdot y!<z!<\frac{7}{5}\cdot y!$. Two distinct factorials differ by a factor of at least 2, so $z<y$ would yield $z!\le \frac{1}{2}\cdot y!<\frac{4}{5}\cdot y!<z!$, contradiction. Analogously $z>y$ would yield $z!\ge 2\cdot y!>\frac{7}{5}\cdot y!>z!$, contradiction. Therefore the only option is $z=y$. Then the given equation simplifies to $3\cdot x!=y!$. Here $y>x$ and the product of $x+1,x+2,\dots ,y$ must be 3. This can only happen when $y=x+1=3$, giving the solution $(x,y,z)=(2,3,3)$. If $x>y$, then $3\cdot x!<3\cdot x!+4\cdot y!<7\cdot x!$, so $\frac{3}{5}\cdot x!<z!<\frac{7}{5}\cdot x!$. Analogously to the previous case we get $z=x$. Then the given equation simplifies to $4\cdot y!=2\cdot x!$ or $2\cdot y!=x!$. Analogously to the previous case, this is only possible when $x=y+1=2$, giving the solution $(x,y,z)=(2,1,2)$.

Solution 2: Notice that $x\le z$ and $y\le z$, because if either $x\ge z+1$ or $y\ge z+1$, then $$5\cdot z!=3\cdot x!+4\cdot y!\ge 3\cdot \max (x!,y!)\ge 3\cdot (z+1)!,$$ where dividing by $z!$ gives $5\ge 3(z+1)$, from which $z\le \frac{2}{3}<1$, contradiction.

If $x=z$, then like in Solution 1, we only get the solution $(2,1,2)$. If $y=z$, then like in Solution 1, we only get the solution $(2,3,3)$. * If $x\le z-1$ and $y\le z-1$, then $$5\cdot z!=3\cdot x!+4\cdot y!\le 3\cdot (z-1)!+4\cdot (z-1)!=7\cdot (z-1)!,$$ where dividing by $(z-1)!$ gives $5z\le 7$. This leaves only the option $z=1$, which is impossible by the assumptions $x\le z-1$ and $y\le z-1$.

Solution 3: We divide the sides $3x!+4y!=5z!$ by the smallest factorial present in the equation. This leaves an equation $3a+4b=5c$, where $a,b,c$ are positive integers, of which at least one is equal to 1. If $c=1$, then $3a+4b=5$, giving no solutions. If $a=b=1$, then $7=5c$, also giving no solutions. If $a=1,b>1,c>1$ and $3+4b=5c$, then $y>x$ and $z>x$, meaning that both $b$ and $c$ are divisible by $x+1$. Then 3 is also divisible by $x+1$, which gives $x=2$. Then $z=3$, or else the right hand side of the equation is divisible by 4, whereas the left hand side isn't. Then also $y=3$. If $b=1,a>1,c>1$ and $3a+4=5c$, then $x>y$ and $z>y$, meaning that both $a$ and $c$ are divisible by $y+1$. Then 4 is also divisible by $y+1$, which gives $y=1$ or $y=3$. If $y=1$, then $z=2$, or else the right hand side of the equation is divisible by 3, whereas the left hand side isn't. Then also $x=2$. But if $y=3$, then the left hand side is never divisible by 5, so there are no solutions. Thus the only suitable triples are $(2,1,2)$ and $(2,3,3)$.