# How to find x,y,z such that $\frac{1800}{x}=a$, $\frac{1800}{y}=b$, $\frac{1800}{z}=c$, $\frac{1800}{a+b+c}=d$

I'm really fascinated by how questions and problems are designed in mathematics. So, I was designing a simple word problem, and in the course I fell into this situation:

a,b,c,d are natural numbers. $a>1$, $b>1$, $c>1$

$a+b+c <1800$

x,y,z are natural numbers.

$x\neq y \neq z$

$x >1$, $y>1$, $z>1$

Find all triplets x,y,z (no need for permutations) so that:

$\dfrac{1800}{x}=a$

$\dfrac{1800}{y}=b$

$\dfrac{1800}{z}=c$

$\dfrac{1800}{a+b+c}=d$

My attempt:

$1800$ has $36$ divisors. But by excluding $1$ and $1800$, we have $34$ divisors. Total triplets formed without permutations are $\dfrac{34\cdot33\cdot32}{3\cdot2\cdot1}=5984$

I know by intuition that the solution doesn't probably exceed 8 or 9 triplets, but I'm not able to make any progress. Thank you for your help.

• As long as $a,b,c$ and $a+b+c$ are divisors of $1800$, you will be able to pick natural numbers $d,x,y,z$ to fulfill the equations. Consider adding a condition such as $a,b,c$ are distinct numbers (makes $a+b+c$ more interesting) or maybe even all coprime (if it works out nicely). – Marconius Jul 30 '15 at 2:40
• With a program, I got $371$ triples $(a,b,c)$ such that $a,b,c,a+b+c$ are all divisors of $1800$ and $a < b < c$. – JimmyK4542 Jul 30 '15 at 2:53
• $x\neq y \neq z$ is true for $25\neq 30 \neq 25$ so maybe if you want $x, y, z$ all different you should say that. – jbuddenh Jul 30 '15 at 2:55
• For example, some triplets $(x,y,z)$: $(90, 225, 56)$, $(90, 200, 58)$, $(90, 180, 60)$, $(600, 900, 450)$, etc. I ran my program and I got 7251 results. – GAVD Jul 30 '15 at 2:56
• @JimmyK4542: That's really nice. Thank you. I'm still looking for a mathematical approach, if such approach exists. – Rafiq Jul 30 '15 at 2:58

There are 324 triples $[x,y,z]$ that make all your conditions true. I don't know whether $324=18^2$ is a coincidence or not. These were found by a short maple program. You can see the program and its output here: http://1drv.ms/1H4yqn2