1
$\begingroup$

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.

$\endgroup$
11
  • $\begingroup$ 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). $\endgroup$ – Marconius Jul 30 '15 at 2:40
  • $\begingroup$ With a program, I got $371$ triples $(a,b,c)$ such that $a,b,c,a+b+c$ are all divisors of $1800$ and $a < b < c$. $\endgroup$ – JimmyK4542 Jul 30 '15 at 2:53
  • 1
    $\begingroup$ $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. $\endgroup$ – jbuddenh Jul 30 '15 at 2:55
  • $\begingroup$ 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. $\endgroup$ – GAVD Jul 30 '15 at 2:56
  • $\begingroup$ @JimmyK4542: That's really nice. Thank you. I'm still looking for a mathematical approach, if such approach exists. $\endgroup$ – Rafiq Jul 30 '15 at 2:58
1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ I wanted to upvote you, but I haven't that right yet. Thank you so much,jbuddenh $\endgroup$ – Rafiq Jul 30 '15 at 4:49

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.