0
$\begingroup$

$$\frac1 a + \frac 1 b +\frac 1 c = \frac 34$$

Find number of triplets of $a\ , b\ , c\in \mathbb{Z}^+$

Should it not be infinite since it can be $\frac 34$ or $\frac38$ or $\frac9{12}$ etc. till infinity?

Please help since I am clueless on how to even begin solving this.

$\endgroup$
  • 1
    $\begingroup$ Do you mean integer like $a,b,c \in \mathbb{Z}$ or integral? $\endgroup$ – MrYouMath Jun 18 '16 at 18:03
  • $\begingroup$ No of integral solutions $\endgroup$ – ghostrider Jun 18 '16 at 18:04
  • 1
    $\begingroup$ Do you count 2,5,20 and 2,20,5 and 5,2,20 and 5,20,2 and 20,2,5 and 20,5,2 as one solution or six? $\endgroup$ – almagest Jun 18 '16 at 18:10
  • $\begingroup$ 6 for ordered and 1 for unordered solutions.. Can you please share how did you get to those values in the first place? $\endgroup$ – ghostrider Jun 18 '16 at 18:11
  • 2
    $\begingroup$ So you want the number of ordered and the number of unordered solutions? $\endgroup$ – almagest Jun 18 '16 at 18:12
4
$\begingroup$

Probably the most elementary way to do this is bounding the variables. W.L.O.G. suppose that $a \le b \le c$. Observe that if $a \ge 5$ then $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \le \frac{3}{5}<\frac{3}{4}$$Thus $a \le 4$. $a=1$ doesn't work. Hence we have three cases to check: $a=2, 3, 4$. If $a=2$ equation reduces to$$\frac{1}{b}+\frac{1}{c}=\frac{1}{4}$$Obviously $b>4$ and if $b>8$ then$$\frac{1}{b}+\frac{1}{c} \le \frac{2}{9} < \frac{1}{4} $$which is again a contradiction. So $ 5\le b \le 8$ and from here you get solutions $(a, b, c)=(2, 5, 20), (2, 6, 12), (2, 8, 8)$. For $a=3$ you get$$\frac{1}{b}+\frac{1}{c}=\frac{5}{12}$$Here we must have $b>2$ and if $b>4$ then$$\frac{1}{b}+\frac{1}{c} \le \frac{2}{5} < \frac{5}{12} $$So $ 3\le b \le 4$ and from here you get solutions $(a, b, c)=(3, 3, 12), (3, 4, 6)$. For $a=4$ you get$$\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$$Here we must have $b>3$ and if $b>4$ then$$\frac{1}{b}+\frac{1}{c} \le \frac{2}{5} < \frac{1}{2} $$Hence you must have $ b=4$, which gives one solution $(a, b, c)=(4, 4, 4)$.

$\endgroup$
1
$\begingroup$

Write this as the following: $$\frac 1 a+\frac 1 b+\frac 1 c=\frac{3n}{4n}$$ Then multiply both sides by $4n$: $$\frac {4n} a+\frac {4n} b+\frac {4n} c=3n$$ Now, we need to find three divisors of $4n$ such that they sum to $3n$.

  • $n=1 \implies$ Obviously, we have $1+1+1=3$. This gives us $a=b=c=\frac 4 1=4$.
  • $n=2 \implies$ The divisors of $8$ are $1,2,4,8$, so we have $6=2+2+2$. This gives us $a=b=c=\frac 8 2=4$. (From now on, we know that all of the sums that are just $x+x+x$ will just give us $a=b=c=4$, so we're going to ignore that sum.) We also have $6=1+1+4$ which gives us $a=\frac 8 1=8$, $b=\frac 8 1=8$, and $c=\frac 8 4=2$.
  • $n=3 \implies$ The divisors of $12$ are $1,2,3,4,6,12$. We can sum to $9$ with $1+2+6$, so we get $a=\frac {12} 1=12$, $b=\frac{12}{2}=6$ and $c=\frac{12}{6}=2$. We also have $9=2+3+4$, giving us $a=\frac{12}{2}=6$, $b=\frac{12}{3}=4$ and $c=\frac{12}{4}=3$. Furthermore, we have $9=1+4+4$, giving us $a=\frac{12}{1}=12$, $b=\frac{12}{4}=3$, and $c=\frac{12}{4}=3$

Keep going to find more solutions. Just for recap, here are my five solutions for $(a, b, c)$: $$(4, 4, 4)$$ $$(8, 8, 2)$$ $$(12, 6, 2)$$ $$(6, 4, 3)$$ $$(12, 3, 3)$$ However, there is still one more solution. Can you find it?

$\endgroup$
  • $\begingroup$ How does that help to establish the number of solutions? Are you claiming it is infinite? $\endgroup$ – almagest Jun 18 '16 at 18:38
  • 1
    $\begingroup$ @almagest No, I'm getting them started on how to find solutions. $\endgroup$ – Noble Mushtak Jun 18 '16 at 18:39
  • $\begingroup$ Won't this reach to infinity since we can do it for infinite N values? $\endgroup$ – ghostrider Jun 18 '16 at 18:42
  • 1
    $\begingroup$ @ghostrider No, because eventually, as you keep doing this out, you'll realize that you're getting the same solutions over and over again, so the number of solutions is finite. $\endgroup$ – Noble Mushtak Jun 18 '16 at 18:45
1
$\begingroup$

We start by looking for unordered solutions (ie we regard $(a,b,c)=(2,8,8)$ and $(8,2,8)$ as the same solution.

Suppose $a=b=c$, then clearly there is the unique solution $(4,4,4)$.

Suppose $a=b\ne c$. We have $8c+4a=3ac$, so $(3a-8)c=4a$. The LHS must be positive, so $a\ge 3$. If $a>8$, then $3a-8>2a$ and $3a-8<4a$, so $c=1$ is too small and $c\ge2$ is too big. That leaves $a=3,4,5,6,7,8$. It is easy to check that 3 and 8 give the solutions $(3,3,12),(8,8,2)$.

So it remains to consider the case $a<b<c$. We have $\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}<\frac{45}{60}=\frac{3}{4}$. So we must have $a\le3$. We cannot have $a=1$ because $1>\frac{3}{4}$. So $a=2$ or 3. Consider first $a=3$.

We have $\frac{1}{3}+\frac{1}{5}+\frac{1}{6}=\frac{7}{10}<\frac{3}{4}$. So we must have $b=4$. It is easy to check this gives the solutions $(3,4,6)$ and no others.

Finally consider $a=2$. We have already know that $(2,8,8)$ is a solution, so any solution with $a<b<c$ must have $b<8$. It must have $b>4$ since $\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$. So we check $b=5,6$ and find the solutions $(2,5,20),(2,6,12)$.

That gives a total of 6 unordered solutions. Three have $a,b,c$ unequal, so each generate 6 ordered solutions. Two have just two of $a,b,c$ equal, so they each generate 3 ordered solutions. Finally, the solution with $a=b=c$ gives only one ordered solution, for a total of 25 ordered solutions.

$\endgroup$

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.