# Triples of positive real numbers $(a,b,c)$ such that $\lfloor a\rfloor bc=3,\; a\lfloor b\rfloor c=4,\;ab\lfloor c\rfloor=5$

Find the all ordered triplets of positive real numbers $(a,b,c)$ such that: $$\lfloor a\rfloor bc=3,\quad a\lfloor b\rfloor c=4,\quad ab\lfloor c\rfloor=5,$$ where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.

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What have you tried? –  dtldarek Jun 22 '12 at 23:32
I've edited your question a bit. I assumed that "arranged" in "arranged triplet" meant "ordered", and changed accordingly; if that was mistaken, I apologize, and feel free to correct it. –  Zev Chonoles Jun 22 '12 at 23:36
You are completely right,thank you –  Frank Jun 22 '12 at 23:39
dtldarek ,i found one solution which is $\lfloor a\rfloor\lfloor b\rfloor\lfloor c\rfloor\in\{1,2,4\}$ but not sure how to find the all solutions –  Frank Jun 22 '12 at 23:46

Hint: Note that $a$, $b$, and $c$ are $\ge 1$. By the first condition, all of our numbers are in the interval $[1,2)$, or two of them are in $[1,2)$ and the other is in $[2,3)$. So there are only $4$ cases to consider, and each gives us an explicit system of ordinary equations (no floors). And it's not really even $4$.

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No, it is not $4$,there are $2$ cases . –  Frank Jun 23 '12 at 1:54
@MohammedAl-mubark: I am happy to that you have seen how the rest goes. –  André Nicolas Jun 23 '12 at 1:56

Now we have by dividing equations and putting together the ratios we get the ratio $$\frac a{\lfloor a\rfloor}:\frac b{\lfloor b\rfloor}:\frac c{\lfloor c\rfloor}=20:15:12$$

This shows that $\frac a{\lfloor a\rfloor}\ge\frac53$ since $\frac c{\lfloor c\rfloor}\ge1$. It is checkable that $\frac a{\lfloor a\rfloor}\ge\frac53$ forces $\lfloor a\rfloor=1$, since $a,b,c\ge1$

Similarly, we get $\frac b{\lfloor b\rfloor}\ge\frac54$ This forces $\lfloor b\rfloor\le3$

Now,$$a=\frac a{\lfloor a\rfloor}\ge\frac53$$

we already have $ab\ge\frac{25}{16}$. But $ab\lfloor c\rfloor=5$ so that $\lfloor c\rfloor\le\frac{80}{25}$ whence $\lfloor c\rfloor\le 3$ But if one of $b,c$ is $3$ or more then it forces the other two to be $1$ or less which contradicts either the ratios above or the fact that all of a,b,c are at least 1. So, $\lfloor b\rfloor=\lfloor c\rfloor\le2$ and $\lfloor a\rfloor=1$

The ratios become $$a:\frac b{\lfloor b\rfloor}:\frac c{\lfloor c\rfloor}=20:15:12$$

Case 1: $\lfloor b\rfloor=1$ Then $a:b=4:3$, so $b=\frac{3a}4$. The third equation in data then gives$$\frac{3a^2}4\lfloor c\rfloor=5$$ So that $3a^2\lfloor c\rfloor=20$. But $a<2$ implies $\lfloor c\rfloor=2$ Now we get $a=\sqrt\frac{10}3$ $b=\frac{\sqrt{30}}4$ Since $\lfloor a\rfloor=1$, we have by the first equation in data that $\frac{\sqrt{30}c}4=3$ That is, $c=\frac25{\sqrt{30}}$

Case 2: $\lfloor b\rfloor=2$ Then $a:b=2:3$, so $b=\frac{3a}2$. The third equation in data then gives$$\frac{3a^2}2\lfloor c\rfloor=5$$ So that $3a^2\lfloor c\rfloor=10$.

Suppose $\lfloor c\rfloor=1$, we get $3a^2=10$ again and as above using ratios we get the same $a$ and $b$, which contradicts that $\lfloor b\rfloor\ne1$ here! This contradiction gives $\lfloor c\rfloor=2$ and we get $6a^2=10$ and $a=\sqrt\frac53$ But we saw earlier that $a\ge\frac53$. So case 2 is contradictory and we have only 1 solution!

$$(a,b,c)=\left(\sqrt\frac{10}3,\frac{\sqrt{30}}4,\frac25{\sqrt{30}}\right)$$

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