# 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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Since you are new, I want to let you know some things about the site. In order to get the best possible answers, it is helpful if you say what your thoughts on the problem are so far; this will prevent people from telling you things you already know, and help them write answers at an appropriate level. Also, people are much happier to help those who show they've tried the problem themselves first. Also, many see commands ("Find") as rude when asking for help; please consider rewriting your post. – Zev Chonoles Jun 22 '12 at 23:41
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
Mo, I appreciate the work that you have shown in this ^^ most recent comment. To emphasize (edit: and add my personal commentary to) a part of Zev's comment, you should include the work that you have done in the question itself. – The Chaz 2.0 Jun 23 '12 at 0:25

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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