Solve the equation $7t+[2t] =52 $ ,where $[x]$ denotes the floor function for $x$.

Solve the equation $7t+\left\lfloor 2t\right\rfloor =52 $.

My effort

Using the fact that for any number $x$ we have that $x=\left\lfloor x\right\rfloor+\{x\}$ (where $\{x\}$ is the fractional part of $x$) for $7t$ ,I have that:

\begin{array}{c} 7t+\left\lfloor 2t\right\rfloor &=52 \\ \left\lfloor 7t\right\rfloor+\{7t\} +\left\lfloor 2t\right\rfloor &=52 \end{array}

where $\{7t\}=0$ ,since we have no fractional part, and from this it also follows that $\left\lfloor 7t\right\rfloor=7t$

So the equation breaks down to \begin{array}{c} 7t+\left\lfloor 2t\right\rfloor=52 \\ \left\lfloor 2t\right\rfloor =52-7t \\ \end{array}

Now, applying the definition of the floor function, I have that \begin{array}{c} 52-7t \le 2t <53-7t \\ 52\le 9t <53 \\ 52/9 \le t < 53/9 \\ \end{array}

Question

Is my effort correct? Are there other ways to approach the problem?

Solution 1:

Outline

You are absolutely right in your calculations, you only forgot to apply the other condition, that the fractional part $\{7t\} = 0$. So when you got your interval for $t$, you have to choose a $t$ satisfying this condition too.

It is easy to see that $t = \color{blue}{\frac{41}{7}}$ is the only $t$ which will be in the interval of length $\frac{1}{9}$

Note - added explanation : $\frac{52}{9} = 5\frac{7}{9}$ and $\frac{53}{9} = 5\frac{8}{9}$. So choose $t = 5\frac{6}{7}$.

Solution 2:

The work so far shows that any solution lies in that interval, but not that every value in that interval is a solution.

On the other hand, since $\lfloor 2t \rfloor$ and $52$ are integers, if $t$ is a solution, then $7t$ must be an integer, too, that is, we can write $t = \frac{a}{7}$ for some integer $a$. Since the interval has length $\frac{1}{9}$, there is at most one such value in the interval (in fact, there turns out be exactly one), so we can solve the problem just by checking it.