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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?

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    $\begingroup$ The solution should certainly not be an interval, as the l.h.s. of the equation has derivative $7$ wherever the derivative is defined (i.e., everywhere except the half-integers), but the r.h.s is constant. $\endgroup$ – Travis Feb 10 '16 at 16:01
  • $\begingroup$ @Travis Where's the mistake then ? $\endgroup$ – Mr. Y Feb 10 '16 at 16:02
  • $\begingroup$ I don't think your sign for floor function is right $\endgroup$ – Archis Welankar Feb 10 '16 at 16:03
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    $\begingroup$ The beginning was good. From what you wrote you can conclude that $t=n+\frac{k}{7}$ for some integer $n$ and some integer $k$ between $0$ and $6$. Then the end should come quickly. $\endgroup$ – André Nicolas Feb 10 '16 at 16:04
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    $\begingroup$ It's not true that 7{x} = {7x}, I think. That seems like a mistake. $\endgroup$ – peter.petrov Feb 10 '16 at 16:07
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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 $t = 5\frac{6}{7}$.

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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.

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  • $\begingroup$ I did not understand why derivative of LHS = 77 as you have mentioned in the first line. $\endgroup$ – Shailesh Feb 10 '16 at 16:33
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    $\begingroup$ @Shailesh: It should be $7$ for the derivative, a typo. $\endgroup$ – Ross Millikan Feb 10 '16 at 17:18
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    $\begingroup$ @Shailesh Ross is right, thanks to both of you for pointing this out. $\endgroup$ – Travis Feb 10 '16 at 19:55
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Are there other ways to approach the problem?

Sure. Let $t = m + n$, where $m$ is a multiple of $1/2$ and $0 \le n < 1/2$. For example $\underbrace{12/5}_t = \underbrace{5/2}_m + \underbrace{1/10}_n$. There is always 1 unique $(m,n)$ pair for each $t$. Then you have:

$$7t + \lfloor 2t \rfloor = 52$$ $$7(m+n) + \lfloor 2(m+n) \rfloor = 52$$ $$7m+7n + \lfloor 2m+2n \rfloor = 52$$

$2m$ is an integer and $0 \le 2n < 1$ so

$$7m + 7n + 2m = 52$$ $$9m + 7n = 52$$

The largest value $7n$ can be is $3.5$, so $52 - 3.5 < 9m \le 52$, so $5.3\bar 8 < m \le 5.\bar 7$, so $m = 5.5$.

$$49.5 + 7n = 52$$ $$n = 5/14$$

$$t = m + n = 11/5 + 5/14 = 41/7$$

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