Find all solutions of the equation $13[x]+25\{x\}=271$

Question

For a real number $x$, let $[x]$ denote the largest integer $\le x$, and let $\{x \}$ denote $x-[x]$. Find all solutions of the equation: $$13[x]+25\{x\}=271.$$

I tried to simplify the equation by:

$$13[x]+25(x-[x])=271$$

$$\implies 13[x]+25x-25[x]=271$$

$$\implies 25x-12[x]=271$$

$$\implies x= \dfrac{271+12[x]}{25}.$$

Now I saw that for $[x]=17$, we get $x=19$, but this is not a solution.

Do I have to go on manually checking for such solution, if it exists? Is there any generic/empirical way to solve this?

Some more facts I have observed (I don't know whether I am hitting at the right point or not).

If $x=[x]$,

$$13x=271$$

$$\implies x=\dfrac{271}{13}\approx 2.84.$$

Thus $[x]=2.$

But if $x>[x]$, then $$13[x]+25\{x\}=271,$$

where $\{x\}$ is the fractional part of $x$.

Thus $25\{x\}=271-[x]$.

But as $[x]$ is always an integer, $271-[x] \ne 25\{x\}.$

So we have only solution for $x=[x]$.

Possibly I am wrong at this. Please show me the correct way to solve it.

Answer 1

$\displaystyle \frac{25x-271}{12}=\lfloor x \rfloor$.

\begin{align*} x-1<\frac{25x-271}{12}&\le x\\ 19+\frac{12}{13}< x&\le20+\frac{11}{13}\\ \lfloor x \rfloor&=19 \textrm{ or }20 \end{align*}

If $\lfloor x \rfloor=19$, $\displaystyle x=\frac{12(19)+271}{25}=19.96$.

If $\lfloor x \rfloor=20$, $\displaystyle x=\frac{12(20)+271}{25}=20.44$

The only answers are $19.96$ and $20.44$.

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The other way to solve this is to write $\displaystyle x=\frac{12\lfloor x\rfloor+271}{25}$. Then

\begin{align*} \lfloor x\rfloor\le \frac{12\lfloor x\rfloor+271}{25}&<\lfloor x\rfloor+1\\ 18+\frac{12}{13}<\lfloor x\rfloor&\le 20+\frac{11}{13} \end{align*}

As $\lfloor x\rfloor$ is an integer, $\lfloor x\rfloor=19$ or $20$.

When $\lfloor x\rfloor=19$, $\displaystyle x=\frac{12(19)+271}{25}=19.96$.

When $\lfloor x\rfloor=20$, $\displaystyle x=\frac{12(20)+271}{25}=20.44$.

Answer 2

From 13 [x] + 25 {x} = 271 we get 25 {x} = 271 - 13 [x]

This implies that 25 {x} is an integer since the RHS is an integer

Therefore {x} = K/25 where K can take any integer value from 0 to 24 (inclusive) since 0 < {x} < 1

It can be verified that {x} = 0 does not yield a solution

Now note that 271 ≡ K (mod 13) from the original equation

K can thus equal 11 or 24

Putting K=11 yields [x] = 20 which gives x = 20.44 as one solution.

Putting K=24 yields [x] = 19 which gives x = 19.96 as another solution

Therefore the only solutions of the baove equation are x = {20.44, 19.96}