Integer part in equations

Let $x = n+k$ where $n$ is an integer part and $k$ is a fractional part ( $0 \leq k < 1$ ).

The equation becomes

$3(n+k) - n = 8$

$2n+3k = 8$ $(*)$

Since $0 \leq k < 1$ , we get

$2n \leq 2n+3k = 8 < 2n+3$

Solve compound inequality we get

$\displaystyle \frac{5}{2} < n \leq 4$ .

Since $n$ is an integer, we get $n = 3, 4$ .

If $n = 3$ , from $(*)$ we get $\displaystyle k = \frac{2}{3}$ .

If $n = 4$ , we get $k = 0$ .

Therefore, $x = n+k = \boxed{\displaystyle \frac{2}{3}, 4}$ . ~~~

Rewrite the equation as shown: $3x=8+\lfloor x\rfloor$

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Knowing that $\lfloor x\rfloor$ is an integer $$ $$ $$ $\Rightarrow$ $$ $$ $$ $8+\lfloor x\rfloor$ is an integer $$ $$ $$ $\Rightarrow$ $$ $$ $$ $3x$ is an integer.

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Replacing $x=\frac{k}{3}$ with $k$ integer, the equation became $k-8=\lfloor \frac{k}{3}\rfloor$

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For the definition of integer part $$ $$ $\Rightarrow$ $$ $$ $\frac{k}{3}-1<\lfloor \frac{k}{3}\rfloor \leq \frac{k}{3}$ $$ $$ $\Rightarrow$ $$ $$ $\frac{k}{3}-1<k-8\leq \frac{k}{3}$ $$ $$ $\Rightarrow$ $$ $$ $\frac{21}{2}<k \leq12$ .

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$k=11$ or $k=12$ from which $x=\frac{11}{3}$ or $x=4$ .
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$30*(\frac{11}{3}+4)=230$

x-1≤[x]≤x from definition of [x],then -x≤-[x]≤1-x and 2x≤3x-[x]≤2x+1 ,now from our equation we get that 2x≤8≤2x+1, which means 3.5≤x≤4, let x be x=3.5+a ,0≤a≤0.5. Now our equation become 10.5+3a-[3.5+a]=8,there are two solutions,One when [3.5+a]=3 and than 10.5+3a-3=8 , 3a=0.5,a=1/6 so x= 7/2+1/6=22/6 and other when [3.5+a]=4 and 10.5+3a-4=8 , 3a=1.5 , a=0.5,so x=3.5+0.5=4 is other solution. S=4+22/6=23/3,and 30xS=230

We consider the equation: floor(x) = 3x - 8. Hence we find the rational solutions to the equation since x = a/3 where a is an integer to imply that floor(a/3) = a - 8 since floor x is an integer.

By trial and error for intervals, or considering the lines y = 3x - 8 and y = floor(x), we can infer that the intersections are the points (11/3, 3) and (4, 4). Implying that 30 x S = 230.