solve-in-R-x-2-2x-3-x-0-

Question Number 86009 by M±th+et£s last updated on 26/Mar/20

s o l v e i n R : [\frac{x}{2}] + [\frac{2 x}{3}] - x = 0

Answered by MJS last updated on 26/Mar/20

[\frac{x}{2}] \in Z \land [\frac{2 x}{3}] \in Z \Rightarrow x \in Z

let k, m, n \in Z

[\frac{x}{2}] = {\begin{cases} \frac{x}{2}; x = 2 m \\ \frac{x}{2} - \frac{1}{2}; x = 2 m + 1 \end{cases}

[\frac{2 x}{3}] = {\begin{cases} \frac{2 x}{3}; x = 3 n \\ \frac{2 x}{3} - \frac{1}{3}; x = 3 n + 2 \\ \frac{2 x}{3} - \frac{2}{3}; 2 x = 3 n + 1 \end{cases}

(1) x = 2 m = 3 n \Rightarrow x = 6 k

[\frac{6 k}{2} [+ [\frac{12 k}{3}] = 7 k

7 k = 6 k \Rightarrow k = 0 \Rightarrow x = 0 ∙

(2) x = 2 m = 3 n + 2 \Rightarrow x = 6 k + 2

[\frac{6 k + 2}{2}] + [\frac{12 k + 4}{3}] = 7 k + 2

7 k + 2 = 6 k + 2 \Rightarrow k = 0 \Rightarrow x = 2 ∙

(3) x = 2 m = 3 n + 1 \Rightarrow x = 6 k + 4

[\frac{6 k + 4}{2}] + [\frac{12 k + 8}{3}] = 7 k + 4

7 k + 4 = 6 k + 4 \Rightarrow k = 0 \Rightarrow x = 4 ∙

(4) x = 2 m + 1 = 3 n \Rightarrow x = 6 k + 3

[\frac{6 k + 3}{2}] + [\frac{12 k + 6}{3}] = 7 k + 3

7 k + 3 = 6 k + 3 \Rightarrow k = 0 \Rightarrow x = 3 ∙

(5) x = 2 m + 1 = 3 n + 2 \Rightarrow x = 6 k + 5

[\frac{6 k + 5}{2}] + [\frac{12 k + 10}{3}] = 7 k + 5

7 k + 5 = 6 k + 5 \Rightarrow k = 0 \Rightarrow x = 5 ∙

(6) x = 2 m + 1 = 3 n + 1 \Rightarrow x = 6 k + 1

[\frac{6 k + 1}{2}] + [\frac{12 k + 2}{3}] = 7 k

7 k = 6 k + 1 \Rightarrow k = 1 \Rightarrow x = 7 ∙

answer is x \in {0, 2, 3, 4, 5, 7}

Commented by M±th+et£s last updated on 26/Mar/20

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