no. of positive integral solutions of ||x - 1| - 2| + x = 3

Question

What are the no. of positive integral solutions of ||x - 1| - 2| + x = 3 ?

My effort

||x - 1| - 2| = 3 - x

|x - 1| - 2 = 3 - x OR |x - 1| - 2 = x - 3

|x - 1| = 5 - x OR |x - 1| = x - 1

x - 1 = 5 - x OR x - 1 = x - 5 OR x - 1 $\geq$ 0

2x = 6 OR x $\geq$ 1

Therefore, x $\geq$ 1

But the any value of x greater than 1 except 3 does not satisfy the equation. Where have I gone wrong?

Answer 1

You should start with expanding the innermost absolute value: $$ \vert \vert x-1 \vert -2 \vert = \begin{cases} \vert x-3 \vert & x \geqslant 1 \\ \vert -1-x \vert & x < 1 \end{cases} = \begin{cases} x-3 & x \geqslant 1, x > 3 \\ 3-x & x \geqslant 1, x \leqslant 3 \\ 1+x & x < 1, x \geqslant -1 \\ -1-x & x < 1, x < -1 \end{cases} = \begin{cases} x-3 & x > 3 \\ 3-x & 1 \leqslant x \leqslant 3\\ 1+x & -1 \leqslant x < 1 \\ -1-x & x < -1 \end{cases} $$ You now seek solutions for each of these branches being equal to $3-x$ within the appropriate domain, giving you three integral solutions $x=1, x=2, x=3$.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large\sum_{x\ =\ 0}^{\infty} \delta_{\verts{\vphantom{\large A}\verts{x\ -\ 1}\ -\ 2}\ +\ x,3}} =\sum_{x\ =\ 0}^{\infty}\oint_{\verts{z}\ =\ 1} {1 \over z^{-\verts{\verts{\vphantom{\Large A}x\ -\ 1}\ -\ 2}\ -\ x\ +\ 4}} \,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1}{1 \over z^{4}} \sum_{x\ =\ 0}^{\infty}z^{\verts{\verts{\vphantom{\Large A}x\ -\ 1}\ -\ 2}\ +\ x} \,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ 1}{1 \over z^{4}}\pars{% z + z^{3} + \sum_{x\ =\ 2}^{\infty}z^{\verts{x\ -\ 3}\ +\ x}} \,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1}{1 \over z^{4}}\pars{% z + z^{3} + z^{3} + z^{3} + \sum_{x\ =\ 4}^{\infty}z^{2x - 3}} \,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1}\pars{% {1 \over z^{3}} + {3 \over z} + \sum_{x\ =\ 4}^{\infty}{1 \over z^{7 - 2x}}} \,{\dd z \over 2\pi\ic} \\[5mm]&=\overbrace{\oint_{\verts{z}\ =\ 1}{1 \over z^{3}}\,{\dd z \over 2\pi\ic}} ^{\ds{=\ \dsc{0}}}\ +\ \overbrace{\oint_{\verts{z}\ =\ 1}{3 \over z}\,{\dd z \over 2\pi\ic}} ^{\ds{=\ \dsc{3}}}\ +\ \sum_{x\ =\ 4}^{\infty}\overbrace{% \oint_{\verts{z}\ =\ 1}{1 \over z^{7 - 2x}}\,{\dd z \over 2\pi\ic}} ^{\ds{=\ \dsc{\delta_{x,3}}}}\ =\ \color{#66f}{\LARGE 3} \end{align}