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Question Number 98806 by bemath last updated on 16/Jun/20

for a is integer number such that ∣∣x−1∣ −2∣ ≤ a exactly has 2013 solution

$$\mathrm{for}\:{a}\:\mathrm{is}\:\mathrm{integer}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mid\mid{x}−\mathrm{1}\mid\:−\mathrm{2}\mid\:\leqslant\:{a}\:\:\mathrm{exactly}\:\mathrm{has}\:\mathrm{2013} \\ $$$$\mathrm{solution} \\ $$

Commented by Rasheed.Sindhi last updated on 16/Jun/20

Assuming x an integer: ∣∣x−1∣ −2∣ ≤ a ⇒ { ((∣x−1−2∣≤a⇒∣x−3∣≤a...(i))),((∣−x+1−2∣≤a⇒∣−x−1∣≤a..(ii))) :} (i)⇒ { ((x−3≤a⇒x≤a+3)),((−x+3≤a⇒x≥−a+3)) :} ..(iii) (ii)⇒ { ((−x−1≤a⇒x≥−a−1)),((x+1≤a⇒x≤a−1)) :}..(iv) (iii)⇒−a+3≤x≤a+3......(v) (iv)⇒−a−1≤x≤a−1......(vi) (v)&(vi)⇒−a−1≤x≤a+3 −1004−1≤x≤1004+3 −1005≤x≤1007 x=−1005,−1004,...,−2,−1,0,1,2...,1006,1007 a=1004

$${Assuming}\:{x}\:{an}\:{integer}: \\ $$$$\mid\mid{x}−\mathrm{1}\mid\:−\mathrm{2}\mid\:\leqslant\:{a} \\ $$$$\Rightarrow\begin{cases}{\mid{x}−\mathrm{1}−\mathrm{2}\mid\leqslant{a}\Rightarrow\mid{x}−\mathrm{3}\mid\leqslant{a}...\left({i}\right)}\\{\mid−{x}+\mathrm{1}−\mathrm{2}\mid\leqslant{a}\Rightarrow\mid−{x}−\mathrm{1}\mid\leqslant{a}..\left({ii}\right)}\end{cases} \\ $$$$\left({i}\right)\Rightarrow\begin{cases}{{x}−\mathrm{3}\leqslant{a}\Rightarrow{x}\leqslant{a}+\mathrm{3}}\\{−{x}+\mathrm{3}\leqslant{a}\Rightarrow{x}\geqslant−{a}+\mathrm{3}}\end{cases}\:..\left({iii}\right) \\ $$$$\left({ii}\right)\Rightarrow\begin{cases}{−{x}−\mathrm{1}\leqslant{a}\Rightarrow{x}\geqslant−{a}−\mathrm{1}}\\{{x}+\mathrm{1}\leqslant{a}\Rightarrow{x}\leqslant{a}−\mathrm{1}}\end{cases}..\left({iv}\right) \\ $$$$\left({iii}\right)\Rightarrow−{a}+\mathrm{3}\leqslant{x}\leqslant{a}+\mathrm{3}......\left({v}\right) \\ $$$$\left({iv}\right)\Rightarrow−{a}−\mathrm{1}\leqslant{x}\leqslant{a}−\mathrm{1}......\left({vi}\right) \\ $$$$\left({v}\right)\&\left({vi}\right)\Rightarrow−{a}−\mathrm{1}\leqslant{x}\leqslant{a}+\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1004}−\mathrm{1}\leqslant{x}\leqslant\mathrm{1004}+\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1005}\leqslant{x}\leqslant\mathrm{1007} \\ $$$${x}=−\mathrm{1005},−\mathrm{1004},...,−\mathrm{2},−\mathrm{1},\mathrm{0},\mathrm{1},\mathrm{2}...,\mathrm{1006},\mathrm{1007} \\ $$$$\:\:\:\:\:\:\:{a}=\mathrm{1004} \\ $$

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