求方程 $\left[\dfrac x2\right]+\left[\dfrac x3\right]+\left[\dfrac x7\right]=x$ 的所有解，其中 $[a]$ 表示不超过 $a$ 的最大整数．
【难度】
【出处】
2009年全国高中数学联赛安徽省预赛
【标注】
【答案】
$\{0,-6,-12,-14,-18,-20,-21,-24,-26,-27,-28,-30,-32,-33,-34,-35,-36,-38,-39,-40,-41,-44,-45,-46,-47,-49,-50,-51,-52,-53,-55,-57,-58,-59,-61,-64,-65,-67,-71,-73,-79,-85\}$
【解析】
由方程知解 $x$ 是整数．
设 $x=42p+q$，其中 $p\in \mathbb Z$，$q\in\{0,1,\cdots ,41\}$．
由$$\left[\dfrac x2\right]+\left[\dfrac x3\right]+\left[\dfrac x7\right]=x,$$得$$p=\left[\dfrac q2\right]+\left[\dfrac q3\right]+\left[\dfrac q7\right]-q,$$因此方程的解集为\[\begin{split}& \left\{ 42\left(\left[\dfrac q2\right]+\left[\dfrac q3\right]+\left[\dfrac q7\right]-q\right)+q|q=0,1,\cdots ,41\right\}\\ =&\{0,-6,-12,-14,-18,-20,-21,-24,-26,-27,\\ &-28,-30,-32,-33,-34,-35,-36,-38,-39,\\&-40,-41,-44,-45,-46,-47,-49,-50,-51,\\&-52,-53,-55,-57,-58,-59,-61,-64,-65,\\&-67,-71,-73,-79,-85\}\end{split}\]
设 $x=42p+q$，其中 $p\in \mathbb Z$，$q\in\{0,1,\cdots ,41\}$．
由$$\left[\dfrac x2\right]+\left[\dfrac x3\right]+\left[\dfrac x7\right]=x,$$得$$p=\left[\dfrac q2\right]+\left[\dfrac q3\right]+\left[\dfrac q7\right]-q,$$因此方程的解集为\[\begin{split}& \left\{ 42\left(\left[\dfrac q2\right]+\left[\dfrac q3\right]+\left[\dfrac q7\right]-q\right)+q|q=0,1,\cdots ,41\right\}\\ =&\{0,-6,-12,-14,-18,-20,-21,-24,-26,-27,\\ &-28,-30,-32,-33,-34,-35,-36,-38,-39,\\&-40,-41,-44,-45,-46,-47,-49,-50,-51,\\&-52,-53,-55,-57,-58,-59,-61,-64,-65,\\&-67,-71,-73,-79,-85\}\end{split}\]
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