数论高斯记号作业

求满足 $[\frac{x}{2}]+[\frac{x+1}{3}]+[\frac{x+2}{5}]=x$的所有 $x$ 值之和。

$\because$ gcd(2,3,5)=30
$\therefore x=30k+r,k\in \mathbb{Z},-15\le r\le 14$
带入原方程
$$\begin{gathered} [\frac{x}{2}]+[\frac{x+1}{3}]+[\frac{x+2}{5}] \\ =[\frac{30k+r}{2}]+[\frac{30k+r+1}{3}]+[\frac{30k+r+2}{5}] \\ =[\frac{30k}{2}+\frac{r}{2}]+[\frac{30k}{3}+\frac{r+1}{3}]+[\frac{30k}{5}+\frac{r+2}{5}] \\ =[15k+\frac{r}{2}]+[10k+\frac{r+1}{3}]+[6k+\frac{r+2}{5}] \\ =15k+[\frac{r}{2}]+10k+[\frac{r+1}{3}]+6k+[\frac{r+2}{5}] \\ =31k+[\frac{r}{2}]+[\frac{r+1}{3}]+[\frac{r+2}{5}] \\ =30k+r \end{gathered}$$
$\therefore [\frac{r}{2}]+[\frac{r+1}{3}]+[\frac{r+2}{5}]=r-k$
$\therefore \frac{r}{2}+\frac{r+1}{3}+\frac{r+2}{5}-3<r-k\le \frac{r}{2}+\frac{r+1}{3}+\frac{r+2}{5}$
$\therefore \frac{r}{2}+\frac{r+1}{3}+\frac{r+2}{5}-r-3<-k\le \frac{r}{2}+\frac{r+1}{3}+\frac{r+2}{5}-r$
$\therefore \frac{15r+10r+10+6r+12-30r}{30}-3<-k\le \frac{15r+10r+10+6r+12-30r}{30}$
$\therefore \frac{r-68}{30}<-k\le \frac{r+22}{30}$
$\therefore \frac{68-r}{30}>k\ge \frac{-r-22}{30},-15\le r\le 14$
$\therefore 68-r>30\times k\ge -r-22,-15\le r\le 14$
$\therefore 68-r>30\times k\ge -r-22,15\ge -r\ge -14$
$\therefore 68-15>30\times k\ge -14-22$
$\therefore 53>30\times k\ge -36$
$\therefore 1.8>k\ge -1.2$
$\therefore k=1,0,-1$

1. 当 $k=-1$ 时候，$\therefore 68-r>-30\ge -r-22$
   $68-r>-30\to -r>-98\to r<98$
   $-30\ge -r-22\to -8\ge -r\to 8\le r$
   $\therefore 8\le r<98$
   $\therefore 8\le r\le 14$
   验算 $[\frac{r}{2}]+[\frac{r+1}{3}]+[\frac{r+2}{5}]=r-k$ 得 $r=8,14$ 有解。所以 $x=30k+r=-30+r=-22,-16$。

2. 当 $k=0$ 时候，
   $\therefore 68-r>0\ge -r-22$
   $\therefore -15\le r\le 14$
   验算得有解 $x=-12,-10,-7,-6,-4,-2,-1,2,3,4,5,6,9,10,11,12,13$。

3. 当 $k=1$ 时候，
   $\therefore 68-r>30\ge -r-22$
   $\therefore -15\le r\le 14$
   验算得有解 $x=15,16,17,19,21,22,25,27,31,37$。

全部得 $x=-22,-16,-10,-7,-6,-4,-2,-1,2,3,4,5,6,9,10,11,12,13,15,16,17,19,21,22,2527,31,37$

$\sum x=225$

已知 $0<a<1$，且满足 $[a+\frac{1}{30}]+[a+\frac{2}{30}]+...+[a+\frac{29}{30}]=18$，求 $[10a]$ 的值。

$\because 0<a<1$
$\therefore 0+\frac{1}{30}<a+\frac{1}{30}<1+\frac{1}{30}$
$\therefore 0<a+\frac{1}{30}<2$
$\therefore 0<a+\frac{1}{30}<a+\frac{2}{30}<...<a+\frac{29}{30}<2$
$$\begin{gathered} \therefore 0<[a+\frac{1}{30}]<2, \\ 0<[a+\frac{2}{30}]<2, \\ ..., \\ 0<[a+\frac{29}{30}]<2 \end{gathered}$$
即 $[a+\frac{x}{30}],x\in [1,29]$ 一定是 0 或者 1.
根据题意，其中 $18$ 个等于 $1$。
$$\begin{gathered} \therefore [a+\frac{1}{30}]=[a+\frac{2}{30}]=...=[a+\frac{11}{30}]=0 \\ [a+\frac{12}{30}]=[a+\frac{13}{30}]=...=[a+\frac{29}{30}]=1 \end{gathered}$$
$\therefore 0<a+\frac{11}{30}<1,1\le a+\frac{12}{30}<2$
$\therefore 18\le 30a<19$
$\therefore 6\le 10a<\frac{19}{3}$
$\therefore [10a]=6$

设 $r$ 满足 $[r+\frac{19}{100}]+[r+\frac{20}{100}]+[r+\frac{21}{100}]...+[r+\frac{91}{100}]=546$，求 $[100r]$ 的值

解：
[r+19100]+[r+20100]+[r+21100]...+[r+91100]=[[r]+{r}+19100]+[[r]+{r}+20100]+[[r]+{r}+21100]+...+[[r]+{r}+91100]=[r]+[{r}+19100]+[r]+[{r}+20100]+[r]+[{r}+21100]+...+[r]+[{r}+91100]=73[r]+[{r}+19100]+[{r}+20100]+[{r}+21100]+...+[{r}+91100]=546[r+\frac{19}{100}]+[r+\frac{20}{100}]+[r+\frac{21}{100}]...+[r+\frac{91}{100}]\\ =[[r]+\{r\}+\frac{19}{100}]+[[r]+\{r\}+\frac{20}{100}]+[[r]+\{r\}+\frac{21}{100}]+...+[[r]+\{r\}+\frac{91}{100}]\\ =[r]+[\{r\}+\frac{19}{100}]+[r]+[\{r\}+\frac{20}{100}]+[r]+[\{r\}+\frac{21}{100}]+...+[r]+[\{r\}+\frac{91}{100}]\\ =73[r]+[\{r\}+\frac{19}{100}]+[\{r\}+\frac{20}{100}]+[\{r\}+\frac{21}{100}]+...+[\{r\}+\frac{91}{100}]=546[r+10019​]+[r+10020​]+[r+10021​]...+[r+10091​]=[[r]+{r}+10019​]+[[r]+{r}+10020​]+[[r]+{r}+10021​]+...+[[r]+{r}+10091​]=[r]+[{r}+10019​]+[r]+[{r}+10020​]+[r]+[{r}+10021​]+...+[r]+[{r}+10091​]=73[r]+[{r}+10019​]+[{r}+10020​]+[{r}+10021​]+...+[{r}+10091​]=546
$\because 546=73\times 7+35$
$\therefore [r]=7$
∴[{r}+19100]+[{r}+20100]+[{r}+21100]+...+[{r}+91100]=35\therefore [\{r\}+\frac{19}{100}]+[\{r\}+\frac{20}{100}]+[\{r\}+\frac{21}{100}]+...+[\{r\}+\frac{91}{100}]=35∴[{r}+10019​]+[{r}+10020​]+[{r}+10021​]+...+[{r}+10091​]=35
∵19100≤{r}+19100<2,20100≤{r}+20100<2,...,91100≤{r}+91100<2\because \frac{19}{100} \leq \{r\}+\frac{19}{100}<2, \frac{20}{100} \leq \{r\}+\frac{20}{100}<2,...,\frac{91}{100} \leq \{r\}+\frac{91}{100}<2∵10019​≤{r}+10019​<2,10020​≤{r}+10020​<2,...,10091​≤{r}+10091​<2
∴[{r}+19100]\therefore [\{r\}+\frac{19}{100}]∴[{r}+10019​] 是 $0$ 或者 $1$
$\because$ 数列是严格单调递增
$\therefore$ 有前 $73-35=38$ 项为 $0$，后 $35$ 项为 $1$。
∴43100≤{r}<44100\therefore \frac{43}{100} \leq \{r\} <\frac{44}{100}∴10043​≤{r}<10044​
∴43≤100{r}<100\therefore 43 \leq 100\{r\} <100∴43≤100{r}<100
∴[100×{r}]=43\therefore [100 \times \{r\}]=43∴[100×{r}]=43
∴[100r]=[100×([r]+{r})]=[100×(7+{r})]=700+43=743\therefore [100r]=[100\times([r]+\{r\})]=[100\times(7+\{r\})]=700+43=743∴[100r]=[100×([r]+{r})]=[100×(7+{r})]=700+43=743