求满足 [x2]+[x+13]+[x+25]=x[\\frac{x}{2}]+[\\frac{x+1}{3}]+[\\frac{x+2}{5}]=x[2x​]+[3x+1​]+[5x+2​]=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
[x2]+[x+13]+[x+25]=[30k+r2]+[30k+r+13]+[30k+r+25]=[30k2+r2]+[30k3+r+13]+[30k5+r+25]=[15k+r2]+[10k+r+13]+[6k+r+25]=15k+[r2]+10k+[r+13]+6k+[r+25]=31k+[r2]+[r+13]+[r+25]=30k+r[\\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[2x​]+[3x+1​]+[5x+2​]=[230k+r​]+[330k+r+1​]+[530k+r+2​]=[230k​+2r​]+[330k​+3r+1​]+[530k​+5r+2​]=[15k+2r​]+[10k+3r+1​]+[6k+5r+2​]=15k+[2r​]+10k+[3r+1​]+6k+[5r+2​]=31k+[2r​]+[3r+1​]+[5r+2​]=30k+r
∴[r2]+[r+13]+[r+25]=r−k\\therefore [\\frac{r}{2}]+[\\frac{r+1}{3}]+[\\frac{r+2}{5}]=r-k∴[2r​]+[3r+1​]+[5r+2​]=r−k
∴r2+r+13+r+25−3<r−k≤r2+r+13+r+25\\therefore \\frac{r}{2}+\\frac{r+1}{3}+\\frac{r+2}{5}-3<r-k \\leq \\frac{r}{2}+\\frac{r+1}{3}+\\frac{r+2}{5}∴2r​+3r+1​+5r+2​−3<r−k≤2r​+3r+1​+5r+2​
∴r2+r+13+r+25−r−3<−k≤r2+r+13+r+25−r\\therefore \\frac{r}{2}+\\frac{r+1}{3}+\\frac{r+2}{5}-r-3<-k \\leq \\frac{r}{2}+\\frac{r+1}{3}+\\frac{r+2}{5}-r∴2r​+3r+1​+5r+2​−r−3<−k≤2r​+3r+1​+5r+2​−r
∴15r+10r+10+6r+12−30r30−3<−k≤15r+10r+10+6r+12−30r30\\therefore \\frac{15r+10r+10+6r+12-30r}{30}-3<-k \\leq \\frac{15r+10r+10+6r+12-30r}{30}∴3015r+10r+10+6r+12−30r​−3<−k≤3015r+10r+10+6r+12−30r​
∴r−6830<−k≤r+2230\\therefore \\frac{r-68}{30}<-k \\leq \\frac{r+22}{30}∴30r−68​<−k≤30r+22​
∴68−r30>k≥−r−2230,−15≤r≤14\\therefore \\frac{68-r}{30}>k \\ge \\frac{-r-22}{30},\\ -15\\leq r \\leq 14∴3068−r​>k≥30−r−22​, −15≤r≤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 [r2]+[r+13]+[r+25]=r−k[\\frac{r}{2}]+[\\frac{r+1}{3}]+[\\frac{r+2}{5}]=r-k[2r​]+[3r+1​]+[5r+2​]=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+130]+[a+230]+...+[a+2930]=18[a+\\frac{1}{30}]+[a+\\frac{2}{30}]+...+[a+\\frac{29}{30}]=18[a+301​]+[a+302​]+...+[a+3029​]=18，求 $[10a]$ 的值。

$\because 0<a<1$
∴0+130<a+130<1+130\\therefore 0+\\frac{1}{30}<a+\\frac{1}{30}<1+\\frac{1}{30}∴0+301​<a+301​<1+301​
∴0<a+130<2\\therefore 0<a+\\frac{1}{30}<2∴0<a+301​<2
∴0<a+130<a+230<...<a+2930<2\\therefore 0<a+\\frac{1}{30}<a+\\frac{2}{30}<...<a+\\frac{29}{30}<2∴0<a+301​<a+302​<...<a+3029​<2
∴0<[a+130]<2,0<[a+230]<2,...,0<[a+2930]<2\\therefore 0<[a+\\frac{1}{30}]<2,\\\\ 0<[a+\\frac{2}{30}]<2,\\\\ ...,\\\\ 0<[a+\\frac{29}{30}]<2∴0<[a+301​]<2,0<[a+302​]<2,...,0<[a+3029​]<2
即 [a+x30],x∈[1,29][a+\\frac{x}{30}], \\ x \\in [1,29][a+30x​], x∈[1,29] 一定是 0 或者 1.
根据题意，其中 $18$ 个等于 $1$。
∴[a+130]=[a+230]=...=[a+1130]=0[a+1230]=[a+1330]=...=[a+2930]=1\\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∴[a+301​]=[a+302​]=...=[a+3011​]=0[a+3012​]=[a+3013​]=...=[a+3029​]=1
∴0<a+1130<1,1≤a+1230<2\\therefore 0 < a+\\frac{11}{30}<1, 1\\leq a+\\frac{12}{30}<2∴0<a+3011​<1,1≤a+3012​<2
$\therefore 18\le 30a<19$
∴6≤10a<193\\therefore 6 \\leq 10a < \\frac{19}{3}∴6≤10a<319​
$\therefore [10a]=6$

设 $r$ 满足 [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}]=546[r+10019​]+[r+10020​]+[r+10021​]...+[r+10091​]=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