求解取整方程的x值-优快云博客

本文链接：https://blog.youkuaiyun.com/justidle/article/details/129206761

求满足 $[x2]+[x+13]+[x+25]=x[\frac{x}{2}]+[\frac{x+1}{3}]+[\frac{x+2}{5}]=x$ 的所有 $x$ 值之和。

$∵\because$ gcd(2,3,5)=30
$−15≤r≤14\therefore x=30k+r,\ k \in \mathbb{Z},\ -15 \leq r \leq 14$
带入原方程
$[x2]+[x+13]+[x+25]=[30k+r2]+[30k+r+13]+[30k+r+25]=[30k2+r2]+[30k3+r+13]+[30k5+r+25]=[15k+r2]+[10k+r+13]+[6k+r+25]=15k+[r2]+10k+[r+13]+6k+[r+25]=31k+[r2]+[r+13]+[r+25]=30k+r[\frac{x}{2}]+[\frac{x+1}{3}]+[\frac{x+2}{5}]\\ =[\frac{30k+r}{2}]+[\frac{30k+r+1}{3}]+[\frac{30k+r+2}{5}]\\ =[\frac{30k}{2}+\frac{r}{2}]+[\frac{30k}{3}+\frac{r+1}{3}]+[\frac{30k}{5}+\frac{r+2}{5}]\\ =[15k+\frac{r}{2}]+[10k+\frac{r+1}{3}]+[6k+\frac{r+2}{5}]\\ =15k+[\frac{r}{2}]+10k+[\frac{r+1}{3}]+6k+[\frac{r+2}{5}]\\ =31k+[\frac{r}{2}]+[\frac{r+1}{3}]+[\frac{r+2}{5}]\\ =30k+r$
$∴[r2]+[r+13]+[r+25]=r−k\therefore [\frac{r}{2}]+[\frac{r+1}{3}]+[\frac{r+2}{5}]=r-k$
$∴r2+r+13+r+25−3<r−k≤r2+r+13+r+25\therefore \frac{r}{2}+\frac{r+1}{3}+\frac{r+2}{5}-3<r-k \leq \frac{r}{2}+\frac{r+1}{3}+\frac{r+2}{5}$
$∴r2+r+13+r+25−r−3<−k≤r2+r+13+r+25−r\therefore \frac{r}{2}+\frac{r+1}{3}+\frac{r+2}{5}-r-3<-k \leq \frac{r}{2}+\frac{r+1}{3}+\frac{r+2}{5}-r$
$∴15r+10r+10+6r+12−30r30−3<−k≤15r+10r+10+6r+12−30r30\therefore \frac{15r+10r+10+6r+12-30r}{30}-3<-k \leq \frac{15r+10r+10+6r+12-30r}{30}$
$∴r−6830<−k≤r+2230\therefore \frac{r-68}{30}<-k \leq \frac{r+22}{30}$
$−15≤r≤14\therefore \frac{68-r}{30}>k \ge \frac{-r-22}{30},\ -15\leq r \leq 14$
$−15≤r≤14\therefore 68-r>30\times k \ge -r-22,\ -15\leq r \leq 14$
$15≥−r≥−14\therefore 68-r>30\times k \ge -r-22,\ 15\ge -r \ge -14$
$∴68−15>30×k≥−14−22\therefore 68-15>30\times k \ge -14-22$
$∴53>30×k≥−36\therefore 53>30\times k \ge -36$
$∴1.8>k≥−1.2\therefore 1.8>k \ge -1.2$
$∴k=1,0,−1\therefore k=1,0,-1$

当 $k = - 1$ 时候， $∴68−r>−30≥−r−22\therefore 68-r>-30 \ge -r-22$
$\rightarrow -r>-98 \rightarrow r<98$
$\ge -r-22 \rightarrow -8 \ge -r \rightarrow 8 \leq r$
$∴8≤r<98\therefore 8 \leq r < 98$
$∴8≤r≤14\therefore 8 \leq r \leq 14$
验算 $[r2]+[r+13]+[r+25]=r−k[\frac{r}{2}]+[\frac{r+1}{3}]+[\frac{r+2}{5}]=r-k$ 得 $r = 8, 14$ 有解。所以 $x = 30 k + r = - 30 + r = - 22, - 16$ 。
当 $k = 0$ 时候，
$∴68−r>0≥−r−22\therefore 68-r>0 \ge -r-22$
$∴−15≤r≤14\therefore -15 \leq r \leq 14$
验算得有解 $x = - 12, - 10, - 7, - 6, - 4, - 2, - 1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 13$ 。
当 $k = 1$ 时候，
$∴68−r>30≥−r−22\therefore 68-r>30 \ge -r-22$
$∴−15≤r≤14\therefore -15 \leq r \leq 14$
验算得有解 $x = 15, 16, 17, 19, 21, 22, 25, 27, 31, 37$ 。

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

$∑x=225\sum x=225$

已知 $0 < a < 1$ ，且满足 $[a+130]+[a+230]+...+[a+2930]=18[a+\frac{1}{30}]+[a+\frac{2}{30}]+...+[a+\frac{29}{30}]=18$ ，求 $[10 a]$ 的值。

$∵0<a<1\because 0<a<1$
$∴0+130<a+130<1+130\therefore 0+\frac{1}{30}<a+\frac{1}{30}<1+\frac{1}{30}$
$∴0<a+130<2\therefore 0<a+\frac{1}{30}<2$
$∴0<a+130<a+230<...<a+2930<2\therefore 0<a+\frac{1}{30}<a+\frac{2}{30}<...<a+\frac{29}{30}<2$
$∴0<[a+130]<2,0<[a+230]<2,...,0<[a+2930]<2\therefore 0<[a+\frac{1}{30}]<2,\\ 0<[a+\frac{2}{30}]<2,\\ ...,\\ 0<[a+\frac{29}{30}]<2$
即 $x∈[1,29][a+\frac{x}{30}], \ x \in [1,29]$ 一定是 0 或者 1.
根据题意，其中 $18$ 个等于 $1$ 。
$∴[a+130]=[a+230]=...=[a+1130]=0[a+1230]=[a+1330]=...=[a+2930]=1\therefore [a+\frac{1}{30}]=[a+\frac{2}{30}]=...=[a+\frac{11}{30}]=0\\ [a+\frac{12}{30}]=[a+\frac{13}{30}]=...=[a+\frac{29}{30}]=1$
$∴0<a+1130<1,1≤a+1230<2\therefore 0 < a+\frac{11}{30}<1, 1\leq a+\frac{12}{30}<2$
$∴18≤30a<19\therefore 18 \leq 30a < 19$
$∴6≤10a<193\therefore 6 \leq 10a < \frac{19}{3}$
$∴[10a]=6\therefore [10a] =6$

设 $r$ 满足 $[r+19100]+[r+20100]+[r+21100]...+[r+91100]=546[r+\frac{19}{100}]+[r+\frac{20}{100}]+[r+\frac{21}{100}]...+[r+\frac{91}{100}]=546$ ，求 $[100 r]$ 的值

解：
$[r+19100]+[r+20100]+[r+21100]...+[r+91100]=[[r]+{r}+19100]+[[r]+{r}+20100]+[[r]+{r}+21100]+...+[[r]+{r}+91100]=[r]+[{r}+19100]+[r]+[{r}+20100]+[r]+[{r}+21100]+...+[r]+[{r}+91100]=73[r]+[{r}+19100]+[{r}+20100]+[{r}+21100]+...+[{r}+91100]=546[r+\frac{19}{100}]+[r+\frac{20}{100}]+[r+\frac{21}{100}]...+[r+\frac{91}{100}]\\ =[[r]+\{r\}+\frac{19}{100}]+[[r]+\{r\}+\frac{20}{100}]+[[r]+\{r\}+\frac{21}{100}]+...+[[r]+\{r\}+\frac{91}{100}]\\ =[r]+[\{r\}+\frac{19}{100}]+[r]+[\{r\}+\frac{20}{100}]+[r]+[\{r\}+\frac{21}{100}]+...+[r]+[\{r\}+\frac{91}{100}]\\ =73[r]+[\{r\}+\frac{19}{100}]+[\{r\}+\frac{20}{100}]+[\{r\}+\frac{21}{100}]+...+[\{r\}+\frac{91}{100}]=546$
$∵546=73×7+35\because 546=73 \times 7+35$
$∴[r]=7\therefore [r]=7$
$∴[{r}+19100]+[{r}+20100]+[{r}+21100]+...+[{r}+91100]=35\therefore [\{r\}+\frac{19}{100}]+[\{r\}+\frac{20}{100}]+[\{r\}+\frac{21}{100}]+...+[\{r\}+\frac{91}{100}]=35$
$∵19100≤{r}+19100<2,20100≤{r}+20100<2,...,91100≤{r}+91100<2\because \frac{19}{100} \leq \{r\}+\frac{19}{100}<2, \frac{20}{100} \leq \{r\}+\frac{20}{100}<2,...,\frac{91}{100} \leq \{r\}+\frac{91}{100}<2$
$∴[{r}+19100]\therefore [\{r\}+\frac{19}{100}]$ 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 $1$ 。
$∴43100≤{r}<44100\therefore \frac{43}{100} \leq \{r\} <\frac{44}{100}$
$∴43≤100{r}<100\therefore 43 \leq 100\{r\} <100$
$∴[100×{r}]=43\therefore [100 \times \{r\}]=43$
$∴[100r]=[100×([r]+{r})]=[100×(7+{r})]=700+43=743\therefore [100r]=[100\times([r]+\{r\})]=[100\times(7+\{r\})]=700+43=743$