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 xxx 值之和。
∵\because∵ gcd(2,3,5)=30
∴x=30k+r, k∈Z, −15≤r≤14\therefore x=30k+r,\ k \in \mathbb{Z},\ -15 \leq r \leq 14∴x=30k+r, k∈Z, −15≤r≤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
∴68−r>30×k≥−r−22, −15≤r≤14\therefore 68-r>30\times k \ge -r-22,\ -15\leq r \leq 14∴68−r>30×k≥−r−22, −15≤r≤14
∴68−r>30×k≥−r−22, 15≥−r≥−14\therefore 68-r>30\times k \ge -r-22,\ 15\ge -r \ge -14∴68−r>30×k≥−r−22, 15≥−r≥−14
∴68−15>30×k≥−14−22\therefore 68-15>30\times k \ge -14-22∴68−15>30×k≥−14−22
∴53>30×k≥−36\therefore 53>30\times k \ge -36∴53>30×k≥−36
∴1.8>k≥−1.2\therefore 1.8>k \ge -1.2∴1.8>k≥−1.2
∴k=1,0,−1\therefore k=1,0,-1∴k=1,0,−1
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当 k=−1k=-1k=−1 时候,∴68−r>−30≥−r−22\therefore 68-r>-30 \ge -r-22∴68−r>−30≥−r−22
68−r>−30→−r>−98→r<9868-r>-30 \rightarrow -r>-98 \rightarrow r<9868−r>−30→−r>−98→r<98
−30≥−r−22→−8≥−r→8≤r-30 \ge -r-22 \rightarrow -8 \ge -r \rightarrow 8 \leq r−30≥−r−22→−8≥−r→8≤r
∴8≤r<98\therefore 8 \leq r < 98∴8≤r<98
∴8≤r≤14\therefore 8 \leq r \leq 14∴8≤r≤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=8,14r=8,14r=8,14 有解。所以 x=30k+r=−30+r=−22,−16x=30k+r=-30+r=-22,-16x=30k+r=−30+r=−22,−16。 -
当 k=0k=0k=0 时候,
∴68−r>0≥−r−22\therefore 68-r>0 \ge -r-22∴68−r>0≥−r−22
∴−15≤r≤14\therefore -15 \leq r \leq 14∴−15≤r≤14
验算得有解 x=−12,−10,−7,−6,−4,−2,−1,2,3,4,5,6,9,10,11,12,13x=-12,-10,-7,-6,-4,-2,-1,2,3,4,5,6,9,10,11,12,13x=−12,−10,−7,−6,−4,−2,−1,2,3,4,5,6,9,10,11,12,13。 -
当 k=1k=1k=1 时候,
∴68−r>30≥−r−22\therefore 68-r>30 \ge -r-22∴68−r>30≥−r−22
∴−15≤r≤14\therefore -15 \leq r \leq 14∴−15≤r≤14
验算得有解 x=15,16,17,19,21,22,25,27,31,37x=15,16,17,19,21,22,25,27,31,37x=15,16,17,19,21,22,25,27,31,37。
全部得 x=−22,−16,−10,−7,−6,−4,−2,−1,2,3,4,5,6,9,10,11,12,13,15,16,17,19,21,22,2527,31,37x=-22,-16,-10,-7,-6,-4,-2,-1,2,3,4,5,6,9,10,11,12,13,15,16,17,19,21,22,2527,31,37x=−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∑x=225
已知 0<a<10<a<10<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][10a][10a] 的值。
∵0<a<1\because 0<a<1∵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 181818 个等于 111。
∴[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
∴18≤30a<19\therefore 18 \leq 30a < 19∴18≤30a<19
∴6≤10a<193\therefore 6 \leq 10a < \frac{19}{3}∴6≤10a<319
∴[10a]=6\therefore [10a] =6∴[10a]=6
设 rrr 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][100r][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
∵546=73×7+35\because 546=73 \times 7+35∵546=73×7+35
∴[r]=7\therefore [r]=7∴[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 000 或者 111
∵\because∵ 数列是严格单调递增
∴\therefore∴ 有前 73−35=3873-35=3873−35=38 项为 000,后 353535 项为 111。
∴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