數與座標系/1 數的性質與餘數問題-範例

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範例1

已知

  • a\in N，\frac{5a+12}{2a-3}\in N

求解

  • a\,

答案

  • \left\{ 2,3,8,21 \right\}

詳解

  1. 因為\frac{5a+12}{2a-3}\in N，則2a-3\,是5a+12\,的因數；且2a-3\,也必是自己的因數。所以：
  2. \left\{ \begin{align} & 2a-3|5a+12 \\ & 2a-3|2a-3 \\ \end{align} \right.\Rightarrow 2a-3|2\left( 5a+12 \right)-5\left( 2a-3 \right)，提示：因數線性組合
    \Rightarrow 2a-3|\left( 10a+24 \right)-\left( 10a-15 \right)\Rightarrow 2a-3|24+15\Rightarrow 2a-3|39\Rightarrow \left\{ \begin{align} & 2a-3=1 \\ & 2a-3=3 \\ & 2a-3=13 \\ & 2a-3=39 \\ \end{align} \right.\Rightarrow \left\{ \begin{align} & a=2 \\ & a=3 \\ & a=8 \\ & a=21 \\ \end{align} \right.


範例1類題1

已知

求解

答案

詳解

範例1類題2

已知：a\in Z\,且a|a+8,a-1|a+11\,

求解：a\,

答案

詳解

詳解


已知

  • a\in Z\,且a|a+8,a-1|a+11\,

求解

  • a\,

答案

  • a=-2,-1,2,4\,

詳解

  1. \left\{ \begin{align} & a|a+8 \\ & a|a \\ \end{align} \right.\Rightarrow a|\left( a+8 \right)-\left( a \right)\Rightarrow a|8\Rightarrow \left\{ \begin{align} & a=\pm 1 \\ & a=\pm 2 \\ & a=\pm 4 \\ & a=\pm 8 \\ \end{align} \right.，提示：因數線性組合
  2. \left\{ \begin{align} & \left\{ \begin{align} & a=1 \\ & a-1|a+11 \\ \end{align} \right.\Rightarrow 0|12\left( N/A \right) \\ & \left\{ \begin{align} & a=-1\# \\ & a-1|a+11 \\ \end{align} \right.\Rightarrow -2|10 \\ & \left\{ \begin{align} & a=2\# \\ & a-1|a+11 \\ \end{align} \right.\Rightarrow 1|13 \\ & \left\{ \begin{align} & a=-2\# \\ & a-1|a+11 \\ \end{align} \right.\Rightarrow -3|9 \\ & \left\{ \begin{align} & a=4\# \\ & a-1|a+11 \\ \end{align} \right.\Rightarrow 3|15 \\ & \left\{ \begin{align} & a=-4 \\ & a-1|a+11 \\ \end{align} \right.\Rightarrow -5|7\left( N/A \right) \\ & \left\{ \begin{align} & a=8 \\ & a-1|a+11 \\ \end{align} \right.\Rightarrow 7|19\left( N/A \right) \\ & \left\{ \begin{align} & a=-8 \\ & a-1|a+11 \\ \end{align} \right.\Rightarrow -9|3\left( N/A \right) \\ \end{align} \right.


範例2

已知

  1. a\in N,b\in Z,ab-3a+2b=0
  2. n\in N,p=2n^{2}-19n+35，p\, 為質數

求解

  1. \left( a,b \right)
  2. \left( n,p \right)

答案

  1. \left( a,b \right)=\left( 1,1 \right)或\left( 4,2 \right)
  2. \left( n,p \right)=\left( 2,5 \right)或\left( 8,11 \right)

詳解

  1. 解\left( a,b \right)
    1. ab-3a+2b=0\Rightarrow ab-3a+2b+\left( -3\times 2 \right)=0+\left( -3\times 2 \right)，提示：因式分解配方
      \Rightarrow \left( a+2 \right)\left( b-3 \right)=-6，提示：因式分解
    2. \Rightarrow \left\{ \begin{align} & \left( a+2 \right)=-6,-3,-2,-1,+1,+2,+3,+6 \\ & \left( b-3 \right)=+1,+2,+3,+6,-6,-3,-2,-1 \\ \end{align} \right.，提示：因數分解
    3. \Rightarrow \left\{ \begin{align} & a=-8,-5,-4,-3,-1,0,1,4 \\ & b=+4,+5,+6,+9,-3,0,1,2 \\ \end{align} \right.，提示：一元一次方程式
    4. \because a\in N\therefore \left\{ \begin{align} & a=1,4 \\ & b=1,2 \\ \end{align} \right.\Rightarrow \left( a,b \right)=\left( 1,1 \right)或\left( 4,2 \right)\#
  2. 解\left( n,p \right)
    1. p=2n^{2}-19n+35\Rightarrow p=\left( 2n-5 \right)\left( n-7 \right)，提示：因式分解
    2. \Rightarrow \left( 2n-5 \right)\left( n-7 \right)=\left( \pm 1 \right)\cdot \left( \pm p \right)=p，提示：質數定義
    3. \left\{ \begin{align} & \left\{ \begin{align} & \left( 2n-5 \right)=1 \\ & \left( n-7 \right)=p \\ \end{align} \right.\Rightarrow \left\{ \begin{align} & n=3 \\ & p=-4 \\ \end{align} \right.\left( N/A \right) \\ & \left\{ \begin{align} & \left( 2n-5 \right)=-1 \\ & \left( n-7 \right)=-p \\ \end{align} \right.\Rightarrow \left\{ \begin{align} & n=2 \\ & p=5 \\ \end{align} \right.\Rightarrow \left( n,p \right)=\left( 2,5 \right)\# \\ & \left\{ \begin{align} & \left( n-7 \right)=1 \\ & \left( 2n-5 \right)=p \\ \end{align} \right.\Rightarrow \left\{ \begin{align} & n=8 \\ & p=11 \\ \end{align} \right.\Rightarrow \left( n,p \right)=\left( 8,11 \right)\# \\ & \left\{ \begin{align} & \left( n-7 \right)=-1 \\ & \left( 2n-5 \right)=-p \\ \end{align} \right.\Rightarrow \left\{ \begin{align} & n=6 \\ & p=-7 \\ \end{align} \right.\left( N/A \right) \\ \end{align} \right.

範例2類題1

已知

求解

答案

詳解

範例2類題2

已知

求解

答案

詳解


範例3

已知

求解

答案

詳解

  1. \left\{ \begin{align} & \frac{n}{3}=q_{1}+\frac{2}{3} \\ & \frac{n}{5}=q_{2}+\frac{3}{5} \\ & \frac{n}{7}=q_{3}+\frac{4}{7} \\ \end{align} \right.\Rightarrow \left\{ \begin{align} & n=3q_{1}+2 \\ & n=5q_{2}+3 \\ & n=7q_{3}+4 \\ \end{align} \right.
    \Rightarrow \left\{ \begin{align} & 5q_{2}+3=3q_{1}+2 \\ & 7q_{3}+4=3q_{1}+2 \\ \end{align} \right.\Rightarrow \left\{ \begin{align} & 3q_{1}-1=5q_{2} \\ & 3q_{1}-2=7q_{3} \\ \end{align} \right.
  2. \Rightarrow \left\{ \begin{align} & q_{1}=2,\left( 2+5\times 1 \right),\left( 2+5\times 2 \right),\left( 2+5\times 3 \right),\cdots \Rightarrow q_{2}\in N \\ & q_{1}=3,\left( 3+7\times 1 \right),\left( 3+7\times 2 \right),\cdots \Rightarrow q_{3}\in N \\ \end{align} \right.
    \Rightarrow \left\{ \begin{align} & q_{1}=2,7,12,{\color{Red}17} ,\cdots \Rightarrow q_{2}\in N \\ & q_{1}=3,10,{\color{Red}17} ,\cdots \Rightarrow q_{3}\in N \\ \end{align} \right.
    \left\{ \begin{align} & q_{1}=17\Rightarrow {\color{Red}n_{1}=3\times 17+2=53} \\ & n_{1}=5\times q_{2}+3=53\Rightarrow q_{2}=10 \\ & n_{1}=7\times q_{3}+4=53\Rightarrow q_{3}=7 \\ \end{align} \right.
  3. \left\{ \begin{align} & n_{2}=1\times \left[ 3,5,7 \right]+53=\frac{3\times 1\times \left[ 3,5,7 \right]}{3}+53=3\times 35+53,q_{1}=35 \\ & n_{2}=1\times \left[ 3,5,7 \right]+53=\frac{5\times 1\times \left[ 3,5,7 \right]}{5}+53=5\times 21+53,q_{2}=21 \\ & n_{2}=1\times \left[ 3,5,7 \right]+53=\frac{7\times 1\times \left[ 3,5,7 \right]}{7}+53=7\times 15+53,q_{3}=15 \\ \end{align} \right.
  4. n_{k+1}=k\times \left[ 3,5,7 \right]+53=105k+53,k\in N\cup \left\{ 0 \right\}\#
  5. R_{105}\left( n \right)=53\#
  6. \min \left( n \right)=53\#
  7. \left\{ \begin{matrix} \begin{align} & k=3\Rightarrow n=368\left( N/A \right) \\ & k=4\Rightarrow n=473 \\ \end{align} \\ k=5\Rightarrow n=578 \\ k=6\Rightarrow n=683 \\ k=7\Rightarrow n=788 \\ k=8\Rightarrow n=893 \\ \begin{align} & k=9\Rightarrow n=998 \\ & k=10\Rightarrow n=1103\left( N/A \right) \\ \end{align} \\ \end{matrix} \right.\Rightarrow 6\#

範例3類題1

已知

求解

答案

詳解

範例3類題2

已知

求解

答案

詳解

範例3類題3

已知

求解

答案

詳解


範例4

已知

求解

答案

詳解

  1. \begin{align} & x,y,z\in Z \\ & \left| x-3 \right|+2\left| y-2 \right|+3\left| z+1 \right|=2 \\ \end{align}
  2. \left\{ \begin{align} & \left| x-3 \right|=0 \\ & \left| y-2 \right|=1 \\ & \left| z+1 \right|=0 \\ \end{align} \right.\Rightarrow \left\{ \begin{align} & x=3 \\ & y=3,1 \\ & z=-1 \\ \end{align} \right.\Rightarrow \left( x,y,z \right)=\left( 3,3,-1 \right)or\left( 3,1,-1 \right)\#
    \left\{ \begin{align} & \left| x-3 \right|=2 \\ & \left| y-2 \right|=0 \\ & \left| z+1 \right|=0 \\ \end{align} \right.\Rightarrow \left\{ \begin{align} & x=5,1 \\ & y=2 \\ & z=-1 \\ \end{align} \right.\Rightarrow \left( x,y,z \right)=\left( 5,2,-1 \right)or\left( 1,2,-1 \right)\#


範例4類題

已知

求解

答案

詳解

範例5

已知

求解

答案

詳解

  1. \left\{ \begin{align} & R_{5}\left( 53 \right)=3\Rightarrow R_{5}\left( 53^{4} \right)=R_{5}\left( 3^{4} \right)=1 \\ & R_{5}\left( 981 \right)=1\Rightarrow R_{5}\left( 981^{6} \right)=R_{5}\left( 1^{6} \right)=1 \\ & R_{5}\left( 372 \right)=2\Rightarrow R_{5}\left( 372^{3} \right)=R_{5}\left( 2^{3} \right)=3 \\ & R_{5}\left( 6758 \right)=3\Rightarrow R_{5}\left( 6758^{2} \right)=R_{5}\left( 3^{2} \right)=4 \\ \end{align} \right.
  2. \begin{align} & R_{5}\left( 53^{4}+981^{6}+372^{3}\times 6758^{2} \right)=R_{5}\left[ R_{5}\left( 53^{4} \right)+R_{5}\left( 981^{6} \right)+R_{5}\left( 372^{3}\times 6758^{2} \right) \right] \\ & =R_{5}\left[ R_{5}\left( 53^{4} \right)+R_{5}\left( 981^{6} \right)+R_{5}\left[ R_{5}\left( 372^{3} \right)\times R_{5}\left( 6758^{2} \right) \right] \right] \\ & =R_{5}\left\{ R_{5}\left( 3^{4} \right)+R_{5}\left( 1^{6} \right)+R_{5}\left[ R_{5}\left( 2^{3} \right)\times R_{5}\left( 3^{2} \right) \right] \right\} \\ & =R_{5}\left( 1+1+R_{5}\left( 3\times 4 \right) \right)=R_{5}\left( 1+1+2 \right)=4\# \\ \end{align}
  3. \begin{align} & \left\{ \begin{align} & x,y\in N \\ & R_{11}\left( x \right)=3 \\ & R_{11}\left( y \right)=7 \\ \end{align} \right. \\ & R_{11}\left( xy+2y \right)=R_{11}\left( 3\times 7+2\times 7 \right)=R_{11}\left( 35 \right)=2\# \\ & R_{11}\left( x^{2}-y^{2}+3 \right)=R_{11}\left( 3^{2}-7^{2}+3 \right)=R_{11}\left( -37 \right)=R_{11}\left( -4\times 11+7 \right)=7\# \\ \end{align}

範例5類題1

已知

求解

答案

詳解

範例5類題2

已知

求解

答案

詳解

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