## 2016年10月30日日曜日

### 数学 – 代数学 - 整数 - 1次の合同式(連立合同式の解法(順に解いていく方法))

$\begin{array}{l}x=3+8t\\ 3+8t\equiv 1\left(\mathrm{mod}\text{\hspace{0.17em}}15\right)\\ 8t\equiv -2\left(\mathrm{mod}\text{\hspace{0.17em}}15\right)\\ 7t\equiv 2\left(\mathrm{mod}\text{\hspace{0.17em}}15\right)\\ 2+15t\text{'}\equiv 0\left(\mathrm{mod}\text{\hspace{0.17em}}7\right)\\ 15t\text{'}\equiv -2\left(\mathrm{mod}\text{\hspace{0.17em}}7\right)\\ t\text{'}\equiv -2\left(\mathrm{mod}\text{\hspace{0.17em}}7\right)\\ 7t\equiv 2-2·15\left(\mathrm{mod}\text{\hspace{0.17em}}15\right)\\ 7t\equiv -28\left(\mathrm{mod}\text{\hspace{0.17em}}15\right)\\ t\equiv -4\left(\mathrm{mod}\text{\hspace{0.17em}}15\right)\\ t=-4+15t\text{'}\text{'}\\ x=3+8\left(-4+15t\text{'}\text{'}\right)\\ =-29+120t\text{'}\text{'}\\ -29+120t\text{'}\text{'}\equiv 11\left(\mathrm{mod}\text{\hspace{0.17em}}20\right)\\ 120t\text{'}\text{'}\equiv 40\left(\mathrm{mod}\text{\hspace{0.17em}}20\right)\\ 40\left(3t\text{'}\text{'}-1\right)\equiv 0\left(\mathrm{mod}\text{\hspace{0.17em}}20\right)\\ x=-29+120t\text{'}\text{'}\\ =91+120t\text{'}\text{'}\text{'}\\ x\equiv 91\left(\mathrm{mod}\text{\hspace{0.17em}}120\right)\end{array}$