Выполните деление чисел в столбик:
| \(\displaystyle -\) | \(\displaystyle 7\) | \(\displaystyle 3\) | \(\displaystyle 2\) | \(\displaystyle 6\) | ||
| \(\displaystyle -\) | ||||||
| \(\displaystyle -\) | ||||||
| \(\displaystyle 0\) |
Шаг 1.
1. Делим \(\displaystyle 7\) на \(\displaystyle 6\) с остатком.
Найдем, какое максимальное количество \(\displaystyle \color{blue}{6}\) можно забрать из \(\displaystyle 7{\small .}\)
То есть найдем неполное частное при делении \(\displaystyle 7\) на \(\displaystyle \color{blue}{6}\) с остатком.
Так как
\(\displaystyle 7=\color{green}{1} \cdot \color{blue}{6}+1 {\small ,}\)
то можно забрать \(\displaystyle \color{green}{1}\) шестерку.
Поэтому пишем \(\displaystyle \color{green}{1}\) в частное:
| \(\displaystyle -\) | \(\displaystyle \small 7\) | \(\displaystyle \small 3\) | \(\displaystyle \small 2\) | \(\displaystyle \small 6\) | ||
| \(\displaystyle \small ?\) | \(\displaystyle \small \color{green}{1}\) | \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||
| \(\displaystyle -\) | \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | ||||
| \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||||
| \(\displaystyle -\) | \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | ||||
| \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||||
| \(\displaystyle \small 0\) |
2. Далее, вычитаем в столбик из \(\displaystyle 7\) произведение \(\displaystyle \color{blue}{6}\cdot \color{green}{1}=\color{green}{6}{\small : }\)
| \(\displaystyle -\) | \(\displaystyle \small \color{orange}{ 7}\) | \(\displaystyle \small 3\) | \(\displaystyle \small 2\) | \(\displaystyle \small 6\) | ||
| \(\displaystyle \small \color{green}{6}\) | \(\displaystyle \small 1\) | \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||
| \(\displaystyle -\) | \(\displaystyle \small 1\) | \(\displaystyle \small ?\) | ||||
| \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||||
| \(\displaystyle -\) | \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | ||||
| \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||||
| \(\displaystyle \small 0\) |
3. Сносим цифру из разряда десятков \(\displaystyle 7{\underline3}2{\small : }\)
| \(\displaystyle -\) | \(\displaystyle \small 7\) | \(\displaystyle \small 3\) | \(\displaystyle \small 2\) | \(\displaystyle \small 6\) | ||
| \(\displaystyle \small 6\) | \(\displaystyle \small 1\) | \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||
| \(\displaystyle -\) | \(\displaystyle \small 1\) | \(\displaystyle \small {\bf 3}\) | ||||
| \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||||
| \(\displaystyle -\) | \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | ||||
| \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||||
| \(\displaystyle \small 0\) |
Получили число \(\displaystyle 13{\small .}\)
Шаг 2.
1. Делим \(\displaystyle 13\) на \(\displaystyle 6{\small .}\)
Найдем, какое максимальное количество \(\displaystyle \color{blue}{6}\) можно забрать из \(\displaystyle 13{\small .}\)
То есть найдем неполное частное при делении \(\displaystyle 13\) на \(\displaystyle \color{blue}{6}{\small .}\)
Так как
\(\displaystyle 13=\color{green}{2} \cdot \color{blue}{6} +1{\small ,}\)
то можно забрать \(\displaystyle \color{green}{2}\) шестерки.
Поэтому пишем \(\displaystyle \color{green}{2}\) следующей цифрой в частное:
| \(\displaystyle -\) | \(\displaystyle \small 7\) | \(\displaystyle \small 3\) | \(\displaystyle \small 2\) | \(\displaystyle \small 6\) | ||
| \(\displaystyle \small 6\) | \(\displaystyle \small 1\) | \(\displaystyle \small\color{green}{2}\) | \(\displaystyle \small ?\) | |||
| \(\displaystyle -\) | \(\displaystyle \small 1\) | \(\displaystyle \small 3\) | ||||
| \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||||
| \(\displaystyle -\) | \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | ||||
| \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||||
| \(\displaystyle \small 0\) |
2. Далее, вычитаем в столбик из \(\displaystyle 13\) произведение \(\displaystyle \color{blue}{6}\cdot \color{green}{2}=\color{green}{12}{\small : }\)
| \(\displaystyle -\) | \(\displaystyle \small 7\) | \(\displaystyle \small 3\) | \(\displaystyle \small 2\) | \(\displaystyle \small 6\) | ||
| \(\displaystyle \small 6\) | \(\displaystyle \small 1\) | \(\displaystyle \small 2\) | \(\displaystyle \small ?\) | |||
| \(\displaystyle -\) | \(\displaystyle \small \color{cyan}{1}\) | \(\displaystyle \small \color{cyan}{3}\) | ||||
| \(\displaystyle \small \color{green}{1}\) | \(\displaystyle \small \color{green}{2} \) | |||||
| \(\displaystyle -\) | \(\displaystyle \small 1\) | \(\displaystyle \small ?\) | ||||
| \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||||
| \(\displaystyle \small 0\) |
3. Сносим цифру из разряда единиц \(\displaystyle 73{\underline2}{\small : }\)
| \(\displaystyle -\) | \(\displaystyle \small 7\) | \(\displaystyle \small 3\) | \(\displaystyle \small 2\) | \(\displaystyle \small 6\) | ||
| \(\displaystyle \small 6\) | \(\displaystyle \small 1\) | \(\displaystyle \small 2\) | \(\displaystyle \small ?\) | |||
| \(\displaystyle -\) | \(\displaystyle \small 1\) | \(\displaystyle \small 3\) | ||||
| \(\displaystyle \small 1\) | \(\displaystyle \small 2\) | |||||
| \(\displaystyle -\) | \(\displaystyle \small 1\) | \(\displaystyle \small {\bf 2}\) | ||||
| \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||||
| \(\displaystyle \small 0\) |
Получили число \(\displaystyle 12{\small .}\)
Шаг 3.
1. Делим \(\displaystyle 12\) на \(\displaystyle 6{\small .}\)
Найдем, какое максимальное количество \(\displaystyle \color{blue}{6}\) можно забрать из \(\displaystyle 12{\small .}\)
То есть найдем частное при делении \(\displaystyle 12\) на \(\displaystyle \color{blue}{6}{\small .}\)
Так как
\(\displaystyle 12=\color{green}{2} \cdot \color{blue}{6}{\small ,}\)
то можно забрать \(\displaystyle \color{green}{2}\) шестерки.
Поэтому пишем \(\displaystyle \color{green}{2}\) следующей цифрой в частное:
| \(\displaystyle -\) | \(\displaystyle \small 7\) | \(\displaystyle \small 3\) | \(\displaystyle \small 2\) | \(\displaystyle \small 6\) | ||
| \(\displaystyle \small 6\) | \(\displaystyle \small 1\) | \(\displaystyle \small 2\) | \(\displaystyle \small \color{green}{ 2}\) | |||
| \(\displaystyle -\) | \(\displaystyle \small 1\) | \(\displaystyle \small 3\) | ||||
| \(\displaystyle \small 1\) | \(\displaystyle \small 2\) | |||||
| \(\displaystyle -\) | \(\displaystyle \small 1\) | \(\displaystyle \small 2\) | ||||
| \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||||
| \(\displaystyle \small 0\) |
2. Далее, вычитаем в столбик из \(\displaystyle 12\) произведение \(\displaystyle \color{blue}{6}\cdot \color{green}{2}=\color{green}{12}{\small : }\)
| \(\displaystyle -\) | \(\displaystyle \small 7\) | \(\displaystyle \small 3\) | \(\displaystyle \small 2\) | \(\displaystyle \small 6\) | ||
| \(\displaystyle \small 6\) | \(\displaystyle \small 1\) | \(\displaystyle \small 2\) | \(\displaystyle \small \color{green}{ 2}\) | |||
| \(\displaystyle -\) | \(\displaystyle \small 1\) | \(\displaystyle \small 3\) | ||||
| \(\displaystyle \small 1\) | \(\displaystyle \small 2\) | |||||
| \(\displaystyle -\) | \(\displaystyle \small \color{red}{ 1}\) | \(\displaystyle \small \color{red}{ 2}\) | ||||
| \(\displaystyle \small \color{green}{1}\) | \(\displaystyle \small \color{green}{2}\) | |||||
| \(\displaystyle \small 0\) |
Получили \(\displaystyle 0{\small , }\) процес деления закончен.
Таким образом,
| \(\displaystyle -\) | \(\displaystyle \small \color{orange}{7}\) | \(\displaystyle \small 3\) | \(\displaystyle \small 2\) | \(\displaystyle \small 6\) | ||
| \(\displaystyle \small 6\) | \(\displaystyle \small 1\) | \(\displaystyle \small 2\) | \(\displaystyle \small 2\) | |||
| \(\displaystyle -\) | \(\displaystyle \small \color{cyan}{1}\) | \(\displaystyle \small \color{cyan}{3}\) | ||||
| \(\displaystyle \small 1\) | \(\displaystyle \small 2\) | |||||
| \(\displaystyle -\) | \(\displaystyle \small \color{red}{ 1}\) | \(\displaystyle \small \color{red}{ 2}\) | ||||
| \(\displaystyle \small 1\) | \(\displaystyle \small 2\) | |||||
| \(\displaystyle \small 0\) |