Разделите в столбик:
| \(\displaystyle -\) | \(\displaystyle 5\) | \(\displaystyle 1\) | \(\displaystyle 3\) | |
| \(\displaystyle -\) | ||||
| \(\displaystyle 0\) |
Шаг 1.
1. Делим \(\displaystyle 5\) на \(\displaystyle 3\) с остатком.
Найдем, какое максимальное количество \(\displaystyle \color{blue}{3}\) можно забрать из \(\displaystyle 5{\small .}\)
То есть найдем неполное частное при делении \(\displaystyle 5\) на \(\displaystyle \color{blue}{3}\) с остатком.
Так как
\(\displaystyle 5=\color{green}{1} \cdot \color{blue}{3}+2 {\small ,}\)
то можно забрать \(\displaystyle \color{green}{1}\) (одну) тройку.
Поэтому пишем \(\displaystyle \color{green}{1}\) в частное:
| \(\displaystyle -\) | \(\displaystyle \small 5\) | \(\displaystyle \small 1\) | \(\displaystyle \small 3\) | |
| \(\displaystyle \small ?\) | \(\displaystyle \small \color{green}{1}\) | \(\displaystyle \small ?\) | ||
| \(\displaystyle -\) | \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | ||
| \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||
| \(\displaystyle \small 0\) |
2. Далее, вычитаем в столбик из \(\displaystyle 5\) произведение \(\displaystyle \color{blue}{3}\cdot \color{green}{1}=\color{green}{3}{\small : }\)
| \(\displaystyle -\) | \(\displaystyle \small 5\) | \(\displaystyle \small 1\) | \(\displaystyle \small 3\) | |
| \(\displaystyle \small \color{green}{3}\) | \(\displaystyle \small \color{green}{1}\) | \(\displaystyle \small ?\) | ||
| \(\displaystyle -\) | \(\displaystyle \small 2\) | \(\displaystyle \small ?\) | ||
| \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||
| \(\displaystyle \small 0\) |
3. Сносим единицы числа \(\displaystyle 5{\underline1}{\small : }\)
| \(\displaystyle -\) | \(\displaystyle \small 5\) | \(\displaystyle \small 1\) | \(\displaystyle \small 3\) | |
| \(\displaystyle \small 3\) | \(\displaystyle \small 1\) | \(\displaystyle \small ?\) | ||
| \(\displaystyle -\) | \(\displaystyle \small 2\) | \(\displaystyle \small {\bf 1}\) | ||
| \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||
| \(\displaystyle \small 0\) |
Получили число \(\displaystyle 21{\small .}\)
Шаг 2.
1. Делим \(\displaystyle 21\) на \(\displaystyle 3{\small .}\)
Найдем, какое максимальное количество \(\displaystyle \color{blue}{3}\) можно забрать из \(\displaystyle 21{\small .}\)
То есть найдем частное при делении \(\displaystyle 21\) на \(\displaystyle \color{blue}{3}{\small .}\)
Так как
\(\displaystyle 21=\color{green}{7} \cdot \color{blue}{3} {\small ,}\)
то можно забрать \(\displaystyle \color{green}{7}\) (семь) троек.
Поэтому пишем \(\displaystyle \color{green}{7}\) следующей цифрой в частное:
| \(\displaystyle -\) | \(\displaystyle \small 5\) | \(\displaystyle \small 1\) | \(\displaystyle \small 3\) | |
| \(\displaystyle \small 3\) | \(\displaystyle \small 1\) | \(\displaystyle \small \color{green}{7}\) | ||
| \(\displaystyle -\) | \(\displaystyle \small 2\) | \(\displaystyle \small 1\) | ||
| \(\displaystyle \small ?\) | \(\displaystyle \small ?\) | |||
| \(\displaystyle \small 0\) |
2. Далее, вычитаем в столбик из \(\displaystyle 21\) произведение \(\displaystyle \color{blue}{3}\cdot \color{green}{7}=\color{green}{21}{\small : }\)
| \(\displaystyle -\) | \(\displaystyle \small 5\) | \(\displaystyle \small 1\) | \(\displaystyle \small 3\) | |
| \(\displaystyle \small 3\) | \(\displaystyle \small 1\) | \(\displaystyle \small \color{green}{7}\) | ||
| \(\displaystyle -\) | \(\displaystyle \small 2\) | \(\displaystyle \small 1\) | ||
| \(\displaystyle \small \color{green}{2}\) | \(\displaystyle \small \color{green}{1}\) | |||
| \(\displaystyle \small 0\) |
Получили число \(\displaystyle 0{\small ,}\) процесс деления закончился.
Таким образом,
| \(\displaystyle -\) | \(\displaystyle 5\) | \(\displaystyle 1\) | \(\displaystyle 3\) | |
| \(\displaystyle 3\) | \(\displaystyle 1\) | \(\displaystyle 7\) | ||
| \(\displaystyle -\) | \(\displaystyle 2\) | \(\displaystyle 1\) | ||
| \(\displaystyle 2\) | \(\displaystyle 1\) | |||
| \(\displaystyle 0\) |