Digital Matematika - 5 класс - Деление в столбик на двузначные числа

Задание

Выполните деление чисел в столбик:

\(\displaystyle -\)	\(\displaystyle 1\)	\(\displaystyle 2\)	\(\displaystyle 1\)	\(\displaystyle 2\)	\(\displaystyle 12\)
\(\displaystyle -\)
	\(\displaystyle -\)
	\(\displaystyle -\)
				\(\displaystyle 0\)

Решение

Сначала составим таблицу умножения на \(\displaystyle 12\) для чисел от \(\displaystyle 1 \) до \(\displaystyle 9{\small : } \)

Таблица умножения на \(\displaystyle 12\)

Теперь рассмотрим сам процесс деления \(\displaystyle 1212\) на \(\displaystyle 12{\small . } \)

Заметим, что в числе \(\displaystyle 1212 \) первая цифра \(\displaystyle 1\) меньше \(\displaystyle 12{\small . }\)

Значит, берем сразу две первых цифры вместе, то есть число \(\displaystyle 12{\small . }\)

Шаг 1.

Делим \(\displaystyle \color{orange}{12}\) на \(\displaystyle 12\) с остатком

1. Делим \(\displaystyle 12\) на \(\displaystyle 12\) с остатком.

Найдем, какое максимальное количество \(\displaystyle \color{blue}{12}\) можно забрать из \(\displaystyle 12{\small .}\)

То есть найдем неполное частное при делении \(\displaystyle 12\) на \(\displaystyle \color{blue}{12}\) с остатком.

Так как

\(\displaystyle 12=\color{green}{1} \cdot \color{blue}{12}+0{\small ,}\)

то \(\displaystyle 12 \) можно забрать \(\displaystyle \color{green}{1}\) раз.

Поэтому пишем \(\displaystyle \color{green}{1}\) в частное:

\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small 12\)
\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)			\(\displaystyle \small \color{green}{1}\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)		\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
				\(\displaystyle \small 0\)

2. Далее, вычитаем в столбик из \(\displaystyle 12\) произведение \(\displaystyle \color{blue}{12}\cdot \color{green}{1}=\color{green}{12}{\small : }\)

\(\displaystyle -\)	\(\displaystyle \small \color{orange}{1}\)	\(\displaystyle \small \color{orange}{2}\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small 12\)
\(\displaystyle -\)	\(\displaystyle \small \color{green}{ 1}\)	\(\displaystyle \small \color{green}{ 2}\)			\(\displaystyle \small \color{green}{1}\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small 0\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)		\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
				\(\displaystyle \small 0\)

3. Сносим цифру из разряда десятков \(\displaystyle 12{\underline 1}2{\small : }\)

\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small 12\)
\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)			\(\displaystyle \small 1\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small 0\)	\(\displaystyle \small \color{red}{ 1}\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)		\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
				\(\displaystyle \small 0\)

4. Так как \(\displaystyle \color{red}{ 1}\) меньше, чем \(\displaystyle 12{\small ,}\) то сносим следующую цифру (единицы числа \(\displaystyle 121{\underline 2}{\small ,}\) то есть цифру \(\displaystyle 2\)), а в частном записываем нуль:

\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small 12\)
\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)			\(\displaystyle \small 1\)	\(\displaystyle \small \color{green}{ 0}\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small 0\)	\(\displaystyle \small \color{red}{ 1}\)	\(\displaystyle \small \color{red}{ 2}\)
	\(\displaystyle -\)		\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
				\(\displaystyle \small 0\)

Получили число \(\displaystyle 12{\small .}\)

Шаг 2.

Делим \(\displaystyle \color{cyan}{12}\) на \(\displaystyle 12\) с остатком

1. Делим \(\displaystyle 12\) на \(\displaystyle 12{\small .}\)

Найдем, какое максимальное количество \(\displaystyle \color{blue}{12}\) можно забрать из \(\displaystyle 12{\small .}\)

То есть найдем частное при делении \(\displaystyle 12\) на \(\displaystyle \color{blue}{12}{\small .}\)

Так как

\(\displaystyle 12=\color{green}{1} \cdot \color{blue}{12}{\small ,}\)

то \(\displaystyle 12 \) можно забрать \(\displaystyle \color{green}{1}\) раз.

Поэтому пишем \(\displaystyle \color{green}{1}\) следующей цифрой в частное:

\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small 12\)
\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)			\(\displaystyle \small 1\)	\(\displaystyle \small 0\)	\(\displaystyle \small \color{green}{ 1}\)
	\(\displaystyle -\)	\(\displaystyle \small 0\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)
	\(\displaystyle -\)		\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
				\(\displaystyle \small 0\)

2. Далее, вычитаем в столбик из \(\displaystyle 12\) произведение \(\displaystyle \color{blue}{12}\cdot \color{green}{1}=\color{green}{12}{\small : }\)

\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small 12\)
\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)			\(\displaystyle \small 1\)	\(\displaystyle \small 0\)	\(\displaystyle \small \color{green}{ 1}\)
	\(\displaystyle -\)	\(\displaystyle \small 0\)	\(\displaystyle \small \color{cyan}{1}\)	\(\displaystyle \small \color{cyan}{2}\)
	\(\displaystyle -\)		\(\displaystyle \small \color{green}{ 1}\)	\(\displaystyle \small \color{green}{ 2}\)
				\(\displaystyle \small 0\)

Получили \(\displaystyle 0{\small , }\) процес деления закончен.

Таким образом,

\(\displaystyle -\)	\(\displaystyle \small \color{orange}{1}\)	\(\displaystyle \small \color{orange}{2}\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small 12\)
\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)			\(\displaystyle \small 1\)	\(\displaystyle \small 0\)	\(\displaystyle \small 1\)
	\(\displaystyle -\)	\(\displaystyle \small 0\)	\(\displaystyle \small \color{cyan}{1}\)	\(\displaystyle \small \color{cyan}{2}\)
	\(\displaystyle -\)		\(\displaystyle \small 1\)	\(\displaystyle \small 2\)
				\(\displaystyle \small 0\)

	\(\displaystyle 1\)	\(\displaystyle 2\)	\(\displaystyle 3\)	\(\displaystyle 4\)	\(\displaystyle 5\)	\(\displaystyle 6\)	\(\displaystyle 7\)	\(\displaystyle 8\)	\(\displaystyle 9\)
\(\displaystyle \color{blue}{12}\)	\(\displaystyle 12\)	\(\displaystyle 24\)	\(\displaystyle 36\)	\(\displaystyle 48\)	\(\displaystyle 60\)	\(\displaystyle 72\)	\(\displaystyle 84\)	\(\displaystyle 96\)	\(\displaystyle 12\)

Теория: Деление в столбик на двузначные числа