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

Задание

Выполните деление чисел в столбик:

\(\displaystyle -\)	\(\displaystyle 2\)	\(\displaystyle 1\)	\(\displaystyle 8\)	\(\displaystyle 3\)	\(\displaystyle 6\)	\(\displaystyle 53\)
\(\displaystyle -\)
		\(\displaystyle -\)
		\(\displaystyle -\)
		\(\displaystyle -\)
		\(\displaystyle -\)
					\(\displaystyle 0\)

Решение

Сначала составим таблицу умножения на \(\displaystyle 53\) для чисел от \(\displaystyle 1 \) до \(\displaystyle 9{\small : } \)

Таблица умножения на \(\displaystyle 53\)

Теперь рассмотрим сам процесс деления \(\displaystyle 21836\) на \(\displaystyle 53{\small . } \)

Заметим, что в числе \(\displaystyle 21836\) первая цифра \(\displaystyle 2\) меньше \(\displaystyle 53{\small . }\)

Далее, следующие две цифры дают число \(\displaystyle 21{\small , } \) которое также меньше \(\displaystyle 53{\small . } \)

Значит, берем сразу три первых цифры вместе, то есть число \(\displaystyle 218{\small . }\)

Шаг 1.

Делим \(\displaystyle \color{orange}{218}\) на \(\displaystyle 53\) с остатком

1. Делим \(\displaystyle 218\) на \(\displaystyle 53\) с остатком.

Найдем, какое максимальное количество \(\displaystyle \color{blue}{53}\) можно забрать из \(\displaystyle 218{\small .}\)

То есть найдем неполное частное при делении \(\displaystyle 218\) на \(\displaystyle \color{blue}{53}\) с остатком.

Так как

\(\displaystyle 218=\color{green}{4} \cdot \color{blue}{53}+6{\small ,}\)

то \(\displaystyle 53\) можно забрать \(\displaystyle \color{green}{4}\) раза.

Поэтому пишем \(\displaystyle \color{green}{4}\) в частное:

\(\displaystyle -\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 8\)	\(\displaystyle \small 3\)	\(\displaystyle \small 6\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)			\(\displaystyle \small \color{green}{4}\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
					\(\displaystyle \small 0\)

2. Далее, вычитаем в столбик из \(\displaystyle 218\) произведение \(\displaystyle \color{blue}{53}\cdot \color{green}{4}=\color{green}{212}{\small : }\)

\(\displaystyle -\)	\(\displaystyle \small \color{orange}{2}\)	\(\displaystyle \small \color{orange}{1}\)	\(\displaystyle \small \color{orange}{8}\)	\(\displaystyle \small 3\)	\(\displaystyle \small 6\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small \color{green}{ 2}\)	\(\displaystyle \small \color{green}{ 1}\)	\(\displaystyle \small \color{green}{ 2}\)			\(\displaystyle \small \color{green}{ 4}\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small 6\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
					\(\displaystyle \small 0\)

3. Сносим цифру из разряда десятков \(\displaystyle 218{\underline 3}6{\small : }\)

\(\displaystyle -\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 8\)	\(\displaystyle \small 3\)	\(\displaystyle \small 6\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)			\(\displaystyle \small 4\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small 6\)	\(\displaystyle \small \bf 3\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
					\(\displaystyle \small 0\)

Получили число \(\displaystyle 63{\small .}\)

Шаг 2.

Делим \(\displaystyle \color{cyan}{63}\) на \(\displaystyle 53\) с остатком

1. Делим \(\displaystyle 63\) на \(\displaystyle 53{\small .}\)

Найдем, какое максимальное количество \(\displaystyle \color{blue}{53}\) можно забрать из \(\displaystyle 63{\small .}\)

То есть найдем неполное частное при делении \(\displaystyle 63\) на \(\displaystyle \color{blue}{53}{\small .}\)

Так как

\(\displaystyle 63=\color{green}{1} \cdot \color{blue}{53} +10{\small ,}\)

то \(\displaystyle 53 \) можно забрать \(\displaystyle \color{green}{1}\) раз.

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

\(\displaystyle -\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 8\)	\(\displaystyle \small 3\)	\(\displaystyle \small 6\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)			\(\displaystyle \small 4\)	\(\displaystyle \small \color{green}{ 1}\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small 6\)	\(\displaystyle \small 3\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
					\(\displaystyle \small 0\)

2. Далее, вычитаем в столбик из \(\displaystyle 63\) произведение \(\displaystyle \color{blue}{53}\cdot \color{green}{1}=\color{green}{53}{\small : }\)

\(\displaystyle -\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 8\)	\(\displaystyle \small 3\)	\(\displaystyle \small 6\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)			\(\displaystyle \small 4\)	\(\displaystyle \small \color{green}{ 1}\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small \color{cyan}{6}\)	\(\displaystyle \small \color{cyan}{3}\)
		\(\displaystyle -\)	\(\displaystyle \small \color{green}{ 5}\)	\(\displaystyle \small \color{green}{ 3}\)
		\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 0\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
					\(\displaystyle \small 0\)

3. Сносим цифру из разряда единиц \(\displaystyle 2183{\underline 6}{\small : }\)

\(\displaystyle -\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 8\)	\(\displaystyle \small 3\)	\(\displaystyle \small 6\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)			\(\displaystyle \small 4\)	\(\displaystyle \small 1\)	\(\displaystyle \small ?\)
		\(\displaystyle -\)	\(\displaystyle \small 6\)	\(\displaystyle \small 3\)
		\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)
		\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 0\)	\(\displaystyle \small \bf 6\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
					\(\displaystyle \small 0\)

Получили число \(\displaystyle 106{\small .}\)

Шаг 3.

Делим \(\displaystyle \color{red}{106}\) на \(\displaystyle 53\) с остатком

1. Делим \(\displaystyle 106\) на \(\displaystyle 53{\small .}\)

Найдем, какое максимальное количество \(\displaystyle \color{blue}{53}\) можно забрать из \(\displaystyle 106{\small .}\)

То есть найдем частное при делении \(\displaystyle 106\) на \(\displaystyle \color{blue}{53}{\small .}\)

Так как

\(\displaystyle 106=\color{green}{2} \cdot \color{blue}{53}{\small ,}\)

то \(\displaystyle 53\) можно забрать \(\displaystyle \color{green}{2}\) раза.

Поэтому пишем \(\displaystyle \color{green}{2}\) следующей цифрой в частное:

\(\displaystyle -\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 8\)	\(\displaystyle \small 3\)	\(\displaystyle \small 6\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)			\(\displaystyle \small 4\)	\(\displaystyle \small 1\)	\(\displaystyle \small \color{green}{ 2}\)
		\(\displaystyle -\)	\(\displaystyle \small 6\)	\(\displaystyle \small 3\)
		\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)
		\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 0\)	\(\displaystyle \small 6\)
		\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
					\(\displaystyle \small 0\)

2. Далее, вычитаем в столбик из \(\displaystyle 106\) произведение \(\displaystyle \color{blue}{53}\cdot \color{green}{3}=\color{green}{106}{\small : }\)

\(\displaystyle -\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 8\)	\(\displaystyle \small 3\)	\(\displaystyle \small 6\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)			\(\displaystyle \small 4\)	\(\displaystyle \small 1\)	\(\displaystyle \small \color{green}{ 2}\)
		\(\displaystyle -\)	\(\displaystyle \small 6\)	\(\displaystyle \small 3\)
		\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)
		\(\displaystyle -\)	\(\displaystyle \small \color{red}{ 1}\)	\(\displaystyle \small \color{red}{ 0}\)	\(\displaystyle \small \color{red}{ 6}\)
		\(\displaystyle -\)	\(\displaystyle \small \color{green}{ 1}\)	\(\displaystyle \small \color{green}{ 0}\)	\(\displaystyle \small \color{green}{ 6}\)
					\(\displaystyle \small 0\)

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

Таким образом,

\(\displaystyle -\)	\(\displaystyle \small \color{orange}{2}\)	\(\displaystyle \small \color{orange}{1}\)	\(\displaystyle \small \color{orange}{8}\)	\(\displaystyle \small 3\)	\(\displaystyle \small 6\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small 2\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)			\(\displaystyle \small 4\)	\(\displaystyle \small 1\)	\(\displaystyle \small 2\)
		\(\displaystyle -\)	\(\displaystyle \small \color{cyan}{6}\)	\(\displaystyle \small \color{cyan}{3}\)
		\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)
		\(\displaystyle -\)	\(\displaystyle \small \color{red}{ 1}\)	\(\displaystyle \small \color{red}{ 0}\)	\(\displaystyle \small \color{red}{ 6}\)
		\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 0\)	\(\displaystyle \small 6\)
					\(\displaystyle \small 0\)

	\(\displaystyle 1\)	\(\displaystyle 2\)	\(\displaystyle 3\)	\(\displaystyle 4\)	\(\displaystyle 5\)	\(\displaystyle 6\)	\(\displaystyle 7\)	\(\displaystyle 8\)	\(\displaystyle 9\)
\(\displaystyle \color{blue}{53}\)	\(\displaystyle 53\)	\(\displaystyle 106\)	\(\displaystyle 159\)	\(\displaystyle 212\)	\(\displaystyle 265\)	\(\displaystyle 318\)	\(\displaystyle 371\)	\(\displaystyle 424\)	\(\displaystyle 477\)

Теория: Деление в столбик на двузначные числа