Digital Matematika - 4 класс - Деление в столбик на двузначные числа

Задание

Выполните деление чисел в столбик:

\(\displaystyle -\)	\(\displaystyle 5\)	\(\displaystyle 3\)	\(\displaystyle 1\)	\(\displaystyle 5\)	\(\displaystyle 9\)	\(\displaystyle 53\)
\(\displaystyle -\)
	\(\displaystyle -\)
	\(\displaystyle -\)
					\(\displaystyle 0\)

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 \(\displaystyle 53159\) на \(\displaystyle 53{\small . } \)

Заметим, что в числе \(\displaystyle 53159 \) первая цифра \(\displaystyle 5\) меньше \(\displaystyle 53{\small . }\)

Значит, берем сразу две первых цифры вместе, то есть число \(\displaystyle 53{\small . }\)

Шаг 1.

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

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

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

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

Так как

\(\displaystyle 53=\color{green}{1} \cdot \color{blue}{53}+0{\small ,}\)

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

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

\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)	\(\displaystyle \small 1\)	\(\displaystyle \small 5\)	\(\displaystyle \small 9\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)				\(\displaystyle \small \color{green}{1}\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)		\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
					\(\displaystyle \small 0\)

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

\(\displaystyle -\)	\(\displaystyle \small \color{orange}{5}\)	\(\displaystyle \small \color{orange}{3}\)	\(\displaystyle \small 1\)	\(\displaystyle \small 5\)	\(\displaystyle \small 9\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small \color{green}{ 5}\)	\(\displaystyle \small \color{green}{ 3}\)				\(\displaystyle \small 1\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small 0\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)		\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
					\(\displaystyle \small 0\)

3. Сносим цифру из разряда сотен \(\displaystyle 53{\underline 1}59{\small : }\)

\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)	\(\displaystyle \small 1\)	\(\displaystyle \small 5\)	\(\displaystyle \small 9\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)				\(\displaystyle \small \color{green}{1}\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small 0\)	\(\displaystyle \small \color{red}{ 1}\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)		\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
					\(\displaystyle \small 0\)

4. Так как \(\displaystyle \color{red}{ 1}\) меньше, чем \(\displaystyle 53{\small ,}\) то сносим следующую цифру (десятки числа \(\displaystyle 531{\underline 5}9{\small ,}\) то есть цифру \(\displaystyle 5\)), а в частном записываем нуль:

\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)	\(\displaystyle \small 1\)	\(\displaystyle \small 5\)	\(\displaystyle \small 9\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)				\(\displaystyle \small 1\)	\(\displaystyle \small \color{green}{ 0}\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small 0\)	\(\displaystyle \small \color{red}{ 1}\)	\(\displaystyle \small \color{red}{ 5}\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)		\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
					\(\displaystyle \small 0\)

5. Так как \(\displaystyle \color{red}{ 15}\) меньше, чем \(\displaystyle 53{\small ,}\) то сносим следующую цифру (единицы числа \(\displaystyle 5315{\underline 9}{\small ,}\) то есть цифру \(\displaystyle 9\)), а в частном записываем нуль:

\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)	\(\displaystyle \small 1\)	\(\displaystyle \small 5\)	\(\displaystyle \small 9\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)				\(\displaystyle \small 1\)	\(\displaystyle \small 0\)	\(\displaystyle \small \color{green}{ 0}\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small 0\)	\(\displaystyle \small \color{red}{ 1}\)	\(\displaystyle \small \color{red}{ 5}\)	\(\displaystyle \small \color{red}{ 9}\)
	\(\displaystyle -\)		\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
					\(\displaystyle \small 0\)

Получили число \(\displaystyle 159{\small .}\)

Шаг 2.

Делим \(\displaystyle \color{cyan}{159}\) на \(\displaystyle 53\) с остатком

1. Делим \(\displaystyle 159\) на \(\displaystyle 53{\small .}\)

Найдем, какое максимальное количество \(\displaystyle \color{blue}{53}\) можно забрать из \(\displaystyle 159{\small .}\)

То есть найдем частное при делении \(\displaystyle 159\) на \(\displaystyle \color{blue}{53}{\small .}\)

Так как

\(\displaystyle 159=\color{green}{3} \cdot \color{blue}{53}{\small ,}\)

то \(\displaystyle 53 \) можно забрать \(\displaystyle \color{green}{3}\) раза.

Поэтому пишем \(\displaystyle \color{green}{3}\) следующей цифрой в частное:

\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)	\(\displaystyle \small 1\)	\(\displaystyle \small 5\)	\(\displaystyle \small 9\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)				\(\displaystyle \small 1\)	\(\displaystyle \small 0\)	\(\displaystyle \small 0\)	\(\displaystyle \small \color{green}{ 3}\)
	\(\displaystyle -\)	\(\displaystyle \small 0\)	\(\displaystyle \small 1\)	\(\displaystyle \small 5\)	\(\displaystyle \small 9\)
	\(\displaystyle -\)		\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
					\(\displaystyle \small 0\)

2. Далее, вычитаем в столбик из \(\displaystyle 159\) произведение \(\displaystyle \color{blue}{53}\cdot \color{green}{3}=\color{green}{159}{\small : }\)

\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)	\(\displaystyle \small 1\)	\(\displaystyle \small 5\)	\(\displaystyle \small 9\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)				\(\displaystyle \small 1\)	\(\displaystyle \small 0\)	\(\displaystyle \small 0\)	\(\displaystyle \small \color{green}{ 3}\)
	\(\displaystyle -\)	\(\displaystyle \small 0\)	\(\displaystyle \small \color{cyan}{1}\)	\(\displaystyle \small \color{cyan}{5}\)	\(\displaystyle \small \color{cyan}{9}\)
	\(\displaystyle -\)		\(\displaystyle \small \color{green}{ 1}\)	\(\displaystyle \small \color{green}{ 5}\)	\(\displaystyle \small \color{green}{ 9}\)
					\(\displaystyle \small 0\)

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

Таким образом,

\(\displaystyle -\)	\(\displaystyle \small \color{orange}{5}\)	\(\displaystyle \small \color{orange}{3}\)	\(\displaystyle \small 1\)	\(\displaystyle \small 5\)	\(\displaystyle \small 9\)	\(\displaystyle \small 53\)
\(\displaystyle -\)	\(\displaystyle \small 5\)	\(\displaystyle \small 3\)				\(\displaystyle \small 1\)	\(\displaystyle \small 0\)	\(\displaystyle \small 0\)	\(\displaystyle \small 3\)
	\(\displaystyle -\)	\(\displaystyle \small 0\)	\(\displaystyle \small \color{cyan}{1}\)	\(\displaystyle \small \color{cyan}{5}\)	\(\displaystyle \small \color{cyan}{9}\)
	\(\displaystyle -\)		\(\displaystyle \small 1\)	\(\displaystyle \small 5\)	\(\displaystyle \small 9\)
					\(\displaystyle \small 0\)

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

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Теория: Деление в столбик на двузначные числа