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

Задание

Выполните деление чисел в столбик:

\(\displaystyle -\)	\(\displaystyle 4\)	\(\displaystyle 3\)	\(\displaystyle 0\)	\(\displaystyle 5\)	\(\displaystyle 35\)
\(\displaystyle -\)
	\(\displaystyle -\)
	\(\displaystyle -\)
	\(\displaystyle -\)
	\(\displaystyle -\)
				\(\displaystyle 0\)

Решение

Сначала составим таблицу умножения на \(\displaystyle 35\) для чисел от \(\displaystyle 1 \) до \(\displaystyle 9{\small : } \)

Таблица умножения на \(\displaystyle 35\)

Теперь начнем сам процесс деления \(\displaystyle 4305\) на \(\displaystyle 35{\small . } \)

Так как первая цифра \(\displaystyle 4\) меньше \(\displaystyle 35{\small ,}\) то берем сразу две первых цифры вместе, то есть число \(\displaystyle 43{\small . }\)

Шаг 1.

Делим \(\displaystyle \color{orange}{43}\) на \(\displaystyle 35\) с остатком

1. Делим \(\displaystyle 43\) на \(\displaystyle 35\) с остатком.

Найдем, какое максимальное количество \(\displaystyle \color{blue}{35}\) можно забрать из \(\displaystyle 43{\small .}\)

То есть найдем неполное частное при делении \(\displaystyle 43\) на \(\displaystyle \color{blue}{35}\) с остатком.

Так как

\(\displaystyle 43=\color{green}{1} \cdot \color{blue}{35}+8{\small ,}\)

то \(\displaystyle 35\) можно забрать \(\displaystyle \color{green}{1}\) раз.

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

\(\displaystyle -\)	\(\displaystyle \small 4\)	\(\displaystyle \small 3\)	\(\displaystyle \small 0\)	\(\displaystyle \small 5\)	\(\displaystyle \small 35\)
\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)			\(\displaystyle \small \color{green}{1}\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
				\(\displaystyle \small 0\)

2. Далее, вычитаем в столбик из \(\displaystyle 43\) произведение \(\displaystyle \color{blue}{35}\cdot \color{green}{1}=\color{green}{35}{\small : }\)

\(\displaystyle -\)	\(\displaystyle \small \color{orange}{4}\)	\(\displaystyle \small \color{orange}{3}\)	\(\displaystyle \small 0\)	\(\displaystyle \small 5\)	\(\displaystyle \small 35\)
\(\displaystyle -\)	\(\displaystyle \small \color{green}{ 3}\)	\(\displaystyle \small \color{green}{ 5}\)			\(\displaystyle \small \color{green}{ 1}\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small 8\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
				\(\displaystyle \small 0\)

3. Сносим цифру из разряда десятков \(\displaystyle 43{\underline 0}5{\small : }\)

\(\displaystyle -\)	\(\displaystyle \small 4\)	\(\displaystyle \small 3\)	\(\displaystyle \small 0\)	\(\displaystyle \small 5\)	\(\displaystyle \small 35\)
\(\displaystyle -\)	\(\displaystyle \small 3\)	\(\displaystyle \small 5\)			\(\displaystyle \small 1\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small 8\)	\(\displaystyle \small \bf 0\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
				\(\displaystyle \small 0\)

Получили число \(\displaystyle 80{\small .}\)

Шаг 2.

Делим \(\displaystyle \color{cyan}{80}\) на \(\displaystyle 35\) с остатком

1. Делим \(\displaystyle 80\) на \(\displaystyle 35{\small .}\)

Найдем, какое максимальное количество \(\displaystyle \color{blue}{35}\) можно забрать из \(\displaystyle 80{\small .}\)

То есть найдем неполное частное при делении \(\displaystyle 80\) на \(\displaystyle \color{blue}{35}{\small .}\)

Так как

\(\displaystyle 80=\color{green}{2} \cdot \color{blue}{35} +10{\small ,}\)

то \(\displaystyle 35 \) можно забрать \(\displaystyle \color{green}{2}\) раза.

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

\(\displaystyle -\)	\(\displaystyle \small 4\)	\(\displaystyle \small 3\)	\(\displaystyle \small 0\)	\(\displaystyle \small 5\)	\(\displaystyle \small 35\)
\(\displaystyle -\)	\(\displaystyle \small 3\)	\(\displaystyle \small 5\)			\(\displaystyle \small 1\)	\(\displaystyle \small \color{green}{ 2}\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small 8\)	\(\displaystyle \small 0\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
				\(\displaystyle \small 0\)

2. Далее, вычитаем в столбик из \(\displaystyle 80\) произведение \(\displaystyle \color{blue}{35}\cdot \color{green}{2}=\color{green}{70}{\small : }\)

\(\displaystyle -\)	\(\displaystyle \small 4\)	\(\displaystyle \small 3\)	\(\displaystyle \small 0\)	\(\displaystyle \small 5\)	\(\displaystyle \small 35\)
\(\displaystyle -\)	\(\displaystyle \small 3\)	\(\displaystyle \small 5\)			\(\displaystyle \small 1\)	\(\displaystyle \small \color{green}{ 2}\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small \color{cyan}{8}\)	\(\displaystyle \small \color{cyan}{0}\)
	\(\displaystyle -\)	\(\displaystyle \small \color{green}{ 7}\)	\(\displaystyle \small \color{green}{ 0}\)
	\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 0\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
				\(\displaystyle \small 0\)

3. Сносим цифру из разряда единиц \(\displaystyle 430{\underline 5}{\small : }\)

\(\displaystyle -\)	\(\displaystyle \small 4\)	\(\displaystyle \small 3\)	\(\displaystyle \small 0\)	\(\displaystyle \small 5\)	\(\displaystyle \small 35\)
\(\displaystyle -\)	\(\displaystyle \small 3\)	\(\displaystyle \small 5\)			\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small ?\)
	\(\displaystyle -\)	\(\displaystyle \small 8\)	\(\displaystyle \small 0\)
	\(\displaystyle -\)	\(\displaystyle \small 7\)	\(\displaystyle \small 0\)
	\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 0\)	\(\displaystyle \small \bf 5\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
				\(\displaystyle \small 0\)

Получили число \(\displaystyle 105{\small .}\)

Шаг 3.

Делим \(\displaystyle \color{red}{105}\) на \(\displaystyle 35\) с остатком

1. Делим \(\displaystyle 105\) на \(\displaystyle 35{\small .}\)

Найдем, какое максимальное количество \(\displaystyle \color{blue}{35}\) можно забрать из \(\displaystyle 105{\small .}\)

То есть найдем частное при делении \(\displaystyle 105\) на \(\displaystyle \color{blue}{35}{\small .}\)

Так как

\(\displaystyle 105=\color{green}{3} \cdot \color{blue}{35}{\small ,}\)

то \(\displaystyle 35\) можно забрать \(\displaystyle \color{green}{3}\) раза.

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

\(\displaystyle -\)	\(\displaystyle \small 4\)	\(\displaystyle \small 3\)	\(\displaystyle \small 0\)	\(\displaystyle \small 5\)	\(\displaystyle \small 35\)
\(\displaystyle -\)	\(\displaystyle \small 3\)	\(\displaystyle \small 5\)			\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small \color{green}{ 3}\)
	\(\displaystyle -\)	\(\displaystyle \small 8\)	\(\displaystyle \small 0\)
	\(\displaystyle -\)	\(\displaystyle \small 7\)	\(\displaystyle \small 0\)
	\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 0\)	\(\displaystyle \small 5\)
	\(\displaystyle -\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)	\(\displaystyle \small ?\)
				\(\displaystyle \small 0\)

2. Далее, вычитаем в столбик из \(\displaystyle 105\) произведение \(\displaystyle \color{blue}{35}\cdot \color{green}{3}=\color{green}{105}{\small : }\)

\(\displaystyle -\)	\(\displaystyle \small 4\)	\(\displaystyle \small 3\)	\(\displaystyle \small 0\)	\(\displaystyle \small 5\)	\(\displaystyle \small 35\)
\(\displaystyle -\)	\(\displaystyle \small 3\)	\(\displaystyle \small 5\)			\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small \color{green}{ 3}\)
	\(\displaystyle -\)	\(\displaystyle \small 8\)	\(\displaystyle \small 0\)
	\(\displaystyle -\)	\(\displaystyle \small 7\)	\(\displaystyle \small 0\)
	\(\displaystyle -\)	\(\displaystyle \small \color{red}{ 1}\)	\(\displaystyle \small \color{red}{ 0}\)	\(\displaystyle \small \color{red}{ 5}\)
	\(\displaystyle -\)	\(\displaystyle \small \color{green}{ 1}\)	\(\displaystyle \small \color{green}{ 0}\)	\(\displaystyle \small \color{green}{ 5}\)
				\(\displaystyle \small 0\)

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

Таким образом,

\(\displaystyle -\)	\(\displaystyle \small \color{orange}{4}\)	\(\displaystyle \small \color{orange}{3}\)	\(\displaystyle \small 0\)	\(\displaystyle \small 5\)	\(\displaystyle \small 35\)
\(\displaystyle -\)	\(\displaystyle \small 3\)	\(\displaystyle \small 5\)			\(\displaystyle \small 1\)	\(\displaystyle \small 2\)	\(\displaystyle \small 3\)
	\(\displaystyle -\)	\(\displaystyle \small \color{cyan}{8}\)	\(\displaystyle \small \color{cyan}{0}\)
	\(\displaystyle -\)	\(\displaystyle \small 7\)	\(\displaystyle \small 0\)
	\(\displaystyle -\)	\(\displaystyle \small \color{red}{ 1}\)	\(\displaystyle \small \color{red}{ 0}\)	\(\displaystyle \small \color{red}{ 5}\)
	\(\displaystyle -\)	\(\displaystyle \small 1\)	\(\displaystyle \small 0\)	\(\displaystyle \small 5\)
				\(\displaystyle \small 0\)

	\(\displaystyle 1\)	\(\displaystyle 2\)	\(\displaystyle 3\)	\(\displaystyle 4\)	\(\displaystyle 5\)	\(\displaystyle 6\)	\(\displaystyle 7\)	\(\displaystyle 8\)	\(\displaystyle 9\)
\(\displaystyle \color{blue}{35}\)	\(\displaystyle 35\)	\(\displaystyle 70\)	\(\displaystyle 105\)	\(\displaystyle 140\)	\(\displaystyle 175\)	\(\displaystyle 210\)	\(\displaystyle 245\)	\(\displaystyle 280\)	\(\displaystyle 315\)

Теория: Деление в столбик на двузначные числа