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:
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\(\displaystyle -\) |
\(\displaystyle \small 1\) |
\(\displaystyle \small 2\) |
\(\displaystyle \small 1\) |
\(\displaystyle \small 2\) |
\(\displaystyle \small 12\) |
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\(\displaystyle \small ?\) |
\(\displaystyle \small ?\) |
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\(\displaystyle \small \color{green}{1}\) |
\(\displaystyle \small ?\) |
\(\displaystyle \small ?\) |
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\(\displaystyle -\) |
\(\displaystyle \small ?\) |
\(\displaystyle \small ?\) |
\(\displaystyle \small ?\) |
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\(\displaystyle \small ?\) |
\(\displaystyle \small ?\) |
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\(\displaystyle \small 0\) |
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2. Далее, вычитаем в столбик из \(\displaystyle 12\) произведение \(\displaystyle \color{blue}{12}\cdot \color{green}{1}=\color{green}{12}{\small : }\)
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\(\displaystyle -\) |
\(\displaystyle \small \color{orange}{1}\) |
\(\displaystyle \small \color{orange}{2}\) |
\(\displaystyle \small 1\) |
\(\displaystyle \small 2\) |
\(\displaystyle \small 12\) |
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\(\displaystyle \small \color{green}{ 1}\) |
\(\displaystyle \small \color{green}{ 2}\) |
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\(\displaystyle \small \color{green}{1}\) |
\(\displaystyle \small ?\) |
\(\displaystyle \small ?\) |
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\(\displaystyle -\) |
\(\displaystyle \small 0\) |
\(\displaystyle \small ?\) |
\(\displaystyle \small ?\) |
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\(\displaystyle \small ?\) |
\(\displaystyle \small ?\) |
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\(\displaystyle \small 0\) |
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3. Сносим цифру из разряда десятков \(\displaystyle 12{\underline 1}2{\small : }\)
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\(\displaystyle -\) |
\(\displaystyle \small 1\) |
\(\displaystyle \small 2\) |
\(\displaystyle \small 1\) |
\(\displaystyle \small 2\) |
\(\displaystyle \small 12\) |
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\(\displaystyle \small 1\) |
\(\displaystyle \small 2\) |
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\(\displaystyle \small 1\) |
\(\displaystyle \small ?\) |
\(\displaystyle \small ?\) |
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\(\displaystyle -\) |
\(\displaystyle \small 0\) |
\(\displaystyle \small \color{red}{ 1}\) |
\(\displaystyle \small ?\) |
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\(\displaystyle \small ?\) |
\(\displaystyle \small ?\) |
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\(\displaystyle \small 0\) |
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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:
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\(\displaystyle -\) |
\(\displaystyle \small 1\) |
\(\displaystyle \small 2\) |
\(\displaystyle \small 1\) |
\(\displaystyle \small 2\) |
\(\displaystyle \small 12\) |
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\(\displaystyle \small 1\) |
\(\displaystyle \small 2\) |
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\(\displaystyle \small 1\) |
\(\displaystyle \small \color{green}{ 0}\) |
\(\displaystyle \small ?\) |
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\(\displaystyle -\) |
\(\displaystyle \small 0\) |
\(\displaystyle \small \color{red}{ 1}\) |
\(\displaystyle \small \color{red}{ 2}\) |
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\(\displaystyle \small ?\) |
\(\displaystyle \small ?\) |
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\(\displaystyle \small 0\) |
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Получили число \(\displaystyle 12{\small .}\)
