Перемножить числа:
\(\displaystyle 9\) | \(\displaystyle 4\) | \(\displaystyle 5\) | |||
\(\displaystyle \times\) | |||||
\(\displaystyle 2\) | \(\displaystyle 3\) | ||||
\(\displaystyle +\) | |||||
1. Умножаем единицы числа \(\displaystyle 2\underline{3} \) (это цифра \(\displaystyle 3 \)) на число \(\displaystyle 945{\small:}\)
\(\displaystyle 3 \cdot 945=2835 {\small.}\)
\(\displaystyle \bf9\) | \(\displaystyle \bf4\) | \(\displaystyle \bf5\) | |||
\(\displaystyle \times\) | |||||
\(\displaystyle 2\) | \(\displaystyle \bf3\) | ||||
\(\displaystyle \bf2\) | \(\displaystyle \bf8\) | \(\displaystyle \bf3\) | \(\displaystyle \bf5\) | ||
\(\displaystyle +\) | |||||
\(\displaystyle ?\) | \(\displaystyle ?\) | \(\displaystyle ?\) | \(\displaystyle ?\) | ||
\(\displaystyle ?\) | \(\displaystyle ?\) | \(\displaystyle ?\) | \(\displaystyle ?\) | \(\displaystyle ?\) |
2. Умножаем десятки числа \(\displaystyle \underline{2}3 \) (это цифра \(\displaystyle 2 \)) на число \(\displaystyle 945{\small:}\)
\(\displaystyle 2 \cdot 945=1890 {\small.}\)
Записываем результат так, чтобы единицы \(\displaystyle 1890\) шли под той цифрой, на которую мы умножали; в нашем случае это десятки числа \(\displaystyle \underline{2}3 \) (то есть цифра \(\displaystyle 2 \)):
\(\displaystyle \bf9\) | \(\displaystyle \bf4\) | \(\displaystyle \bf5\) | |||
\(\displaystyle \times\) | |||||
\(\displaystyle \bf2\) | \(\displaystyle 3\) | ||||
\(\displaystyle 2\) | \(\displaystyle 8\) | \(\displaystyle 3\) | \(\displaystyle 5\) | ||
\(\displaystyle +\) | |||||
\(\displaystyle \bf1\) | \(\displaystyle \bf8\) | \(\displaystyle \bf9\) | \(\displaystyle \bf0\) | ||
\(\displaystyle ?\) | \(\displaystyle ?\) | \(\displaystyle ?\) | \(\displaystyle ?\) | \(\displaystyle ?\) |
3. Складываем числа:
\(\displaystyle 2\) | \(\displaystyle 8\) | \(\displaystyle 3\) | \(\displaystyle 5\) | ||
\(\displaystyle +\) | |||||
\(\displaystyle 1\) | \(\displaystyle 8\) | \(\displaystyle 9\) | \(\displaystyle 0\) | ||
\(\displaystyle \bf2\) | \(\displaystyle \bf1\) | \(\displaystyle \bf7\) | \(\displaystyle \bf3\) | \(\displaystyle \bf5\) |
В итоге получаем:
\(\displaystyle 9\) | \(\displaystyle 4\) | \(\displaystyle 5\) | |||
\(\displaystyle \times\) | |||||
\(\displaystyle 2\) | \(\displaystyle 3\) | ||||
\(\displaystyle 2\) | \(\displaystyle 8\) | \(\displaystyle 3\) | \(\displaystyle 5\) | ||
\(\displaystyle +\) | |||||
\(\displaystyle 1\) | \(\displaystyle 8\) | \(\displaystyle 9\) | \(\displaystyle 0\) | ||
\(\displaystyle 2\) | \(\displaystyle 1\) | \(\displaystyle 7\) | \(\displaystyle 3\) | \(\displaystyle 5\) |
Ответ: \(\displaystyle 21735{\small.}\)