Digital Matematika - 4 сынып - Екі таңбалы сандарға бағандап бөлу

Тапсырма

Сандарды бағанға бөлуді орындаңыз:

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

Шешім

\(\displaystyle 1 \)-ден \(\displaystyle 9{\small } \)-ға дейінгі сандар үшін \(\displaystyle 53\)-ке көбейту кестесін құрамыз:

\(\displaystyle 53\)-ке көбейту кестесі

Енді \(\displaystyle 21836\)-ні \(\displaystyle 53{\small } \)-ке бөлу процесін қарастырайық.

\(\displaystyle 21836\) санындағы алғашқы \(\displaystyle 2\)саны \(\displaystyle 53{\small }\)-тен кем екенін ескерейік .

Әрі қарай, келесі екі сан \(\displaystyle 21{\small }\)санын береді , ол да \(\displaystyle 53{\small } \)-тен кем.

Демек, бірден алғашқы үш санды, яғни \(\displaystyle 218{\small }\) санын бірге аламыз .

1 қадам.

\(\displaystyle \color{orange}{218}\)-ді \(\displaystyle 53\)-ке қалдықпен бөлейік

1. \(\displaystyle 218\)-ді \(\displaystyle 53\)-ке қалдықпен бөлейік.

\(\displaystyle 218{\small }\)-ден максималды қанша \(\displaystyle \color{blue}{53}\) алуға болатынын табайық .

Яғни, \(\displaystyle 218\)-ді \(\displaystyle \color{blue}{53}\)-ке қалдықпен бөлгендегі толымсыз бөліндіні табамыз.

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

онда \(\displaystyle 53\)-ті \(\displaystyle \color{green}{4}\)рет алуға болады.

Сондықтан \(\displaystyle \color{green}{4}\)-ті бөліндіге жазамыз:

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

2. Әрі қарай, бағанда \(\displaystyle 218\)-ден көбейтінді \(\displaystyle \color{blue}{53}\cdot \color{green}{4}=\color{green}{212}{\small }\) азайтамыз:

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

3. \(\displaystyle 218{\underline 3}6{\small }\)ондықтар разрядынан санды аламыз:

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

\(\displaystyle 63{\small }\)санын алдық.

2 қадам.

\(\displaystyle \color{cyan}{63}\)-ті на \(\displaystyle 53\)-ке қалдықпен бөлейік

1. \(\displaystyle 63\)-ті \(\displaystyle 53{\small}\)-ке бөлейік

\(\displaystyle 63{\small}\)-тен максималды қанша \(\displaystyle \color{blue}{53}\) алуға болатынын табайық.

Яғни, \(\displaystyle 63\)-ті \(\displaystyle \color{blue}{53}{\small }\)-ке бөлгендегі бөліндіні табамыз.

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

онда \(\displaystyle 53 \)-ті \(\displaystyle \color{green}{1}\) рет алуға болады.

Сондықтан \(\displaystyle \color{green}{1}\)-ді келесі санмен бөліндіге жазамыз:

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

2. Әрі қарай, бағанда \(\displaystyle 63\)-тен көбейтінді \(\displaystyle \color{blue}{53}\cdot \color{green}{1}=\color{green}{53}{\small }\) азайтамыз:

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

3. \(\displaystyle 2183{\underline 6}{\small }\)бірліктер разрядынан санды аламыз

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

\(\displaystyle 106{\small }\)санын алдық

Шаг 3.

\(\displaystyle \color{red}{106}\)-ны на \(\displaystyle 53\)-ке қалдықпен бөлейік

1. \(\displaystyle 106\)-ны \(\displaystyle 53{\small}\)-ке бөлейік.

\(\displaystyle 106{\small }\)-дан максималды қанша \(\displaystyle \color{blue}{53}\) алуға болатынын табайық.

Яғни, \(\displaystyle 106\)-ны \(\displaystyle \color{blue}{53}{\small }\)-ке бөлгендегі бөліндіні табамыз.

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

онда \(\displaystyle 53\)-ті \(\displaystyle \color{green}{2}\)рет алуға болады.

Сондықтан \(\displaystyle \color{green}{2}\)-ні келесі санмен бөліндіге жазамыз:

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

2. Әрі қарай, бағанда \(\displaystyle 106\)-дан көбейтінді \(\displaystyle \color{blue}{53}\cdot \color{green}{3}=\color{green}{106}{\small }\) азайтамыз:

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

\(\displaystyle 0{\small }\) алдық, бөлу процесі аяқталды.

Осылайша,

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

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

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Теориясы: Екі таңбалы сандарға бағандап бөлу