Matematică, întrebare adresată de robertrazvan1010, 8 ani în urmă

Determină numărul abc stiind, că abc + bc + cb + 6•b = 379​

Răspunsuri la întrebare

Răspuns de Triunghiul1
20

Răspuns:

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\bf \boxed{\boxed{\bf \;\; \overline{abc}=193\;\; }}

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Explicație pas cu pas:

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\bf \overline{abc} +\overline{bc} +\overline{cb} + 6b=379

\bf \;\bf \;

\left.\begin{aligned} \bf \overline{abc} =100a+10b+c  \\ \\  \bf \overline{bc}=10b+c \;\;\;\;\;\;\;\;\;\;   \\   \\ \bf \overline{cb}=10c+b \;\;\;\;\;\;\;\;\;\; \end{aligned} \; \;  \right\} \implies \bf  \overline{abc} +\overline{bc} +\overline{cb} + 6b =

\bf  \;

\bf 100a+10b+c+10b+c+10c+b+6b=379

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\bf 100a + 27b+12c=379 \implies \overline{abc}  <379

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\bf \implies a \in \{1;2;3\} \;\;  \mathsf{Luam \; pe\;  cazuri \; in \; functie \; de \; a:}

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\bf a=1 \;  \Bigg| \implies 100a+27b+12c=100+27b+12c=379\;  \Bigg|_{-100 }

\bf 27b+12c=279\;  \Bigg| : 3 \iff 9b+4c=93   \; \;-  convine

\bf \Big(27b\; ;12c\Big)  \;  \vdots \; \: 3 \implies \Big(9b+4c  \Big) \; \vdots \;\: 3

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\left.\begin{aligned} \bf 4c\; \vdots \;\: 3 \implies 3 \in M_{4c}    \\ \\\bf  (4\; ; 3)=1-prime  \end{aligned}\;\;  \right\}\bf  \implies c \: \; \vdots \;\: 3 \iff c \in M_{3}=\{3;6;9\}

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\bf c=3 \; \Bigg| \implies 9b+4c=9b+12=93 \Big| _{-12} \iff b=\dfrac{81}{9} =9

\bf \implies \boxed{\bf \overline{abc}=193}

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\bf c=6 \; \Bigg| \implies 9b+4c=9b+24=93\;\Big|_{-24} \iff b= \dfrac{69}{9} \notin \mathbb{N}

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\bf c=9\; \Bigg| \implies 9b+4c=9b+36=93\;\Big|_{-36} \iff b=\dfrac{57}{9} \notin \mathbb{N}

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\bf a=2 \; \Bigg| \implies 100a+27b+12c=200+27b+12c=379 \; \Bigg|_{-200}

\bf 27b+12c=179 \; \Bigg| : 3 \iff 9b+4c= \dfrac{179}{3} \; -nu \; convine

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\bf a=3 \; \Bigg| \implies 100a+27b+12c=300+27b+12c=379\; \Bigg|_{-300}

\bf 27b+12c=79\; \Bigg| : 3 \iff 9b+4c = \dfrac{79}{3} \; -nu\; convine

\bf \;

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