Chủ đề này có 1 trả lời

[ChuDong2008]

Tìm các chữ số a, b, c biết:
$\overline{abc}\vdots 7$ và $ (a+b+c) \vdots 7 $.

[perfectstrong]

Lời giải
\[
\begin{array}{l}
\left. \begin{array}{l}
7|a + b + c \\
1 \le a + b + c \le 27 \\
\end{array} \right\} \Rightarrow a + b + c \in \left\{ {7;14;21} \right\} \\
\overline {abc} = 100a + 10b + c = 98a + 7b + 2a + 3b + c = 98a + 7b + \left( {a + b + c} \right) + a + 2b \Rightarrow 7|a + 2b \\
1 \le a + 2b \le 27 \Rightarrow a + 2b \in \left\{ {7;14;21} \right\} \\
TH1:a + b + c = 7 \\
*a + 2b = 7 = a + b + c \Rightarrow b = c \Rightarrow \overline {abc} \in \left\{ {133;322;511} \right\} \\
*a + 2b = 14 = 2a + 2b + 2c \Rightarrow a + 2c = 0:False \\
*a + 2b = 21 = 3a + 3b + 3c \Rightarrow 2a + b + 3c = 0:False \\
TH2:a + b + c = 14 \\
*a + 2b = 7 = \frac{{a + b + c}}{2} \Rightarrow a + 3b = c \Rightarrow 2a + 4b = a + b + c = 7:VN\left( {a;b} \right) \in N^* \times N^* \\
*a + 2b = 14 = a + b + c \Rightarrow b = c \Rightarrow \overline {abc} \in \left\{ {266;455;644;833} \right\} \\
*a + 2b = 21 = \frac{3}{2}\left( {a + b + c} \right) \Rightarrow b = a + 3c \Rightarrow 2a + 4c = a + b + c = 14 \Rightarrow \overline {abc} \in \left\{ {392;581} \right\} \\
TH3:a + b + c = 21 \\
*a + 2b = 7 = \frac{{a + b + c}}{3} \Rightarrow 2a + 5b = c \Rightarrow 3a + 6b = a + b + c = 21 \Rightarrow \left( {a;c} \right) = \left( {1;3} \right);\left( {3;2} \right);\left( {5;1} \right) \Rightarrow b > 10:False \\
*a + 2b = 14 = \frac{2}{3}\left( {a + b + c} \right) \Rightarrow a + 4b = 2c \Rightarrow 2a + 5b = a + b + c = 21 \Rightarrow \left( {a;c} \right) = \left( {8;1} \right);\left( {3;3} \right) \Rightarrow c > 10:False \\
*a + 2b = 21 = a + b + c \Rightarrow b = c \Rightarrow \overline {abc} \in \left\{ {399;588;777;966} \right\} \\
\overline {abc} \in \left\{ {133;266;322;392;399;455;511;581;588;644;777;833;966} \right\} \\
\end{array}
\]