Toán tim số có 3 chữ số abc thỏa mãn abc = (a+b)^2 nhân 4c 27/09/2021 By Abigail tim số có 3 chữ số abc thỏa mãn abc = (a+b)^2 nhân 4c
Ta có: \(\overline {abc} = {\left( {a + b} \right)^2}.4c \Rightarrow c\) là số chẵn và \(x \ne 0 \Rightarrow c \ge 2;\,\,\,\,\left( {c;\,\,5} \right) = 1.\) \(\begin{array}{l} \Rightarrow \overline {abc} < 1000 \Rightarrow 1000 > {\left( {a + b} \right)^2}.4c \ge 8{\left( {a + b} \right)^2}\\ \Rightarrow {\left( {a + b} \right)^2} < 125 \Rightarrow \left( {a + b} \right) \le 11\end{array}\) Lại có: \(\overline {abc} = {\left( {a + b} \right)^2}.4c \Rightarrow 10\overline {ab} = c\left[ {4{{\left( {a + b} \right)}^2} - 1} \right] \vdots \,\,5 \Rightarrow \left[ {4{{\left( {a + b} \right)}^2} - 1} \right] \vdots \,\,5\,\,\,\,\left( {do\,\,\,\left( {c;\,\,5} \right) = 1} \right).\) \( \Rightarrow 5{\left( {a + b} \right)^2} - 5 - \left[ {4{{\left( {a + b} \right)}^2} - 1} \right] = \left( {a + b + 2} \right)\left( {a + b - 2} \right)\,\, \vdots \,\,5\) \( \Rightarrow a + b\) chia 5 dư 2 hoặc 3. Mà \(a + b \le 11 \Rightarrow \left[ \begin{array}{l}a + b = 2\\a + b = 3\\a + b = 7\\a + b = 8\end{array} \right.\) \( + )\,\,\,a + b = 8 \Rightarrow \overline {abc} = 256c < 1000,\,\,\,c > 0,\,\,\,c\) chẵn \( \Rightarrow c = 2 \Rightarrow \overline {abc} = 512\,\,\,\left( {ktm} \right).\) \( + )\,\,\,a + b = 7 \Rightarrow \overline {abc} = 196c < 1000,\,\,\,c > 0,\,\,\,c\) chẵn \( \Rightarrow \left[ \begin{array}{l}c = 2 \Rightarrow \overline {abc} = 392\,\,\,\left( {ktm} \right)\\c = 4 \Rightarrow \overline {abc} = 784\,\,\,\,\left( {ktm} \right)\end{array} \right..\) \( + )\,\,\,a + b = 2 \Rightarrow \overline {abc} = 16 > 99,\,\,\,c > 0,\,\,\,c\) chẵn \( \Rightarrow c = 8 \Rightarrow \overline {abc} = 16.8 = 128\,\,\,\left( {ktm} \right).\) \( + )\,\,\,a + b = 3 \Rightarrow \overline {abc} = 36c \Rightarrow 10\overline {ab} = 35c\, \vdots \,\,7 \Rightarrow \overline {ab} \,\, \vdots \,\,7\) \(\begin{array}{l}a + b = 3 \Rightarrow \left[ \begin{array}{l}\overline {ab} = 30\\\overline {ab} = 21\\\overline {ab} = 12\end{array} \right. \Rightarrow 21\,\, \vdots \,\,7 \Rightarrow \overline {abc} = 216.\\ \Rightarrow \overline {abc} = 216.\end{array}\) Trả lời
Ta có: \(\overline {abc} = {\left( {a + b} \right)^2}.4c \Rightarrow c\) là số chẵn và \(x \ne 0 \Rightarrow c \ge 2;\,\,\,\,\left( {c;\,\,5} \right) = 1.\)
\(\begin{array}{l} \Rightarrow \overline {abc} < 1000 \Rightarrow 1000 > {\left( {a + b} \right)^2}.4c \ge 8{\left( {a + b} \right)^2}\\ \Rightarrow {\left( {a + b} \right)^2} < 125 \Rightarrow \left( {a + b} \right) \le 11\end{array}\) Lại có: \(\overline {abc} = {\left( {a + b} \right)^2}.4c \Rightarrow 10\overline {ab} = c\left[ {4{{\left( {a + b} \right)}^2} - 1} \right] \vdots \,\,5 \Rightarrow \left[ {4{{\left( {a + b} \right)}^2} - 1} \right] \vdots \,\,5\,\,\,\,\left( {do\,\,\,\left( {c;\,\,5} \right) = 1} \right).\) \( \Rightarrow 5{\left( {a + b} \right)^2} - 5 - \left[ {4{{\left( {a + b} \right)}^2} - 1} \right] = \left( {a + b + 2} \right)\left( {a + b - 2} \right)\,\, \vdots \,\,5\) \( \Rightarrow a + b\) chia 5 dư 2 hoặc 3. Mà \(a + b \le 11 \Rightarrow \left[ \begin{array}{l}a + b = 2\\a + b = 3\\a + b = 7\\a + b = 8\end{array} \right.\) \( + )\,\,\,a + b = 8 \Rightarrow \overline {abc} = 256c < 1000,\,\,\,c > 0,\,\,\,c\) chẵn \( \Rightarrow c = 2 \Rightarrow \overline {abc} = 512\,\,\,\left( {ktm} \right).\)
\( + )\,\,\,a + b = 7 \Rightarrow \overline {abc} = 196c < 1000,\,\,\,c > 0,\,\,\,c\) chẵn \( \Rightarrow \left[ \begin{array}{l}c = 2 \Rightarrow \overline {abc} = 392\,\,\,\left( {ktm} \right)\\c = 4 \Rightarrow \overline {abc} = 784\,\,\,\,\left( {ktm} \right)\end{array} \right..\)
\( + )\,\,\,a + b = 2 \Rightarrow \overline {abc} = 16 > 99,\,\,\,c > 0,\,\,\,c\) chẵn \( \Rightarrow c = 8 \Rightarrow \overline {abc} = 16.8 = 128\,\,\,\left( {ktm} \right).\)
\( + )\,\,\,a + b = 3 \Rightarrow \overline {abc} = 36c \Rightarrow 10\overline {ab} = 35c\, \vdots \,\,7 \Rightarrow \overline {ab} \,\, \vdots \,\,7\)
\(\begin{array}{l}a + b = 3 \Rightarrow \left[ \begin{array}{l}\overline {ab} = 30\\\overline {ab} = 21\\\overline {ab} = 12\end{array} \right. \Rightarrow 21\,\, \vdots \,\,7 \Rightarrow \overline {abc} = 216.\\ \Rightarrow \overline {abc} = 216.\end{array}\)
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