Gọi số đó là \(\overline{abc}\left(a,b,c\in N\right)\)
Ta có \(b^2=ac;\overline{abc}-\overline{cba}=495\)
\(\Rightarrow100a+10b+c-100c-10b-a=495\\ \Rightarrow99a-99c=495\\ \Rightarrow a-c=5\)
| a | 5 | 6 | 7 | 8 | 9 |
| c | 0 | 1 | 2 | 3 | 4 |
| b | 0 | \(\sqrt{6}\) | \(\sqrt{14}\) | \(\sqrt{21}\) | 6 |
Vậy số thỏa mãn là 500;964
Ta có:
\(\overline{abc}-\overline{cba}=495\)
\(\Rightarrow\) 100a + 10b + c - (100c + 10b + a) = 495
100a + 10b + c - 100c - 10b - a = 495
(100a - a) + (10b - 10b) + (c - 100c) = 495
99a + 0 + (-99c) = 495
\(\Rightarrow\) 99a - 99c = 495
99 . (a - c) = 495
a - c = 495 : 99
a - c = 5
\(\Rightarrow\) a; c \(\in\) {(5; 0); (6; 1); (7; 2); (8; 3); (9; 4)}
Mà ta có b2 = a.c
\(\Rightarrow\) a; b; c \(\in\) {(5; 0; 0); (9; 6; 4)}
\(\Rightarrow\) \(\overline{abc}\text{}\in\left\{500;964\right\}\)
