Đáp án: $\overline{abcd}=2178$
Giải thích các bước giải:
Ta có:
$\overline{abcd}\cdot 4=\overline{dcba}(*)$
$\to (1000a+100b+10c+d)\cdot 4=1000d+100c+10b+a$
$\to (1000a+100b+10c+d)\cdot 4-(1000d+100c+10b+a)=0$
$\to 3999a+390b-60c-996d=0$
$\to 1333a+130b-20c-332d=0$
$\to 1333a=20c+332d-130b$
Ta có $abcd$ là chữ số $\to 1\le a, d\le 9, 0\le b, c\le 9$
$\to 20\cdot 0+332\cdot 1-130\cdot 9\le 20c+332d-130b\le 20\cdot 9+332\cdot 9-130\cdot 0$
$\to 20\cdot 0+332\cdot 1-130\cdot 9\le 1333a\le 20\cdot 9+332\cdot 9-130\cdot 0$
$\to 1\le a\le 2$ vì $1\le a\le 9$
Mà từ $(*)\to a$ chẵn
$\to a=2$
$\to 1333\cdot 2=20c+332d-130b$
$\to 20c+332d-130b=2666$
$\to 332d=130b+2666-20c$
Ta có $130\cdot 0+2666-20\cdot 9\le 130b+2666-20c\le 130\cdot 9+2666-20\cdot 0$
$\to 130\cdot 0+2666-20\cdot 9\le 332d\le 130\cdot 9+2666-20\cdot 0$
$\to 8\le d\le 11$
Mà $d\cdot 4 $ có tận cùng là $2\to d=8$
$\to 332\cdot 8=130b+2666-20c$
$\to 130b-20c=-10$
$\to 13b-2c=-1$
$\to 13b=2c-1$
$\to b$ lẻ
Mà $2\cdot 0-1\le 2c-1\le 2\cdot 9-1$
$\to 2\cdot 0-1\le 13b\le 2\cdot 9-1$
$\to 0\le b\le 1$
$\to b=1\to c=7$
$\to \overline{abcd}=2178$ thỏa mãn
