Basic descriptions

Self-descriptive numbers in base 10,7,5,4 respectively are 6210001000, 3211000, 21200, 2020.

Suppose $\overline{x_0x_1x_2x_3x_4x_5}$ is a self-descriptive number in base 6. As a consequence of being a self-descriptive number, $x_0+x_1+x_2+x_3+x_4+x_5=6.$

If $x_5\neq 0$ , then some $x_i=5$ . But then four $x_j$ must be $0$ , so $x_0=4$ and $x_4=0$ , a contradiction. Thus, $x_5=0$ .

If $x_4\neq 0$ , then some $x_i=4$ . But then at least three $x_j$ must be $0$ , so $x_0=3$ or $x_0=4$ . Since the sum of the $x_j$ is $6$ , $x_0=4$ . Then $x_1=x_2=x_3=0,$ so $x_4$ cannot count the number of 4's appearing as digits in $\overline{x_0x_1x_2x_3x_4x_5}$ . By contradiction, $x_4=0$ .

If $x_3\neq 0$ , then since the sum of the $x_j$ is $6$ (and $x_4=x_5=0$ ), $x_0=2$ or $x_0=3$ . Assume $x_0=2$ . Then $\{x_1,x_2,x_3\}=\{1,1,2\}$ , so $x_3=0,$ a contradiction. If $x_0=3$ , then similarly $\{x_1,x_2,x_3\}=\{0,1,2\}$ , but then $x_1=x_2=1$ , a contradiction. Therefore, $x_3=0$ .

Thus, we have shown $x_0\ge 3$ , but $x_3=x_4=x_5=0$ , yielding a contradiction. Therefore, there is no delf-descriptive number in base 6.