295. Find the number that is three times less than the square of the sum of its digits.
295. $3 N=\underbrace{(a+b+c+\ldots+l)^{2}}_{n \text { digits }}$. Since a digit does not exceed 9, then $3 N \leqslant 81 n^{2}, \quad N \leqslant 27 n^{2} ;$ from this it is clear that the desired number $N$ should contain no more than three digits, and for $n=3 N \leqslant 243$.

Let $n=1$, then $3 a=a^{2}$, hence $a=3$, therefore, $N=3$.

Let $n=2$, then $(a+b)^{2}=3(10 a+b)$, hence $\left\{\begin{aligned} 10 a+b & =3 m^{2} \\ a+b & =3 m\end{aligned}\right.$

By subtraction we find $3 a=m(m-1)$, hence $a$ is an even digit and at least one of the numbers $m$ and $m-1$ is divisible by 3. Considering that $3 a \leqslant 24$, we conclude that $m \leqslant 5$, hence $m=3,4 ; a=2,4 ; b=7,8$. The desired numbers are 27 and 48.

Let $n=3$, then $a=1,2$.

If $a=1$, then we have:

$$
\left\{\begin{array}{c}
100+10 b+c=3 m^{2} \\
1+b+c=3 m
\end{array}\right.
$$

from which $\left\{\begin{array}{l}3 m^{2} \geqslant 100 \\ 3 m \leqslant 19\end{array}\right.$; hence, $6 \leqslant m \leqslant 6, m=6$. Substituting $m=6$ into the system (1), we get $\left\{\begin{array}{c}10 b+c=8 \\ b+c=17\end{array}\right.$; from this it follows that there are no valid values for $b$ and $c$.

If $a=2$, then $\left\{\begin{array}{c}200+10 b+c=3 m^{2} \\ 2+b+c=3 m\end{array}\right.$, from which

66
$\left\{\begin{array}{l}3 m^{2} \geqslant 200 \\ 3 m \leqslant 20\end{array} ;\right.$ hence $\left\{\begin{array}{l}m>8 \\ m \leqslant 6\end{array}\right.$, which is impossible.

Answer. $3,27,48$.

$3 b=m(m-1)-66$. By trial we find $m=9, b=2$; $m=10, b=8$; but then we will not get valid values for $c$.

Answer. $3,27,48$.