Find the number

$H + U = D^2$ and $H + U = 2D$ , which means that $D^2 = 2D$ and $D = 2$ .

So, $H + U = 4$ . Now, we have to look for the smallest number. So, the value of $H$ has to be the smallest possible. The value of $H$ cannot be $0$ because then the number would be a 2-digit number. So, $H = 1$ and $U = 3$ and the number is $\boxed{123}$

Let $10 > H > 0$ , $10 > D \geq 0$ and $10 > U \geq 0$

$H + U = D^2$

$H + U = 2D$

$H + D + U$ is also a multiple of three

The first two equations can be combined to create

$D^2 = 2D$

This can be re-arranged to find $D$

$D = 2$

Now we know that

$H + U = 4$

Since $H$ s lowest possible value is one, having $H$ as one will help us find the lowest possible number.

$1 + U = 4$

This makes $U$ equal to three.

$H + D + U = 1 + 2 + 3 = 6$

Six is divisible by three so

$\overline {HDC} = 123$

i know if (h+d+u) is divided by 3 then (hdu) is divided by 3

The smallest integer solution is -529, not 123.

At first I tried 024 i.e. 24. I think 24 should be the min. integer. Satisfying all conditions.