Square digits

Let $x = 100a + 10b + c$ , where a, b, and c are single digit positive integers.

What do we know? $x^{2}$ has to end in $396$ , so it has to end in $96$ . Let's focus on $b$ and $c$ for now and come back to $a$ later.

Whatever $a$ is, $(10b+c)^{2}$ must end in 96. Expanding, we get $100b^{2}$ + $20bc + c^{2} = ...96$ . We know that $100b^{2}$ is not going to affect the last two digits as it is already at least 100. That gives us $20bc + c^{2} = ...96$ . Because $c$ is a single digit, it must be either $4$ or $6$ for the expression to end in 6 because $20bc$ will only affect the tens digit. Casework:

$c = 4$ ,

$80b + 16 = ...96$ ,

$80b = ...80$ . From this, we can see that $b$ is either 1 or 6, giving us 2 cases.

$c=6$ ,

$120b + 36 = ...96$ ,

$120b = ...60$ . From this we can see that $b$ can be 3 or 8, giving us 2 more cases.

We know that the three digit number $x$ has to end in 14, 16, 36 or 86. Now we use $a$ , so more casework:

$(100a + 14)^{2} = ...396$ ,

$10000a^{2} + 2800a + 196 = ...396$ .

$10000a^{2}$ clearly won't affect the hundreds, tens, or ones digit, so we can throw that out.

$2800a + 196 = ...396$ ,

$2800a = ...200$ . From this, we can see that $a$ must be $4$ , giving us one answer, $414$ .

We should still check the others to make sure we have the smallest, though.

Similar expansions with $100a + 64$ and $100a + 36$ give:

$12800a = ...700$ , from which we can see no value of $a$ will make 8 into 7, and

$7200a = ...100$ , from which we can also see no value of $a$ will make 2 into 1.

Expansion with $(100a + 86)^{2}$ gives $10000a^{2} + 1720a + 7396 = ...396$ ,

so $1720a = ...000$ , so $a$ can be 0 or 5. But wait! $x$ has to be 3-digit, so $a$ cannot be 0. That gives us only one more answer, 586.

414 is clearly less than 586. Therefore, our answer is $414$ and the sum of the digits is $\boxed{9}$ .