Square Root a Number my Way

Question

Now that we know how to Square a Number my Way, we need an inverse operation, a way to Square Root a Number my Way. To square a number my way, you simply stack it on top of itself a number of times equal to the number of digits it contains, and then take read off every number that is formed both vertically and horizontally, and then add them together. More information about squaring a number in this manner can be found here. To square root a number my way, you simply take the number that has the least decimal digits that squares (my way) to make the number you are square rooting. For example, since 12 can be formed by squaring 6, and there are no numbers with fewer decimal digits that can be squared to form 12, the square root of 12 is 6.

Your Task:

Write a program or function that takes an integer and square roots it as outlined above.

Input:

An integer or string

Output:

A float or string square rooted as outlined above.

Test Cases:

1263 -> 125
57   -> 12

Scoring:

This is code-golf, lowest score in bytes wins!

Answer 1

Python 2, 101 bytes

Ports from my answer to square the numbers my way.

x,z=1,input()
while sum(float(i*len(z))for z in[[i for i in`x`if"/"<i]]for i in[x]+z)!=z:x+=1
print x

Try it online!

Answer 2

Python 3, 112 bytes

f=lambda x,k=0:k if(lambda x:int(x)*len(x)+sum(map(int,map(''.join,zip(*[x]*len(x))))))(str(k))==x else f(x,k+1)

Try it online!

Answer 3

Mathematica, 130 bytes

(For[i=1,(t=Length[s=#&@@RealDigits[#]//.{a___,0}:>{a}];If[IntegerPart@#==0,t++];t#+Tr[FromDigits@Table[#,t]&/@s])&[i]!=#,i++];i)&

other test cases

5994 -> 999
79992 -> 9999
999990 -> 99999

Answer 4

Pyth, 20 bytes

xms+RdsM*RF_lB@jkUT`

Test Suite.

As all the other answers, this assumes the input is a "perfect square (my way)". This also uses the fact that the square (my way) of an integer is always higher than or equal to itself. Also, this is a port of Erik's solution to that challenge. Porting my solution from that challenge would result in a longer submission.

Answer 5

PHP, 92 90 89 84+1 bytes

brute force (assuming that the input is a perfect square)
using my solution from Square a Number my Way:

while((0|$k/=10)||$s-$argn&&$k=++$n+$s=0)$s+=$n+str_repeat($k%10,strlen($n));echo$n;

Run as pipe with -nR or try it online.