Let's find a six-digit square number where the product of the digits is 190512.
Among the digits, none is zero, each is a single-digit divisor of 190,512, which is the product of powers of different prime numbers $2^{4}, 3^{5}, 7^{2}$. According to this, all digits except 5 can appear in the desired $N$ square number, since every other digit can be formed as a product from the factors 2, 3, and 7, and 1 can also be among the digits. We also see that $N$ has two 7s, since $7 \cdot 2=14$ is already not a digit; Thus, the product of the other four digits is $a b c d=2^{4} \cdot 3^{5}$.

Among these, at least one is divisible by 9, because even if each of the four had one 3 as a prime factor, two 3s would end up in one of them, making it 9. And since $9 \cdot 2=18$ is already not a digit, one of the other four digits, say $d=9$. The four 2s as factors are distributed among $a, b, c$, so at least one of them gets at least two, and since $2^{2} \cdot 3>9$, that digit cannot also have a 3 as a factor, so another digit is $3^{2}=9$. Let this be $c$, then $a b=2^{4} \cdot 3=48$, which can only be factored into two digits in one way: $6 \cdot 8$, so the digits of $N$ are, in some order, $6,7,7,8$, 9,9.

Writing down the smallest and largest numbers that can be formed from the found digits,

$$
677899 \leq N \leq 998776,
$$

from which the base $n$ of the square

$$
824 \leq \sqrt{N}=n \leq 999
$$

The sum of the digits of $N$ is 46, which gives a remainder of 1 when divided by 9. Therefore, $N$ is not divisible by 9, $n$ is not divisible by 3, and more precisely, $n=9 \pm 1$ in form, and the sum of the digits of $n$ gives a remainder of 1 or 8 when divided by 9, since $(9 k \pm 2)^{2}=9 m+4$ and $(9 k \pm 4)^{2}=9 m+7$ in form, so both themselves and the sum of their digits give a remainder of 4 or 7 when divided by 9.

The last digit of $N$ among the found ones can only be 6 and 9. Moving on to the last two digits, only an odd digit can stand before a 6, and only an even digit before a 9, because if 6 stands at the end of $N$, then $n$ is even, so $N$ is divisible by 4, so its last two digits are also, $10 k+6=10(k-1)+16$, which only works if $k-1$ is even, i.e., the tens digit $k$ is odd; similarly, if 9 stands at the end of $N$, then $n$ is of the form $2 k+1$, and $N$ is $4 k(k+1)+1=8 m+1=8(m-1)+9$ in form, leaving a number divisible by 8 when the last 9 is omitted, so the tens digit of $N$ is even.

If the last two digits of a square number are given, the last two digits of the base can only be four types (except when the square ends in 00 or 25, which does not apply here). If $A^{2}$ and $B^{2}$ have the same last two digits - where $A$ and $B$ are integers and neither is divisible by 5 - then their difference is divisible by 100:

$$
A^{2}-B^{2}=(A-B)(A+B)=100 k=4 \cdot 25 \cdot k \text {. }
$$

Here, only one of $A-B$ and $A+B$ can be divisible by 25 (if both were divisible by 5, then $A$ and $B$ would also be, and their squares would end in 00 or 25). Moreover, both are even (since $A-B$ and $A+B$ have the same parity, and their product is divisible by $2^{2}$), so either $A+B$ or $A-B$ is divisible by 50. Therefore, if a 5-divisible $c$ number has a last two digits between 1 and 24, then $c^{2},(c+50)^{2},(50-c)^{2}$ and $(100-c)^{2}$ have the same last two digits, and a different ending of the base results in a different ending of the square.

According to this, the following two-digit endings correspond:

| In $N$ | In $n$ | $\left(2^{\prime}\right)$ |
| :---: | :---: | ---: |
| 76 | $24,26,74,76 ;$ | $n \geq 889 ;$ |
| 96 | $14,36,64,86 ;$ | $n \geq 883 ;$ |
| 69 | $13,37,63,87 ;$ | $n \geq 883 ;$ |
| 89 | $17,33,67,83 ;$ | $n \leq 988$. |

We immediately noted the (2') inequality, which replaces one side of (2) after the limits of (1) become narrower due to the digits at the end of $N$.

According to this, if $N$ ends in 76, then $n$ can only be $924,926,974$ and 976, but none of them are solutions, because their remainders when divided by 9 are 6, 8, 2, and 4, respectively, and in the case of the 8 remainder, $926^{2}=857476$, disallowed digits appear.

Similarly, for the 96, 69, and 89 square endings, (2') gives 5, 5, and 7 possible $n$ values, respectively, of which only the following 4 have a remainder of 1 or 8 when divided by 9:

964; $937 ; 883$ and 917,

and calculating their squares, only two of them are valid:

$$
937^{2}=877969, \quad 883^{2}=779689
$$

This solves the problem.

Remarks. 1. After determining the digits $6,7,7,8,9,9$ of the sought number, we can also use the square table found in the school function table for its search. Next to the possible four-digit beginnings of $N$, we write the three-digit number uniquely determined by them (i.e., those whose four digits are among the allowed digits of $N$, as well as those that can result from rounding a number composed of the digits of $N$):

| First four digits of $N$ | $n$ | Evaluation | First four digits of $N$ | $n$ | Evaluation |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $6790 .$. | 824 | $-(1)$ | $8780 .$. | 937 | to try |
| $6889 .$. | 830 | no 0 | $8798 .$. | 938 | $-(2)$ |
| $6989 .$. | 836 | $-(1)$ | $8968 .$. | 947 | to try |
| $7797 .$. | 883 | to try | $8987 .$. | 948 | $-(2)$ |
|  |  |  |  |  |  |

(1) means: $4^{2}$ or $6^{2}$ ends in 6, and we only have one 6.

(2) means: $8^{2}$ ends in 4, and we have no 4.

Of these, five can be immediately discarded due to the last digit of their squares, and from the remaining four numbers, the correct ones, $883^{2}$ and $937^{2}$, can be selected by precisely calculating their squares.

As a sufficiently justified systematic trial, this is also acceptable, as