23. Find a three-digit number, the square of which is a six-digit number such that each subsequent digit, counting from left to right, is greater than the previous one.
23. Since a square can end with the digits $0,1,4,5,6$ or 9, and each subsequent digit of a six-digit number is greater than the previous one, a six-digit number that is the square of a three-digit number can end with either the digit 6 or 9. If it ends with 6, it would be written as 123456. But this number is not a perfect square. Therefore, it must end with the digit 9, which means the sought three-digit number ends with the digit 3 or 7. The leftmost digit of the six-digit number cannot be 4, as the number 456789 is not a perfect square. Therefore, it is less than 4. Let's consider $1 * * * * 9$, then the three-digit number is either $4 * 7$ or $3 * 7$. But 407, $417,427,437,447$ do not satisfy the condition, and $457^{2}=208849$. (We get a number with the leftmost digit not 1, but 2.) We will look for numbers of the form $3 * 7$. Checking shows that such a number is $367\left(367^{2}=134689\right)$.