Let the two-digit number be $N$. The question states that the digits of $N$ are non-zero. We can represent the number $N$ as $10a + b$, where $a$ and $b$ are integers from 1 to 9.

The square of this number, $N^2$, is described as the number $N$ itself preceded by a digit $C$. This means that $N^2$ can be written in the form $C \times 100 + N$, where $C$ is a single digit. Since $N$ is a two-digit number ($10 \le N \le 99$), its square $N^2$ can be a three-digit number (e.g., $10^2=100$) or a four-digit number (e.g., $32^2=1024$). The structure $C \times 100 + N$ implies that $N^2$ is a three-digit number where the hundreds digit is $C$ and the number formed by the last two digits is $N$. This is only possible if $N^2$ is a three-digit number.

So, we have the equation:

$$N^2 = 100C + N$$

We can rearrange this equation:

$$N^2 - N = 100C$$

Factor out $N$ from the left side:

$$N(N-1) = 100C$$

Here, $N$ is a two-digit number with non-zero digits (so $N \in \{11, 12, ..., 19, 21, ..., 99\}$, excluding numbers ending in 0). $N-1$ is the integer immediately preceding $N$. $N$ and $N-1$ are consecutive integers, which means they are coprime, i.e., $\gcd(N, N-1) = 1$.

The right side of the equation is $100C$. Since $C$ is a digit, $C$ can be any integer from 1 to 9 (as the leading digit $C$ cannot be zero for a number preceded by a digit). Thus, $100C$ can be $100, 200, 300, 400, 500, 600, 700, 800, 900$.

So, we are looking for a two-digit number $N$ (with non-zero digits) such that the product of $N$ and $N-1$ is one of these multiples of 100.

The product $N(N-1)$ must be a multiple of $100 = 2^2 \times 5^2$. Since $N$ and $N-1$ are coprime, the factors $2^2=4$ and $5^2=25$ must be distributed between $N$ and $N-1$. This means either $N$ is divisible by 25 and $N-1$ is divisible by 4, or $N$ is divisible by 4 and $N-1$ is divisible by 25.

Check Possible Two Digit Numbers

Let's consider the possible values for $N$ and check the condition $N(N-1) = 100C$ for some digit $C$. $N$ must have non-zero digits.

• If $N$ is a multiple of 25: The possible two-digit numbers with non-zero digits are 25 and 75. (50 is excluded as it has a zero digit).
  • Let $N=25$. The digits are 2 and 5, which are non-zero. Calculate $N(N-1)$: $25 \times (25-1) = 25 \times 24 = 600$. We set this equal to $100C$: $600 = 100C \implies C = 600 / 100 = 6$. Since $C=6$ is a single digit (and non-zero), this is a potential solution. Let's check the original condition: The number is 25. Its square is $25^2 = 625$. Is 625 the number (25) preceded by the digit C (6)? Yes, 625 is 25 preceded by 6 ($600 + 25$). The condition is satisfied.
  • Let $N=75$. The digits are 7 and 5, which are non-zero. Calculate $N(N-1)$: $75 \times (75-1) = 75 \times 74 = 5550$. Set this equal to $100C$: $5550 = 100C \implies C = 5550 / 100 = 55.5$. This value of $C$ is not a single digit, so $N=75$ is not a solution.
• If $N-1$ is a multiple of 25: Possible values for $N-1$ are 25, 50, 75 (assuming $N$ is a two-digit number, $N-1$ would be between 9 and 98).
  • Let $N-1=25 \implies N=26$. The digits are 2 and 6, which are non-zero. Calculate $N(N-1)$: $26 \times 25 = 650$. Set this equal to $100C$: $650 = 100C \implies C = 650 / 100 = 6.5$. This value of $C$ is not a single digit, so $N=26$ is not a solution.
  • Let $N-1=50 \implies N=51$. The digits are 5 and 1, which are non-zero. Calculate $N(N-1)$: $51 \times 50 = 2550$. Set this equal to $100C$: $2550 = 100C \implies C = 2550 / 100 = 25.5$. This value of $C$ is not a single digit, so $N=51$ is not a solution.
  • Let $N-1=75 \implies N=76$. The digits are 7 and 6, which are non-zero. Calculate $N(N-1)$: $76 \times 75 = 5700$. Set this equal to $100C$: $5700 = 100C \implies C = 5700 / 100 = 57$. This value of $C$ is not a single digit, so $N=76$ is not a solution.

Conclusion

The only two-digit number with non-zero digits that satisfies the condition is $N=25$, which results in $C=6$. The square of 25 is 625, which is indeed 25 preceded by the digit 6.

Therefore, the value of $C$ is 6.