Question - Finding a Specific Number Based on Conditions

Let the three-digit number be represented as \( xyz \), where \( x, y, z \) are its digits.

According to the problem, if we append the digit 2 to the right of \( xyz \), we form the number \( xyz2 \).

We require that \( xyz2 > 4106 \).

Express \( xyz2 \) as \( 1000x + 100y + 10z + 2 \).

Thus, we want to find \( 1000x + 100y + 10z + 2 > 4106 \).

By simplifying, we have \( 1000x + 100y + 10z > 4104 \).

Dividing through by 10 gives \( 100x + 10y + z > 410.4 \).

Since \( 100x + 10y + z \) is an integer, it must be at least 411 for the condition to hold.

Therefore, we need to find all three-digit combinations \( xyz \) such that \( 100x + 10y + z \geq 411 \).

Based on the constraints of digit values from 0 to 9, we can deduce suitable values for \( x, y, z \).