In your comment, you said that the given solution was 9 • 9 + 1 • 9 + 8 • 9 = 162. I'll attempt to explain a logic that yields that calculation.
Consider the 3-digit numbers that start with two identical digits. There are 9 choices of the first digit (and inherently the second digit): 11, 22, 33, 44, 55, 66, 77, 88, and 99 (not 00 because the number is in the range 100-999). For each of these, there are 9 choices of the final digit (0-9, except whatever digit was already chosen for the first two). So, there are 9 • 9 such numbers.
Now, suppose that the number ends with two identical digits. There are 10 choices for the last digit (and inherently the second-to-last digit): 00, 11, 22, 33, 44, 55, 66, 77, 88, and 99, but we need to treat 00 separately from the rest. If the number ends with 00, then the first digit can be 1-9, so 9 choices, so 1 • 9. If the number ends with 11-99, there are 8 choices of first digit (1-9 except the digit already chosen), so 9 • 8.
While I have the 8 and 9 transposed in the final term, this is term-by-term the same expression as in the solution you gave.
You misunderstood the problem.
Just assume encountering any three-digit number is equally likely regardless of history. You can completely ignore the letters.
(I believe I'm not doing any wrong by saying this here.)
Best Answer
Your thinking is correct. You simply messed up the algebra (and the inequality sign). Fixing the result we have
$$100 \leq 2k+1 \leq 999$$ $$99 \leq 2k \leq 998$$ $$ \frac{99}{2} \leq k \leq 499$$