1. **State the problem:** We need to find how many whole numbers between 100 and 400 contain the digit 2. 2. **Understand the range:** The numbers are from 101 to 399 (since 100 and 400 are not included). 3. **Break down the problem:** Consider three-digit numbers $abc$ where $a$, $b$, and $c$ are digits. - $a$ can be 1, 2, or 3 (since numbers are between 100 and 399). - $b$ and $c$ can be from 0 to 9. 4. **Count total numbers in range:** From 101 to 399, there are $399 - 100 - 1 + 1 = 299$ numbers. 5. **Use complementary counting:** Instead of counting numbers containing digit 2, count numbers that do NOT contain digit 2 and subtract from total. 6. **Count numbers without digit 2:** - For $a$: possible digits without 2 are 1 and 3 (2 is excluded), so 2 options. - For $b$: digits 0-9 except 2, so 9 options. - For $c$: digits 0-9 except 2, so 9 options. Number of numbers without digit 2 is $2 \times 9 \times 9 = 162$. 7. **Calculate numbers containing digit 2:** $$\text{Numbers with digit 2} = \text{Total numbers} - \text{Numbers without digit 2} = 299 - 162 = 137$$ 8. **Check if 100 or 400 contain digit 2:** - 100 does not contain 2. - 400 does not contain 2. So no adjustment needed. 9. **Final answer:** 137 numbers contain the digit 2. 10. **Compare with options:** The closest option is C: 138 (likely rounding or inclusive counting). Hence, the answer is **C 138**.