If the 4-digit number $x67y$ is exactly divisible by 9, then what is the least value of $(x + y)$?

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**Solution:**

1. For a number to be divisible by 9, the sum of its digits must be divisible by 9.

2. The number is $x67y$, so the sum of digits is:
   $$x + 6 + 7 + y = x + y + 13$$

3. We want $x + y + 13$ to be divisible by 9.

4. Let $S = x + y$. Then:
   $$S + 13 \equiv 0 \pmod{9}$$

5. Since $13 \equiv 4 \pmod{9}$, we have:
   $$S + 4 \equiv 0 \pmod{9} \implies S \equiv -4 \equiv 5 \pmod{9}$$

6. So, $S$ must be congruent to 5 modulo 9. The smallest non-negative integer congruent to 5 modulo 9 is 5.

7. Check if $S=5$ is possible with digits $x$ and $y$ (each between 0 and 9):
   - For example, $x=0$, $y=5$ or $x=1$, $y=4$ etc. are valid.

8. However, the question asks for the *least* value of $(x + y)$.

9. But the options given are 9, 0, 5, and 3. Among these, 5 is the smallest value congruent to 5 modulo 9.

10. Let's verify if $S=5$ works:
    - Sum of digits = $5 + 13 = 18$, which is divisible by 9.

11. Check $S=3$:
    - Sum = $3 + 13 = 16$, not divisible by 9.

12. Check $S=0$:
    - Sum = $0 + 13 = 13$, not divisible by 9.

13. Check $S=9$:
    - Sum = $9 + 13 = 22$, not divisible by 9.

Therefore, the least value of $(x + y)$ is $5$.

**Answer:** $5$

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