The number $823p2q$ is exactly divisible by 7, 11, and 13. What is the value of $(p - q)$?

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**Solution:**

1. The number is $823p2q$ where $p$ and $q$ are digits.

2. The number is divisible by 7, 11, and 13. Since 7, 11, and 13 are prime and pairwise coprime, their product is $7 	imes 11 	imes 13 = 1001$.

3. Therefore, the number $823p2q$ must be divisible by 1001.

4. Dividing the number by 1001 is equivalent to checking:
   $$823p2q \equiv 0 \pmod{1001}$$

5. A known property is that for any number $N$,
   $$N \equiv (	ext{last 3 digits}) + 1000 	imes (	ext{remaining digits}) \pmod{1001}$$
   But more simply, since 1001 = 1000 + 1, the remainder when dividing by 1001 is the same as the remainder when dividing by 1000 + 1.

6. Another approach is to split the number into two parts:
   - First part: $823p$ (the first 4 digits)
   - Second part: $2q$ (the last 2 digits)

But since the number has 6 digits, let's write it as:
   $$N = 823000 + 100p + 20 + q = 823000 + 100p + 20 + q$$

7. Since $N$ is divisible by 1001, we have:
   $$N \equiv 0 \pmod{1001}$$

8. Calculate $823000 \mod 1001$:
   - Divide 823000 by 1001:
     $$823000 \div 1001 \approx 821.1788$$
   - Multiply 1001 by 821:
     $$1001 	imes 821 = 821821$$
   - Remainder:
     $$823000 - 821821 = 1179$$

Since 1179 > 1001, subtract 1001 again:
   $$1179 - 1001 = 178$$

So,
   $$823000 \equiv 178 \pmod{1001}$$

9. Now, the number modulo 1001 is:
   $$N \equiv 178 + 100p + 20 + q = 198 + 100p + q \pmod{1001}$$

10. For $N$ to be divisible by 1001,
    $$198 + 100p + q \equiv 0 \pmod{1001}$$

11. Since $p$ and $q$ are digits (0 to 9), the value of $198 + 100p + q$ must be either 0 or 1001 (or multiples, but only these are possible given the size).

12. Check if it equals 1001:
    $$198 + 100p + q = 1001$$
    $$100p + q = 1001 - 198 = 803$$

13. Since $p$ and $q$ are digits, $100p + q = 803$ implies $p=8$ and $q=3$.

14. Calculate $(p - q)$:
    $$8 - 3 = 5$$

**Answer:** $5$

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