The problem is asking for the smallest number $x$ that satisfies the following system of congruences:

$x \equiv 4 \ (\text{mod} \ 7)$ $x \equiv 5 \ (\text{mod} \ 8)$ $x \equiv 6 \ (\text{mod} \ 9)$

This is a classic case of solving a system of simultaneous congruences. We can approach this using the Chinese Remainder Theorem (CRT) or by solving step-by-step through substitution.

Step-by-step method:

1. Restate the conditions:

   • $x \equiv 4 \ (\text{mod} \ 7)$ means $x = 7k + 4$ for some integer $k$.
   • $x \equiv 5 \ (\text{mod} \ 8)$ means $x = 8m + 5$ for some integer $m$.
   • $x \equiv 6 \ (\text{mod} \ 9)$ means $x = 9n + 6$ for some integer $n$.

2. Substitute the first congruence into the second one: From $x = 7k + 4$, substitute into the second congruence $x \equiv 5 \ (\text{mod} \ 8)$: $7k + 4 \equiv 5 \ (\text{mod} \ 8)$ Simplifying: $7k \equiv 1 \ (\text{mod} \ 8)$ The multiplicative inverse of 7 modulo 8 is 7 (since $7 \times 7 = 49 \equiv 1 \ (\text{mod} \ 8)$). Therefore, multiply both sides by 7: $k \equiv 7 \ (\text{mod} \ 8)$ This means $k = 8p + 7$ for some integer $p$.

3. Substitute $k = 8p + 7$ into $x = 7k + 4$: $x = 7(8p + 7) + 4 = 56p + 49 + 4 = 56p + 53$ So now we have $x = 56p + 53$.

4. Substitute $x = 56p + 53$ into the third congruence $x \equiv 6 \ (\text{mod} \ 9)$: $56p + 53 \equiv 6 \ (\text{mod} \ 9)$ Simplifying modulo 9: $56p + 53 \equiv 2p + 8 \ (\text{mod} \ 9)$ $2p + 8 \equiv 6 \ (\text{mod} \ 9)$ $2p \equiv -2 \equiv 7 \ (\text{mod} \ 9)$ The multiplicative inverse of 2 modulo 9 is 5 (since $2 \times 5 = 10 \equiv 1 \ (\text{mod} \ 9)$). Multiply both sides by 5: $p \equiv 35 \ (\text{mod} \ 9)$ Simplifying $35 \ (\text{mod} \ 9)$, we get: $p \equiv 8 \ (\text{mod} \ 9)$ So $p = 9q + 8$ for some integer $q$.

5. Substitute $p = 9q + 8$ into $x = 56p + 53$: $x = 56(9q + 8) + 53 = 504q + 448 + 53 = 504q + 501$ Therefore, the general solution is $x = 504q + 501$.

6. Find the smallest positive $x$: For $q = 0$, we get $x = 501$.

Thus, the smallest number that satisfies all the given conditions is $\boxed{501}$.

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Would you like more details or have any questions on this? Here are some related questions for further exploration:

1. What is the Chinese Remainder Theorem and how does it apply to solving systems of congruences?
2. How do you find the multiplicative inverse modulo a number?
3. How would the solution change if the moduli were not coprime (e.g., 7, 8, and 10)?
4. How do you solve a system of linear congruences with different moduli step by step?
5. What happens if we add a constant to each congruence (e.g., $x \equiv 5 \ (\text{mod} \ 8)$ becomes $x \equiv 7 \ (\text{mod} \ 8)$)?

Tip: When solving congruences, always check for the multiplicative inverse when working with modular arithmetic—this is a key tool to simplify and solve such problems.