Smallest number that satisfies all the following conditions. Divided by 7 gives remainder of 4. Divided by 8 gives remainder of 5. Divided by 9 gives remainder of 6

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 \( \boxed{501} \).

---

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.

Learn how to solve a system of congruences to find the smallest number that satisfies the conditions: x ≡ 4 (mod 7), x ≡ 5 (mod 8), and x ≡ 6 (mod 9). This problem involves using the Chinese Remainder Theorem and multiplicative inverses, suitable for advanced high school students.

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