Find the last 2 digits of the smallest 5-digit number that when divided by 8, 11, and
12 gives a remainder of 1, 2, and 5 respectively.

To solve this problem, we need to find the smallest 5-digit number \( N \) such that:

\[
\begin{aligned}
N &\equiv 1 \ (\text{mod} \ 8), \\
N &\equiv 2 \ (\text{mod} \ 11), \\
N &\equiv 5 \ (\text{mod} \ 12).
\end{aligned}
\]

We'll use the Chinese Remainder Theorem (CRT) to solve these congruences.

### Step 1: Rewrite the congruences
Given:

\[
\begin{aligned}
N &\equiv 1 \ (\text{mod} \ 8), \\
N &\equiv 2 \ (\text{mod} \ 11), \\
N &\equiv 5 \ (\text{mod} \ 12).
\end{aligned}
\]

Since \( 12 = 3 \times 4 \), and 8 is divisible by 4, we can consider the system:

\[
\begin{aligned}
N &\equiv 5 \ (\text{mod} \ 3), \\
N &\equiv 5 \ (\text{mod} \ 4).
\end{aligned}
\]

The first congruence \( N \equiv 5 \ (\text{mod} \ 3) \) reduces to \( N \equiv 2 \ (\text{mod} \ 3) \) since \( 5 \equiv 2 \ (\text{mod} \ 3) \).

Thus, the new system is:

\[
\begin{aligned}
N &\equiv 1 \ (\text{mod} \ 8), \\
N &\equiv 2 \ (\text{mod} \ 11), \\
N &\equiv 2 \ (\text{mod} \ 3), \\
N &\equiv 5 \ (\text{mod} \ 4).
\end{aligned}
\]

### Step 2: Solve the system of congruences
Let's solve \( N \) modulo 8, 11, and 12 directly.

1. **Congruence 1 and 2**: Combine \( N \equiv 1 \ (\text{mod} \ 8) \) and \( N \equiv 2 \ (\text{mod} \ 11) \).

Assume \( N = 8k + 1 \). Substitute this into the second congruence:

\[
8k + 1 \equiv 2 \ (\text{mod} \ 11)
\]

Simplifying:

\[
8k \equiv 1 \ (\text{mod} \ 11)
\]

To solve for \( k \), find the multiplicative inverse of 8 modulo 11. The inverse of 8 modulo 11 is 7 (since \( 8 \times 7 = 56 \equiv 1 \ (\text{mod} \ 11) \)).

Thus:

\[
k \equiv 7 \ (\text{mod} \ 11)
\]

Let \( k = 11m + 7 \), so:

\[
N = 8(11m + 7) + 1 = 88m + 57
\]

2. **Congruence 3**: Now use \( N = 88m + 57 \) in \( N \equiv 5 \ (\text{mod} \ 12) \):

\[
88m + 57 \equiv 5 \ (\text{mod} \ 12)
\]

Since \( 88 \equiv 4 \ (\text{mod} \ 12) \) and \( 57 \equiv 9 \ (\text{mod} \ 12) \), we get:

\[
4m + 9 \equiv 5 \ (\text{mod} \ 12)
\]

\[
4m \equiv -4 \equiv 8 \ (\text{mod} \ 12)
\]

Since the inverse of 4 modulo 12 is 3:

\[
m \equiv 8 \times 3 = 24 \equiv 0 \ (\text{mod} \ 3)
\]

Let \( m = 3n \), so:

\[
N = 88(3n) + 57 = 264n + 57
\]

We want the smallest 5-digit number, so:

\[
264n + 57 \geq 10000
\]

Solving for \( n \):

\[
264n \geq 9943
\]
\[
n \geq \left\lceil \frac{9943}{264} \right\rceil = 38
\]

Substitute \( n = 38 \) into the equation for \( N \):

\[
N = 264(38) + 57 = 10032 + 57 = 10089
\]

### Conclusion
The smallest 5-digit number \( N \) that satisfies the given conditions is 10089. The last two digits of \( N \) are **89**.

---

Would you like more details, or have any questions?

Here are some related questions:
1. How does the Chinese Remainder Theorem apply to solving congruences?
2. What are multiplicative inverses in modular arithmetic?
3. Can we solve systems of congruences with non-coprime moduli?
4. How can we check if our solution is correct using direct substitution?
5. How does the step-by-step solution process work for congruences involving large numbers?

**Tip:** When solving modular equations, finding the smallest solution often requires combining congruences systematically to simplify the problem.

Learn how to find the last two digits of the smallest 5-digit number that satisfies specific modular conditions using the Chinese Remainder Theorem. This problem involves modular arithmetic and systematic solving of congruences. Suitable for advanced high school students and above.

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