To solve the problem, we need to find all three-digit numbers $n$ that satisfy the following conditions:

1. $n \equiv 1 \pmod{4}$
2. $n \equiv 3 \pmod{5}$

Step 1: Use the Chinese Remainder Theorem

We solve the simultaneous congruences: $n \equiv 1 \pmod{4}$ $n \equiv 3 \pmod{5}.$

The moduli 4 and 5 are coprime, so we can use the Chinese Remainder Theorem. The combined modulus is: $\text{LCM}(4, 5) = 20.$

Thus, we rewrite the problem as: $n \equiv x \pmod{20}.$

To find $x$, solve: $x \equiv 1 \pmod{4},$ $x \equiv 3 \pmod{5}.$

Step 2: Solve the congruences

Let $x = 4k + 1$ (from $x \equiv 1 \pmod{4}$). Substitute into the second congruence: $4k + 1 \equiv 3 \pmod{5}.$

Simplify: $4k \equiv 2 \pmod{5}.$

Since $4 \equiv -1 \pmod{5}$, we rewrite the congruence: $-k \equiv 2 \pmod{5} \quad \implies \quad k \equiv -2 \equiv 3 \pmod{5}.$

Thus, $k = 5m + 3$ for some integer $m$. Substitute back: $x = 4k + 1 = 4(5m + 3) + 1 = 20m + 13.$

Therefore: $x \equiv 13 \pmod{20}.$

So, the solution to the congruences is: $n \equiv 13 \pmod{20}.$

Step 3: Find all three-digit numbers

The general form of $n$ is: $n = 20k + 13,$ where $n$ is a three-digit number ($100 \leq n \leq 999$).

Solve for $k$: $100 \leq 20k + 13 \leq 999.$

Subtract 13: $87 \leq 20k \leq 986.$

Divide by 20: $4.35 \leq k \leq 49.3.$

Since $k$ must be an integer: $k = 5, 6, \dots, 49.$

Count the integers: $k = 49 - 5 + 1 = 45.$

Final Answer:

There are $\boxed{45}$ three-digit numbers that satisfy the given conditions.

Let me know if you'd like detailed step-by-step explanations or have questions!

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1. What is the Chinese Remainder Theorem and when is it applicable?
2. How do we find solutions to simultaneous congruences involving coprime moduli?
3. Can this approach work if the moduli (4 and 5) were not coprime?
4. How do you determine the range of $k$ in problems like this?
5. What are some other applications of modular arithmetic in mathematics?

Tip: When solving modular arithmetic problems, always verify that the moduli are coprime if you use the Chinese Remainder Theorem.