In a study of ancient bird nesting sites, a zoologist models population cycles using a number \( N \) such that \( N \equiv 3 \pmod{4} \), \( N \equiv 2 \pmod{5} \), and \( N \equiv 4 \pmod{7} \). What is the smallest positive \( N \) satisfying all three conditions?

["Finding the Smallest Ancient Nesting Population: Using Modular Arithmetic to Model Bird Population Cycles", "In studies of ancient avian populations, zoologists often seek to understand recurring cycles in nesting behaviors by analyzing historical site data. A recent model developed by a researcher integrates modular arithmetic to uncover patterns in bird nesting site usage. The model hinges on solving a system of congruences that describe population recurrence over time:", "We are given:\n[\n\begin{align}\nN &\equiv 3 \pmod{4}, \\nN &\equiv 2 \pmod{5}, \\nN &\equiv 4 \pmod{7}.\n\end{align}\n]", "The goal is to find the smallest positive integer ( N ) satisfying all three conditions—a problem perfectly suited to the Chinese Remainder Theorem (CRT), a cornerstone of number theory with real-world applications in ecological modeling.", "### Step 1: Solve the First Two Congruences", "We begin by solving:\n[\n\begin{cases}\nN \equiv 3 \pmod{4}, \\nN \equiv 2 \pmod{5}.\n\end{cases}\n]", "Let ( N = 4k + 3 ) for some integer ( k ). Substitute into the second congruence:\n[\n4k + 3 \equiv 2 \pmod{5} \Rightarrow 4k \equiv -1 \pmod{5} \Rightarrow 4k \equiv 4 \pmod{5}.\n]", "Multiply both sides by the modular inverse of 4 modulo 5. Since ( 4 \cdot 4 = 16 \equiv 1 \pmod{5} ), the inverse is 4:\n[\nk \equiv 4 \cdot 4 = 16 \equiv 1 \pmod{5}.\n]", "Thus, ( k = 5m + 1 ) for some integer ( m ), and substituting back:\n[\nN = 4(5m + 1) + 3 = 20m + 7.\n]", "So, ( N \equiv 7 \pmod{20} ).", "### Step 2: Incorporate the Third Congruence", "Now solve:\n[\n\begin{cases}\nN \equiv 7 \pmod{20}, \\nN \equiv 4 \pmod{7}.\n\end{cases}\n]", "Let ( N = 20m + 7 ). Substitute into the second congruence:\n[\n20m + 7 \equiv 4 \pmod{7} \Rightarrow 20m \equiv -3 \pmod{7}.\n]", "Reduce coefficients modulo 7: ( 20 \equiv 6 \pmod{7} ), and ( -3 \equiv 4 \pmod{7} ), so:\n[\n6m \equiv 4 \pmod{7}.\n]", "Find the inverse of 6 modulo 7. Since ( 6 \cdot 6 = 36 \equiv 1 \pmod{7} ), the inverse is 6:\n[\nm \equiv 6 \cdot 4 = 24 \equiv 3 \pmod{7}.\n]", "Thus, ( m = 7n + 3 ) for some integer ( n ), and:\n[\nN = 20(7n + 3) + 7 = 140n + 60 + 7 = 140n + 67.\n]", "The smallest positive ( N ) occurs when ( n = 0 ), so:\n[\nN = 67.\n]", "### Step 3: Verify the Solution", "Check that ( N = 67 ) satisfies all original congruences:", "- ( 67 \div 4 = 16 ) remainder ( 3 \Rightarrow 67 \equiv 3 \pmod{4} ) ✓\n- ( 67 \div 5 = 13 ) remainder ( 2 \Rightarrow 67 \equiv 2 \pmod{5} ) ✓\n- ( 67 \div 7 = 9 ) remainder ( 4 \Rightarrow 67 \equiv 4 \pmod{7} ) ✓", "All conditions are satisfied.", "### Conclusion", "This solves the zoologist’s model: the smallest possible population count ( N ) consistent with all three modular constraints—possibly representing the first year in a recurring nesting cycle—minus constraints tied to ancient ecological patterns. Using modular arithmetic, we consistently reduce complex biological signals into precise mathematical structure.", "Answer: The smallest positive integer ( N ) satisfying the system is (\boxed{67})."]