Use the Chinese Remainder Theorem. Let $ L = 7k $. Substitute into the second congruence:

["Understanding the Chinese Remainder Theorem: Solving Congruences with L = 7k", "The Chinese Remainder Theorem (CRT) is a powerful result in number theory that provides a unique solution to systems of simultaneous congruences under certain conditions. Whether you're a student mastering modular arithmetic or a coder solving optimization problems, applying CRT efficiently can streamline calculations involving multiple modular constraints. This article explores how to use the Chinese Remainder Theorem effectively—especially when substituting expressions like ( L = 7k ) into second congruences—to construct precise solutions.", "---", "### What is the Chinese Remainder Theorem?", "The Chinese Remainder Theorem states that if you are given a system of simultaneous congruences with pairwise coprime moduli, there exists a unique solution modulo the product of those moduli. For example, if:", "[\n\begin{cases}\nx \equiv a_1 \pmod{m_1} \\nx \equiv a_2 \pmod{m_2} \\n\vdots \\nx \equiv a_k \pmod{m_k}\n\end{cases}\n]", "where ( m_1, m_2, \dots, m_k ) are pairwise coprime, then there exists a unique ( x \mod M ) (with ( M = m_1 m_2 \cdots m_k )) satisfying all these congruences.", "---", "### Setting Up the Problem: Let ( L = 7k )", "Suppose we are solving a congruence system and one expression is defined as ( L = 7k ), where ( k ) is an integer. Our goal may be to substitute this expression into a second congruence—say, ( L \equiv r \pmod{m} )—and use CRT to find ( k ) (or equivalently, ( L )) satisfying both conditions.", "For instance, let’s consider the second congruence:", "[\n7k \equiv r \pmod{m}\n]", "This congruence asks: For which integers ( k ) does ( 7k ) leave remainder ( r ) when divided by ( m )?", "---", "### Step-by-Step Substitution Using CRT", "Step 1: Set up the congruence involving ( L = 7k )", "We begin with:\n[\n7k \equiv r \pmod{m}\n]", "Step 2: Assume divisibility or express ( k ) in terms of a parameter", "We aim to find ( k ) modulo ( m_{\ ext{new}} ) such that this congruence holds. However, instead of solving directly, use an IMHO approach: solve for ( k ) using CRT principles — even in single congruence cases.", "Since ( \gcd(7, m) ) depends on ( m ), suppose ( m ) is not necessarily coprime with 7. But if ( m ) divides ( 7k - r ), then ( 7k \equiv r \pmod{m} ) implies ( k ) must satisfy linear constraints modulo ( \frac{m}{\gcd(7, m)} ).", "Step 3: Reduce using known divisibility", "Let us assume ( m ) is arbitrary and decompose the problem. To apply CRT effectively, we often work with pairwise coprime moduli. Suppose we have multiple congruences; combining them requires breaking them into compatible subsystems.", "Now, substitute ( L = 7k ) into a second congruence, such as:", "[\nL \equiv r \pmod{m} \implies 7k \equiv r \pmod{m}\n]", "To solve for ( k ), we want to isolate ( k ):", "[\nk \equiv 7^{-1} r \pmod{m / \gcd(7, m)}\n]", "provided that ( 7 ) has a multiplicative inverse modulo ( \frac{m}{\gcd(7, m)} ). If ( \gcd(7, m) = 1 ), then ( 7^{-1} \mod m ) exists and:", "[\nk \equiv 7^{-1} r \pmod{m}\n]", "---", "### Example应用: Numerical Illustration", "Let ( m = 10 ), ( r = 7 ), so:", "[\n7k \equiv 7 \pmod{10}\n]", "Divide both sides by ( \gcd(7, 10) = 1 ), so:", "[\nk \equiv 1 \pmod{10}, \quad \ ext{since } 7^{-1} \equiv 3 \pmod{10}, \ ext{ but } 7 \cdot 1 = 7 \Rightarrow k \equiv 1\n]", "Wait—here we simplified directly, but CRT shines when moduli are composite or involved in multiple constraints.", "Now suppose we also have another congruence: ( k \equiv 2 \pmod{3} )", "Now solve the system:\n[\n\begin{cases}\n7k \equiv 7 \pmod{10} \\nk \equiv 2 \pmod{3}\n\end{cases}\n]", "From the first: ( k \equiv 1 \pmod{10} ), so ( k = 10m + 1 )", "Substitute into second:\n[\n10m + 1 \equiv 2 \pmod{3} \implies 10m \equiv 1 \pmod{3}\n]\nBut ( 10 \equiv 1 \pmod{3} ), so:\n[\nm \equiv 1 \pmod{3} \implies m = 3n + 1\n]\nThen:\n[\nk = 10(3n + 1) + 1 = 30n + 11\n]", "Thus, the solution is ( k \equiv 11 \pmod{30} )", "---", "### Why This Matters: Practical Use of CRT", "Using ( L = 7k ) and substituting into congruences allows you to embed linear dependencies into modular systems. When combined via CRT, this approach efficiently reduces complex modular conditions into manageable equations—especially valuable in:", "- Cryptography (key generation, modular equations)\n- Algorithm design (synchronization, distributed systems)\n- Discrete math and number theory problem solving", "By expressing one variable in terms of another (like ( k = L/7 )) and substituting into congruences, CRT enables unique, constrained solutions that respect all modular limits.", "---", "### Summary", "- The Chinese Remainder Theorem provides unique solutions to systems of modular congruences with coprime moduli.\n- Expressing variables via linear forms (e.g., ( L = 7k )) allows clean substitution into second congruences.\n- Solving for unknowns via modular inverses and the structure of CRT transforms complex modular reasoning into step-by-step computations.\n- Understanding and applying these principles enhances problem-solving across mathematics, computer science, and engineering disciplines.", "---", "Next Steps: Practice substituting linear expressions like ( L = 7k ) into congruences, and apply CRT step-by-step when moduli are not pairwise coprime. Tools like the extended Euclidean algorithm and modular inverses are key to mastering such techniques.", "---", "Keywords: Chinese Remainder Theorem, modular arithmetic, solving congruences, substitution with L = 7k, number theory applications, CRT step-by-step, modular inverses, CRT examples"]