Let $x = 60k$. Then we solve $60k \equiv 1 \pmod{7}$. Since $60 \equiv 4 \pmod{7}$, we solve:

["Title: Solving the Congruence $60k \equiv 1 \pmod{7}$: A Step-by-Step Guide", "Meta Description:\nLearn how to solve the modular equation $60k \equiv 1 \pmod{7}$ using basic number theory concepts. This step-by-step guide explains key techniques like modular reduction and finding multiplicative inverses. Perfect for students and math enthusiasts.", "---", "## Solving $60k \equiv 1 \pmod{7}$: A Step-by-Step Explanation", "Modular arithmetic is a powerful tool in number theory and cryptography, and solving linear congruences like $60k \equiv 1 \pmod{7}$ is a common task. In this article, we’ll break down the solution clearly—whether you’re studying for a math exam or exploring basic modular equations.", "---", "### Step 1: Simplify the Coefficient Using Modulo 7", "The equation is:\n$$\n60k \equiv 1 \pmod{7}\n$$", "Instead of working directly with 60, simplify the coefficient modulo 7.\nCompute:\n$$\n60 \div 7 = 8 \ ext{ remainder } 4 \quad \Rightarrow \quad 60 \equiv 4 \pmod{7}\n$$", "So the congruence becomes:\n$$\n4k \equiv 1 \pmod{7}\n$$", "This simplification makes the problem easier to solve.", "---", "### Step 2: Understand What the Congruence Means", "We are looking for an integer $k$ such that:\n$4k \equiv 1 \pmod{7}$", "This means we want the multiplicative inverse of 4 modulo 7 — a value $k$ such that $4k \bmod 7 = 1$.", "---", "### Step 3: Solve $4k \equiv 1 \pmod{7}$", "We solve this by testing small integer values of $k$ from 0 to 6 (since modulo 7 repeats every 7):", "- $k = 0$: $4 \cdot 0 = 0 <br/>\not\equiv 1$\n- $k = 1$: $4 \cdot 1 = 4 <br/>\not\equiv 1$\n- $k = 2$: $4 \cdot 2 = 8 \equiv 1 \pmod{7}$ (since $8 - 7 = 1$)", "Success! When $k = 2$, we have:\n$$\n4 \cdot 2 = 8 \equiv 1 \pmod{7}\n$$", "---", "### Step 4: Verify the Solution", "Check:\n$$\n60 \cdot 2 = 120\n$$\nNow compute $120 \mod 7$:\n$$\n120 \div 7 = 17 \ ext{ remainder } 1 \quad \Rightarrow \quad 120 \equiv 1 \pmod{7}\n$$\n✔️ The solution satisfies the original congruence.", "---", "### Bonus: General Insight — Modular Inverses", "In modular arithmetic, the congruence $ax \equiv 1 \pmod{m}$ has a solution if and only if $a$ and $m$ are coprime (i.e., $\gcd(a, m) = 1$).", "Here, $\gcd(4, 7) = 1$, so an inverse exists — which explains why $k = 2$ is a valid solution.", "---", "### Summary", "We solved:\n$$\n60k \equiv 1 \pmod{7}\n\quad \ ext{by reducing } 60 \mod 7 \ ext{ to find } 4k \equiv 1 \pmod{7}\n$$", "Testing values or using inverse techniques confirms that:\n$$\nk \equiv 2 \pmod{7}\n$$", "This means $k = 2 + 7n$ for any integer $n$ is the general solution.", "---", "Want to master more modular arithmetic?\nPractice solving equations like $ak \equiv 1 \pmod{n}$, explore Euler’s theorem, or dive into cryptography applications — all rooted in these fundamentals.", "---", "Keywords: modular arithmetic, solve $60k \equiv 1 \pmod{7}$, inverse of 4 modulo 7, linear congruence, step-by-step modular solving, $k$ modulo 7, mathematical problem solving."]