Math Problem Statement
Find the number of solutions to \begin{align*} N &\equiv 2 \pmod{6}, \ N &\equiv 1 \pmod{7}, \ N &\equiv 9 \pmod{15} \end{align*}in the interval $0 \le N < 2520.$
Solution
To solve the system of congruences
N &\equiv 2 \pmod{6}, \\ N &\equiv 1 \pmod{7}, \\ N &\equiv 9 \pmod{15}, \end{align*}$$ we will use the method of successive substitutions. ### Step 1: Solve the first two congruences First, consider the congruences $$N \equiv 2 \pmod{6}$$ and $$N \equiv 1 \pmod{7}.$$ Let $$N = 6k + 2$$ for some integer $$k$$. Substituting this into the second congruence gives: $$6k + 2 \equiv 1 \pmod{7}.$$ Simplifying this, we have: $$6k \equiv -1 \pmod{7} \quad \Rightarrow \quad 6k \equiv 6 \pmod{7} \quad \Rightarrow \quad k \equiv 1 \pmod{7}.$$ Thus, $$k = 7m + 1$$ for some integer $$m$$. Substituting back, we get: $$N = 6(7m + 1) + 2 = 42m + 6 + 2 = 42m + 8.$$ This means: $$N \equiv 8 \pmod{42}.$$ ### Step 2: Combine with the third congruence Now we have: $$N \equiv 8 \pmod{42}$$ and $$N \equiv 9 \pmod{15}.$$ Let $$N = 42n + 8$$ for some integer $$n$$. Substituting this into the third congruence gives: $$42n + 8 \equiv 9 \pmod{15}.$$ Calculating $$42 \mod 15$$: $$42 \equiv 12 \pmod{15}.$$ Thus, we have: $$12n + 8 \equiv 9 \pmod{15} \quad \Rightarrow \quad 12n \equiv 1 \pmod{15}.$$ To solve $$12n \equiv 1 \pmod{15}$$, we need to find the multiplicative inverse of 12 modulo 15. The extended Euclidean algorithm shows: $$12 \cdot 3 \equiv 36 \equiv 6 \pmod{15}.$$ Since $$6$$ and $$15$$ are not coprime, we continue with the coefficients, leading to: $$12 \cdot 6 \equiv 72 \equiv 12 \pmod{15}.$$ The multiplicative inverse of 12 mod 15 can be tested through trials or systematic substitutions. We find: $$12n \equiv 1 \Rightarrow n \equiv 8 \pmod{15}.$$ Thus, $$n = 15p + 8$$ for some integer $$p$$. Substituting back for $$N$$: $$N = 42(15p + 8) + 8 = 630p + 336 + 8 = 630p + 344.$$ Thus: $$N \equiv 344 \pmod{630}.$$ ### Step 3: Counting the solutions Now we need to find how many solutions $$N = 630p + 344$$ are there in the range $$0 \leq N < 2520$$. Setting up the inequality: $$0 \leq 630p + 344 < 2520.$$ Solving the left side: $$630p + 344 \geq 0 \quad \Rightarrow \quad p \geq -\frac{344}{630} \quad \Rightarrow \quad p \geq 0 \quad \text{(since $$p$$ must be a non-negative integer)}.$$ Solving the right side: $$630p + 344 < 2520 \quad \Rightarrow \quad 630p < 2520 - 344 = 2176 \quad \Rightarrow \quad p < \frac{2176}{630} \approx 3.45.$$ Thus, $$p$$ can take the values $$0, 1, 2, 3$$ (four integer values). ### Conclusion The number of solutions $$N$$ in the interval $$0 \leq N < 2520$$ is: $$\boxed{4}.$$Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Number Theory
Modular Arithmetic
Systems of Congruences
Formulas
N ≡ a (mod m)
N = km + b
Theorems
Chinese Remainder Theorem
Suitable Grade Level
Grades 10-12
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