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 $0 \le N < 2520.$

To solve the system of congruences

\[
\begin{align*}
N &\equiv 2 \pmod{6}, \\
N &\equiv 1 \pmod{7}, \\
N &\equiv 9 \pmod{15},
\end{align*}
\]

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:

\[
\boxed{4}.
\]

Find the number of solutions to N ≡ 2 (mod 6), N ≡ 1 (mod 7), N ≡ 9 (mod 15) in the interval 0 ≤ N < 2520.

Discover how to find the number of solutions to the system of congruences N ≡ 2 (mod 6), N ≡ 1 (mod 7), N ≡ 9 (mod 15) within the interval 0 ≤ N < 2520. This detailed guide covers essential number theory concepts and modular arithmetic techniques.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation