Willie finds that he can arrange bills in sets of seven with six left over, sets of 8 with one left over, or sets of 15 with three left over. If there are fewer than 200 bills how many are there?

We are tasked with finding how many bills Willie has, given that:

- When arranged in sets of 7, there are 6 left over.
- When arranged in sets of 8, there is 1 left over.
- When arranged in sets of 15, there are 3 left over.
- There are fewer than 200 bills.

We can represent these conditions using the following system of congruences:

\[
\begin{aligned}
x &\equiv 6 \pmod{7}, \\
x &\equiv 1 \pmod{8}, \\
x &\equiv 3 \pmod{15}.
\end{aligned}
\]

### Step 1: Solve the first two congruences

We begin with the system:

\[
\begin{aligned}
x &\equiv 6 \pmod{7}, \\
x &\equiv 1 \pmod{8}.
\end{aligned}
\]

Let \( x = 7k + 6 \) for some integer \( k \). Substituting this into the second congruence:

\[
7k + 6 \equiv 1 \pmod{8}.
\]

Simplifying:

\[
7k \equiv -5 \pmod{8}.
\]

Since \( -5 \equiv 3 \pmod{8} \), this becomes:

\[
7k \equiv 3 \pmod{8}.
\]

Now, we solve for \( k \). The inverse of 7 modulo 8 can be found by trial and error, or by using the extended Euclidean algorithm. We find that:

\[
7 \times 7 = 49 \equiv 1 \pmod{8}.
\]

Thus, multiplying both sides of \( 7k \equiv 3 \pmod{8} \) by 7:

\[
k \equiv 21 \pmod{8}.
\]

Since \( 21 \equiv 5 \pmod{8} \), we have:

\[
k \equiv 5 \pmod{8}.
\]

Therefore, \( k = 8m + 5 \) for some integer \( m \). Substituting this back into \( x = 7k + 6 \):

\[
x = 7(8m + 5) + 6 = 56m + 35 + 6 = 56m + 41.
\]

Thus, we now have:

\[
x \equiv 41 \pmod{56}.
\]

### Step 2: Incorporate the third congruence

We now solve the system:

\[
\begin{aligned}
x &\equiv 41 \pmod{56}, \\
x &\equiv 3 \pmod{15}.
\end{aligned}
\]

Let \( x = 56n + 41 \) for some integer \( n \). Substituting this into the second congruence:

\[
56n + 41 \equiv 3 \pmod{15}.
\]

Since \( 56 \equiv 11 \pmod{15} \) and \( 41 \equiv 11 \pmod{15} \), this simplifies to:

\[
11n + 11 \equiv 3 \pmod{15}.
\]

Simplifying further:

\[
11n \equiv -8 \pmod{15}.
\]

Since \( -8 \equiv 7 \pmod{15} \), we have:

\[
11n \equiv 7 \pmod{15}.
\]

Now, we find the inverse of 11 modulo 15, which is 11 (since \( 11 \times 11 = 121 \equiv 1 \pmod{15} \)). Multiplying both sides by 11:

\[
n \equiv 77 \pmod{15}.
\]

Since \( 77 \equiv 2 \pmod{15} \), we have:

\[
n \equiv 2 \pmod{15}.
\]

Thus, \( n = 15p + 2 \) for some integer \( p \). Substituting this back into \( x = 56n + 41 \):

\[
x = 56(15p + 2) + 41 = 840p + 112 + 41 = 840p + 153.
\]

Thus, \( x \equiv 153 \pmod{840} \).

### Step 3: Find the solution under 200

We are looking for \( x < 200 \). The smallest solution to \( x \equiv 153 \pmod{840} \) is \( x = 153 \).

Thus, Willie has **153 bills**.

---

Would you like more details or have any further questions?

Here are five related questions:
1. How do we find the inverse of a number modulo another number?
2. What is the significance of the Chinese Remainder Theorem in solving systems of congruences?
3. How can we check if the solution \( x = 153 \) satisfies all three initial conditions?
4. What if the total number of bills was restricted to a different range, say below 500?
5. How does modular arithmetic apply in cryptographic systems?

**Tip:** When solving systems of congruences, it's useful to work step-by-step, reducing the system progressively with each pair of congruences.

Learn how to solve a modular arithmetic problem involving sets of bills with remainders when divided by 7, 8, and 15. Using the Chinese Remainder Theorem, we find the total number of bills under 200. Suitable for Grades 9-12.

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