We are given two conditions about the number $n$:

1. When $n$ is divided by 8, the remainder is 1. This means: $n \equiv 1 \pmod{8}$
2. When $n$ is divided by 5, the remainder is 2. This means: $n \equiv 2 \pmod{5}$

We are asked to determine the remainder when $3n$ is divided by 40.

Step 1: Solving the system of congruences

We need to solve the system: $n \equiv 1 \pmod{8}$ $n \equiv 2 \pmod{5}$ This is a system of simultaneous congruences, which can be solved using the Chinese Remainder Theorem (CRT).

Start with the first congruence:

$n = 8k + 1 \quad \text{for some integer } k$

Now, substitute this expression for $n$ into the second congruence: $8k + 1 \equiv 2 \pmod{5}$

Simplifying this: $8k \equiv 1 \pmod{5}$ Since $8 \equiv 3 \pmod{5}$, the congruence becomes: $3k \equiv 1 \pmod{5}$ We now need to solve this congruence for $k$. The multiplicative inverse of 3 modulo 5 is 2 (since $3 \times 2 = 6 \equiv 1 \pmod{5}$), so multiply both sides of the equation by 2: $k \equiv 2 \pmod{5}$ Thus, $k = 5m + 2$ for some integer $m$.

Step 2: Substitute back to find $n$

Now substitute this expression for $k$ back into $n = 8k + 1$: $n = 8(5m + 2) + 1 = 40m + 17$ Thus, $n = 40m + 17$.

Step 3: Find $3n \mod 40$

Now, we need to calculate $3n \mod 40$. Since $n = 40m + 17$, we have: $3n = 3(40m + 17) = 120m + 51$ Now, take this modulo 40: $3n \equiv 51 \pmod{40}$ $51 - 40 = 11$ Thus, $3n \equiv 11 \pmod{40}$.

Final Answer:

The remainder when $3n$ is divided by 40 is $\boxed{11}$.

Would you like more details on any of the steps or have any further questions?

Here are 5 related questions you might find useful:

1. How does the Chinese Remainder Theorem work in solving modular systems?
2. How do you find multiplicative inverses in modular arithmetic?
3. What are some applications of the Chinese Remainder Theorem?
4. How can we generalize this method for larger moduli?
5. How does modular arithmetic simplify problems in number theory?

Tip: When solving modular arithmetic problems, break them down step-by-step and use known properties like inverses to simplify the process.