The problem states:

• A number when divided by 119 leaves a remainder of 19.
• If the same number is divided by 17, the remainder is to be determined.

Let's denote the unknown number as $N$. From the given information:

1. $N \equiv 19 \pmod{119}$, meaning $N = 119k + 19$ for some integer $k$.

Now, we need to find the remainder when $N$ is divided by 17.

Since $N = 119k + 19$, we will check how $N$ behaves modulo 17:

$N = 119k + 19$ Now, reduce 119 modulo 17: $119 \div 17 = 7 \quad \text{remainder} \, 0 \quad \text{so}, \, 119 \equiv 0 \pmod{17}$ Thus, $N = 119k + 19 \equiv 0k + 19 \equiv 19 \pmod{17}$.

Now reduce $19 \mod 17$: $19 \div 17 = 1 \quad \text{remainder} \, 2$

Therefore, the remainder when the number is divided by 17 is 2.

The correct answer is $\boxed{2}$ (option (d)).

Would you like more details or have any other questions?

Here are 5 related questions to explore this topic further:

1. What is the method of solving modular arithmetic problems systematically?
2. How does the Chinese Remainder Theorem work for solving simultaneous congruences?
3. Can you solve for the remainder if the number is divided by another number, say 13?
4. What are some real-life applications of modular arithmetic?
5. How do you simplify larger numbers in modular arithmetic?

Tip: When working with modular arithmetic, reducing large numbers step by step helps avoid errors and makes calculations simpler.