To solve this problem, let's determine the remainders when the order of divisors (5, 6, and 3) is reversed to (3, 6, 5). Here's how we can work through the solution step by step.

Step-by-Step Solution:

1. Original Divisor Order (5, 6, 3):

We are given:

• The number $N$ gives a remainder of 2 when divided by 5.
• The quotient from the first division is then divided by 6, giving a remainder of 3.
• The resulting quotient is divided by 3, giving a remainder of 1.

Let:

• $N = 5a + 2$ (from the first division, remainder 2),
• The quotient $a$ satisfies $a = 6b + 3$ (from the second division, remainder 3),
• Finally, $b$ satisfies $b = 3c + 1$ (from the third division, remainder 1).

Substituting these equations:

• $b = 3c + 1$,
• $a = 6(3c + 1) + 3 = 18c + 6 + 3 = 18c + 9$,
• $N = 5(18c + 9) + 2 = 90c + 45 + 2 = 90c + 47$.

Thus: $N = 90c + 47.$

2. Reversed Divisor Order (3, 6, 5):

Now, let's compute the remainders when $N = 90c + 47$ is divided by 3, 6, and 5 in reverse order.

• First Division (by 3): $90c + 47 \mod 3 = (90c \mod 3) + (47 \mod 3).$ Since $90c \mod 3 = 0$ (90 is divisible by 3), and $47 \mod 3 = 2$, the remainder is: $\text{Remainder} = 2.$

• Second Division (by 6): Divide $\frac{90c + 47}{3}$, which is $30c + 15$. The remainder when $30c + 15$ is divided by 6: $(30c \mod 6) + (15 \mod 6) = 0 + 3.$ The remainder is: $\text{Remainder} = 3.$

• Third Division (by 5): Divide $\frac{30c + 15}{6}$, which simplifies to $5c + 2.5$. We focus only on the integer part $5c + 2$. When $5c + 2$ is divided by 5: $(5c \mod 5) + (2 \mod 5) = 0 + 2.$ The remainder is: $\text{Remainder} = 2.$

Final Answer:

The remainders when the order is reversed (3, 6, 5) are 2, 3, 2.

Correct Option:

a. 2, 3, 2

Would you like further clarification or a breakdown of any specific part? Here are some related questions to deepen understanding:

1. How do modular arithmetic properties help in solving remainder problems?
2. Can the solution be generalized for other divisor and remainder combinations?
3. What is the mathematical significance of reversing the order of operations?
4. How would the result change if the divisors were not relatively prime?
5. What strategies can simplify handling such chained remainder problems?

Tip: Modular arithmetic is powerful in solving these problems. Always simplify step by step while keeping track of intermediate results to avoid errors.