a number being successively divided by 3, 5 and 8 leaves remainders 1, 4 and 7 respectively find the respective remainders if the order of divisors be reversed

Let's denote the number as \( N \). We are given the following conditions:

1. \( N \) when divided by 3 gives a remainder of 1.
2. The quotient from the first division, when divided by 5, gives a remainder of 4.
3. The quotient from the second division, when divided by 8, gives a remainder of 7.

Let's represent these conditions mathematically:

\[
N = 3a + 1
\]

Where \( a \) is the quotient when \( N \) is divided by 3.

Next, \( a \) divided by 5 leaves a remainder of 4:

\[
a = 5b + 4
\]

Where \( b \) is the quotient when \( a \) is divided by 5.

Next, \( b \) divided by 8 leaves a remainder of 7:

\[
b = 8c + 7
\]

Where \( c \) is the quotient when \( b \) is divided by 8.

Now, we substitute these equations back to find \( N \):

\[
N = 3a + 1 = 3(5b + 4) + 1 = 15b + 12 + 1 = 15b + 13
\]

Substitute \( b = 8c + 7 \) into the equation:

\[
N = 15(8c + 7) + 13 = 120c + 105 + 13 = 120c + 118
\]

So the number \( N \) can be expressed as:

\[
N = 120c + 118
\]

### Reverse the Order of Divisors:

Now, if we reverse the order of divisors (8, 5, 3), we need to find the remainders when \( N \) is divided by these divisors.

1. **First, divide by 8**: 

   \[
   N = 120c + 118
   \]
   \[
   N \mod 8 = (120c + 118) \mod 8
   \]

   Since \( 120 \equiv 0 \mod 8 \) and \( 118 \equiv 6 \mod 8 \):

   \[
   N \mod 8 = 6
   \]

   So, the remainder when \( N \) is divided by 8 is 6.

2. **Next, divide by 5**:

   \[
   \text{Let } N_1 = 6 \times \text{(some integer)} + 6
   \]
   We don't need the exact multiple; instead, we check \( N_1 \mod 5 \):

   \( 6 \equiv 1 \mod 5 \)

   So, the remainder when \( N \) is divided by 5 is 3.

3. **Finally, divide by 3**:

   Using the similar process:
   \[
   N \mod 3 = (120c + 118) \mod 3
   \]

   \( 120 \equiv 0 \mod 3 \) and \( 118 \equiv 2 \mod 3 \):

   So, the remainder when \( N \) is divided by 3 is 2.

### Conclusion:

The respective remainders when the divisors are reversed (8, 5, 3) are:

- **Remainder when divided by 8:** 6
- **Remainder when divided by 5:** 3
- **Remainder when divided by 3:** 2

---

Would you like further details or explanations?

### Related Questions:
1. What is the general method to solve such remainder problems?
2. How do the properties of congruences help in solving these problems?
3. What are the uses of the Chinese Remainder Theorem in problems like this?
4. How can you determine the original number \( N \) using the conditions given?
5. What would the remainders be if different divisors were used in reverse?
6. Can you find \( N \) using a direct approach without solving step by step?
7. How does the order of operations affect the result?
8. What are some common divisibility rules that might simplify problems like this?

### Tip:
Understanding how congruences work and how they combine is crucial for solving complex remainder problems efficiently.

Explore how to solve a modular arithmetic problem where a number, when successively divided by 8, 5, and 3, leaves remainders of 7, 4, and 1 respectively. Learn about modular arithmetic, congruences, and the Chinese Remainder Theorem. Suitable for high school students.

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