A number lying between 1000 and 2000 is such that on division by 2,3,4,5,6,7 and 8 leaves remainder 1,2,3,4,5,6 and 7 respectively. The number is

Options

A. 518
B. 416
C. 364
D. 1679

Answer

To solve the problem step by step, we need to find a number between 1000 and 2000 that leaves specific remainders when divided by the numbers 2 through 8.

### Step-by-Step Solution:

1. **Understanding the Remainders**:
- The problem states that when the number \( N \) is divided by:
- 2, it leaves a remainder of 1
- 3, it leaves a remainder of 2
- 4, it leaves a remainder of 3
- 5, it leaves a remainder of 4
- 6, it leaves a remainder of 5
- 7, it leaves a remainder of 6
- 8, it leaves a remainder of 7
- We can observe that for each divisor \( d \), the remainder is \( d - 1 \).

2. **Setting Up the Equation**:
- This can be rewritten as:
\[ N \equiv -1 \mod 2 \]
\[ N \equiv -1 \mod 3 \]
\[ N \equiv -1 \mod 4 \]
\[ N \equiv -1 \mod 5 \]
\[ N \equiv -1 \mod 6 \]
\[ N \equiv -1 \mod 7 \]
\[ N \equiv -1 \mod 8 \]
- This means \( N + 1 \) is divisible by all these numbers.

3. **Finding the LCM**:
- We need to find the least common multiple (LCM) of the numbers 2, 3, 4, 5, 6, 7, and 8.
- The LCM of these numbers is 840.

4. **Formulating the Number**:
- Since \( N + 1 \) is divisible by 840, we can express \( N \) as:
\[ N = 840k - 1 \]
- Here, \( k \) is a natural number.

5. **Finding the Range**:
- We need \( N \) to be between 1000 and 2000:
\[ 1000 < 840k - 1 < 2000 \]
- Adding 1 to all parts of the inequality:
\[ 1001 < 840k < 2001 \]
- Dividing by 840:
\[ \frac{1001}{840} < k < \frac{2001}{840} \]
- This simplifies to:
\[ 1.191 < k < 2.380 \]
- Since \( k \) must be a natural number, the only possible value for \( k \) is 2.

6. **Calculating the Number**:
- Substituting \( k = 2 \) into the equation for \( N \):
\[ N = 840 \times 2 - 1 = 1680 - 1 = 1679 \]

7. **Final Answer**:
- The required number is **1679**.