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

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 certain numbers. Here’s how we can approach it:

### Step 1: Understand the Remainders
The problem states that the number leaves the following remainders:
- When divided by 2, remainder is 1
- When divided by 3, remainder is 2
- When divided by 4, remainder is 3
- When divided by 5, remainder is 4
- When divided by 6, remainder is 5
- When divided by 7, remainder is 6
- When divided by 8, remainder is 7

This means that if we denote the number as \( N \), we can express these conditions mathematically:
- \( N \equiv 1 \mod 2 \)
- \( N \equiv 2 \mod 3 \)
- \( N \equiv 3 \mod 4 \)
- \( N \equiv 4 \mod 5 \)
- \( N \equiv 5 \mod 6 \)
- \( N \equiv 6 \mod 7 \)
- \( N \equiv 7 \mod 8 \)

### Step 2: Rewrite the Conditions
We can rewrite these conditions in a more useful form:
- \( N - 1 \equiv 0 \mod 2 \)
- \( N - 2 \equiv 0 \mod 3 \)
- \( N - 3 \equiv 0 \mod 4 \)
- \( N - 4 \equiv 0 \mod 5 \)
- \( N - 5 \equiv 0 \mod 6 \)
- \( N - 6 \equiv 0 \mod 7 \)
- \( N - 7 \equiv 0 \mod 8 \)

This means that \( N - 1 \) is a common multiple of all the divisors (2, 3, 4, 5, 6, 7, 8).

### Step 3: Find the LCM
To find \( N - 1 \), we need to calculate the least common multiple (LCM) of the numbers 2, 3, 4, 5, 6, 7, and 8.

**Finding the LCM:**
- The prime factorization of the numbers is:
- 2 = \( 2^1 \)
- 3 = \( 3^1 \)
- 4 = \( 2^2 \)
- 5 = \( 5^1 \)
- 6 = \( 2^1 \times 3^1 \)
- 7 = \( 7^1 \)
- 8 = \( 2^3 \)

The LCM is found by taking the highest power of each prime:
- \( 2^3 \) (from 8)
- \( 3^1 \) (from 3)
- \( 5^1 \) (from 5)
- \( 7^1 \) (from 7)

So, the LCM is:
\[ LCM = 2^3 \times 3^1 \times 5^1 \times 7^1 = 8 \times 3 \times 5 \times 7 \]
Calculating this gives:
\[ 8 \times 3 = 24 \]
\[ 24 \times 5 = 120 \]
\[ 120 \times 7 = 840 \]

### Step 4: Express the Number
Since \( N - 1 \) is a multiple of 840, we can express \( N \) as:
\[ N = 840k + 1 \]
where \( k \) is a positive integer.

### Step 5: Find the Value of \( k \)
We need \( N \) to be between 1000 and 2000:
\[ 1000 < 840k + 1 < 2000 \]
Subtracting 1 from all parts:
\[ 999 < 840k < 1999 \]
Dividing by 840:
\[ \frac{999}{840} < k < \frac{1999}{840} \]
Calculating the bounds:
\[ 1.188 < k < 2.380 \]
Since \( k \) must be an integer, the only possible value for \( k \) is 2.

### Step 6: Calculate \( N \)
Substituting \( k = 2 \) into the equation for \( N \):
\[ N = 840 \times 2 + 1 = 1680 + 1 = 1681 \]

### Conclusion
Thus, the number that satisfies all the conditions is:
\[ \boxed{1681} \]