A number between 1000 and 2000 which when divided by 30, 36 and 80 gives a remainder 11 in each case is:

Options

A. 11523
B. 1451
C. 1641
D. 1712

Answer

To find a number between 1000 and 2000 that gives a remainder of 11 when divided by 30, 36, and 80, we can follow these steps:

### Step 1: Understand the condition
We need a number \( N \) such that:
- \( N \mod 30 = 11 \)
- \( N \mod 36 = 11 \)
- \( N \mod 80 = 11 \)

This means that if we subtract 11 from \( N \), the resulting number \( N - 11 \) should be divisible by 30, 36, and 80.

### Step 2: Set up the equation
Let \( M = N - 11 \). Then we need:
- \( M \mod 30 = 0 \)
- \( M \mod 36 = 0 \)
- \( M \mod 80 = 0 \)

This means \( M \) is a common multiple of 30, 36, and 80.

### Step 3: Find the LCM of the divisors
To find \( M \), we calculate the least common multiple (LCM) of 30, 36, and 80.

1. **Prime factorization**:
- \( 30 = 2 \times 3 \times 5 \)
- \( 36 = 2^2 \times 3^2 \)
- \( 80 = 2^4 \times 5 \)

2. **Take the highest power of each prime**:
- For \( 2 \): \( 2^4 \) (from 80)
- For \( 3 \): \( 3^2 \) (from 36)
- For \( 5 \): \( 5^1 \) (from 30)

3. **Calculate the LCM**:
\[ \text{LCM} = 2^4 \times 3^2 \times 5^1 = 16 \times 9 \times 5 \]
\[ = 144 \times 5 = 720 \]

### Step 4: Find possible values of \( N \)
Since \( M = 720k \) for some integer \( k \), we can express \( N \) as:
\[ N = M + 11 = 720k + 11 \]

We need \( N \) to be between 1000 and 2000:
\[ 1000 < 720k + 11 < 2000 \]

### Step 5: Solve the inequalities
1. **Lower bound**:
\[ 720k + 11 > 1000 \implies 720k > 989 \implies k > \frac{989}{720} \approx 1.373 \]
So, \( k \geq 2 \).

2. **Upper bound**:
\[ 720k + 11 < 2000 \implies 720k < 1989 \implies k < \frac{1989}{720} \approx 2.7625 \]
So, \( k \leq 2 \).

### Step 6: Determine the valid value of \( k \)
The only integer value for \( k \) that satisfies both conditions is \( k = 2 \).

### Step 7: Calculate \( N \)
Substituting \( k = 2 \) into the equation for \( N \):
\[ N = 720 \times 2 + 11 = 1440 + 11 = 1451 \]

### Conclusion
The number between 1000 and 2000 that gives a remainder of 11 when divided by 30, 36, and 80 is **1451**.