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

Options

A. 1451
B. 1641
C. 1712
D. 1523

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: Find the LCM of 30, 36, and 80
To solve the problem, we first need to find the least common multiple (LCM) of the three numbers: 30, 36, and 80.

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

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

Therefore, the LCM is:
\[ LCM = 2^4 \times 3^2 \times 5^1 = 16 \times 9 \times 5 = 720 \]

### Step 2: Find the number in the range 1000 to 2000
Next, we need to find multiples of 720 that fall between 1000 and 2000.

- The first multiple of 720 is \(720 \times 1 = 720\) (too low).
- The second multiple is \(720 \times 2 = 1440\) (within range).
- The third multiple is \(720 \times 3 = 2160\) (too high).

Thus, the only multiple of 720 that lies between 1000 and 2000 is **1440**.

### Step 3: Adjust for the remainder
Since we need a number that gives a remainder of 11 when divided by 30, 36, and 80, we can express our desired number as:
\[ \text{Number} = 720k + 11 \]
where \(k\) is a positive integer.

For \(k = 2\):
\[ \text{Number} = 720 \times 2 + 11 = 1440 + 11 = 1451 \]

### Step 4: Verify the result
Now, we check if 1451 gives a remainder of 11 when divided by 30, 36, and 80.

- \(1451 \div 30 = 48\) remainder \(11\)
- \(1451 \div 36 = 40\) remainder \(11\)
- \(1451 \div 80 = 18\) remainder \(11\)

All conditions are satisfied.

### Final Answer
Therefore, the number we are looking for is **1451**.

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