Which is the least four-digit number which when divided by 5, 6 and 8 leaves remainder 2 in each case?

Options

A. 1032
B. 1062
C. 1042
D. 1082

Answer

To solve the problem of finding the least four-digit number which, when divided by 5, 6, and 8, leaves a remainder of 2 in each case, we can follow these steps:

### Step 1: Find the LCM of the Divisors
First, we need to find the least common multiple (LCM) of the numbers 5, 6, and 8.

- The prime factorization of the numbers is:
- 5 = 5^1
- 6 = 2^1 * 3^1
- 8 = 2^3

- The LCM is found by taking the highest power of each prime number:
- LCM = 2^3 * 3^1 * 5^1 = 8 * 3 * 5 = 120

### Step 2: Set Up the Equation
Since we want a number that leaves a remainder of 2 when divided by 5, 6, and 8, we can express this number in the following form:
\[ N = 120k + 2 \]
where \( k \) is a non-negative integer.

### Step 3: Find the Least Four-Digit Number
We need to find the smallest value of \( N \) that is a four-digit number. The smallest four-digit number is 1000. Therefore, we set up the inequality:
\[ 120k + 2 \geq 1000 \]

Subtracting 2 from both sides:
\[ 120k \geq 998 \]

Dividing both sides by 120:
\[ k \geq \frac{998}{120} \]
\[ k \geq 8.3167 \]

Since \( k \) must be a whole number, we take \( k = 9 \).

### Step 4: Calculate the Number
Now we substitute \( k = 9 \) back into the equation for \( N \):
\[ N = 120 \times 9 + 2 \]
\[ N = 1080 + 2 \]
\[ N = 1082 \]

### Step 5: Verify the Result
Finally, we should verify that 1082 leaves a remainder of 2 when divided by 5, 6, and 8.

- Dividing by 5:
\[ 1082 \div 5 = 216 \text{ remainder } 2 \]

- Dividing by 6:
\[ 1082 \div 6 = 180 \text{ remainder } 2 \]

- Dividing by 8:
\[ 1082 \div 8 = 135 \text{ remainder } 2 \]

Since all conditions are satisfied, the least four-digit number that meets the criteria is **1082**.