What is the least four-digit number when divided by 3, 4, 5 and 6 leaves a remainder 2 in each case?

The correct option is 1022.

To find the least four-digit number that leaves a remainder of 2 when divided by 3, 4, 5, and 6, we first need to determine the Least Common Multiple (LCM) of these divisors.

Step 1: Find the LCM of 3, 4, 5, and 6.
- Prime factorization of 3 = 3
- Prime factorization of 4 = 2²
- Prime factorization of 5 = 5
- Prime factorization of 6 = 2 × 3

Taking the highest powers of all prime factors involved:
$\text{LCM}=2^2\times 3\times 5=4\times 3\times 5=60$

Any number that is divisible by 3, 4, 5, and 6 must be a multiple of 60.

Step 2: Find the smallest four-digit multiple of 60.
The smallest four-digit number is 1000.
Now, divide 1000 by 60 to find the required multiple:
$1000÷60=16\text{with a remainder of}40$

To find the next complete multiple of 60 that is greater than or equal to 1000:
$\text{Smallest 4-digit multiple}=1000+(60-40)=1000+20=1020$

Step 3: Add the required remainder.
Since the required number must leave a remainder of 2 in each case, we add 2 to the smallest four-digit multiple of 60:
$\text{Required number}=1020+2=1022$

Therefore, the least four-digit number is 1022.