Find the smallest 4-digit number which when divided by 2, 3 and 5 leaves a remainder of 1 in each case?

Correct option is B

​Given:

the smallest 4-digit number N

 N leaves a remainder of 1 when divided by 2, 3, and 5.

Solution:

The least common multiple (LCM) of 2, 3, and 5 is 30.
Since N leaves a remainder of 1, it can be expressed as: $$N = 30k + 1$$​​

where k is an integer.

The smallest 4-digit number is 1000. Solve for k:

$$30k + 1 \geq 1000 \implies 30k \geq 999 \implies k \geq \frac{999}{30} = 33.3$$​​

Taking : $$k = 34$$​​

​$$N = 30(34) + 1 = 1020 + 1 = 1021$$​​