What is the least number, which when divided by 12, 21 and 35, leaves the same remainder 6.

Correct option is D

Given:
We need to find the least number that, when divided by 12, 21, and 35, leaves the same remainder.
The remainder is given as 6.
Concept Used:
This problem involves finding the least common multiple (LCM) of 12, 21, and 35, and using that LCM to determine the required number.
If a number leaves the same remainder when divided by multiple divisors, the required number can be found using the formula:
Required number = (LCM of divisors) × n + remainder,
where n is a positive integer.
Solution:
Finding the LCM of 12, 21, and 35:
Prime factorization of 12: 12 = 2² × 3
Prime factorization of 21: 21 = 3 × 7
Prime factorization of 35: 35 = 5 × 7
To find the LCM, take the highest powers of all the prime factors. So, the LCM is: LCM = 2² × 3 × 5 × 7 = 420.
The required number should be of the form:
420 × n + 6.
For the least number, take n = 1: Required number = 420 × 1 + 67 = 420 + 6 = 426.
However, this number (487) exceeds 420. To adjust the number so that it leaves a remainder of 67 when divided by 12, 21, and 35, we subtract the LCM (420) from 426: 426 - 420 = 6.
Thus, the least number that satisfies the condition is 426.
Final Answer: (d) 426