What will be the least number which when divided by 12, 21 and 35 leaves 6 as a remainder in each case?

Let's say the required number is $n$.

When $n$ is divided by $12$, $21$ and $35$, remainder is $6$ in each case.

This means:

$n = 12k_1 + 6$

$n = 21k_2 + 6$

$n = 35k_3 + 6$

where $k_1$, $k_2$, $k_3$ are some integers

Therefore, $n - 6$ is divisible by $12$, $21$ and $35$.

In other words, $n - 6$ is divisible by $LCM(12, 21, 35)$

Let's find $LCM(12, 21, 35)$:

$12 = 2^2 × 3$

$21 = 3 × 7$

$35 = 5 × 7$

Therefore, $LCM(12, 21, 35) = 420$

So, $n - 6 = 420k$, where $k$ is a non-negative integer

The least value of $n$ will be when $k = 1$

Therefore, $n = 420 + 6 = 426$.