Find the least number which when added to 10 is exactly divisible by 14, 35, 40 and 56.

$$14 = 2 \times 7 = 2^1 \times 7^1$$ $$35 = 5 \times 7 = 5^1 \times 7^1$$ $$40 = 2 \times 2 \times 2 \times 5 = 2^3 \times 5^1$$ $$56 = 2 \times 2 \times 2 \times 7 = 2^3 \times 7^1$$

The LCM is the product of all prime factors raised to their highest powers.
The prime factors are $2, 5, 7$.
Highest power of 2 is $2^3$.
Highest power of 5 is $5^1$.
Highest power of 7 is $7^1$.

$$\text{LCM}(14, 35, 40, 56) = 2^3 \times 5^1 \times 7^1 = 8 \times 5 \times 7 = 40 \times 7 = 280$$

Let the required number be $N$.
According to the problem, when 10 is added to $N$, the result is exactly divisible by $14, 35, 40,$ and 56. This means $N + 10$ must be equal to the LCM.

$$N + 10 = \text{LCM}(14, 35, 40, 56)$$ $$N + 10 = 280$$

$$N = 280 - 10$$ $$N = 270$$

The least number is $\text{270}$.