The least number which when diminished by $5$ is divisible by each of $21$, $28$, $36$, $45$ is:

Let the required least number be $N$. If it is diminished by $5$, the result ($N-5$) is exactly divisible by $21$, $28$, $36$, and $45$. This means $N-5$ must equal the Lowest Common Multiple ($\text{LCM}$) of these numbers.
Find the prime factorizations:

$$21=3\times 7$$

$$28=2^2\times 7$$

$$36=2^2\times 3^2$$

$$45=3^2\times 5$$

Calculate the $\text{LCM}$ by using the maximum power of each prime base:

$$\text{LCM}=2^2\times 3^2\times 5\times 7$$

$$\text{LCM}=4\times 9\times 5\times 7=1260$$

Set up the equation to find $N$:

$$N-5=1260⟹N=1260+5=1265$$