Write the smallest digit and the greatest digit in the blank space of the following number so that the number formed is divisible by 3.
${\text{4765}}....{\text{2}}{\text{.}}$

Hint: -Any number is divisible by 3, if the sum of its digits is divisible by 3.

Given number:
${\text{4765}}....{\text{2}}{\text{.}}$
It is given that the number is divisible by 3.
By the divisibility rule of 3, any number is divisible by 3, if the sum of its digits is divisible by 3.
Let the digit on the blank space be $b$.
\[
  4 + 7 + 6 + 5 + b + 2 \\
   \Rightarrow b + 24 \\
\]
As we see that 24 is divisible by \[{\text{3 = }}\dfrac{{24}}{3} = 8{\text{ times}}{\text{.}}\]
Therefore in order to make the required number divisible by 3, the least value of $b = 0$ and the maximum value of $b = 9$.
If \[b = 0\], then\[0 + 24 = 24\]. This is divisible by 3.
So, the number becomes 476502
If \[b = 9\], then\[9 + 24 = 33\]. This is divisible by 3.
So, the number becomes 476592
Therefore the smallest digit is 0 and the greatest digit is 9.

Note: - In such types of questions the key concept we have to remember is that always remember the divisibility rule of 3 which is stated above, then according to this rule calculate the minimum and maximum value in the black space so that number is exactly divisible by 3.