Question

The number of 5 digits number, which are divisible by 4 with digits from the set \[\left\{ \text{1},\text{ 2},\text{ 3},\text{ 4},\text{ 5} \right\}\] and the repetition is allowed is

Answer 1

Hint: To solve this question, we will first see divisibility rule of 4, which is, that a number is divisible by 4 if its end two digits are divisible by 4. By understanding the possibility of digits possible at the end points using \[\left\{ \text{1},\text{ 2},\text{ 3},\text{ 4},\text{ 5} \right\}\] we will consider cases and count their possibilities. Then, we will look for numbers possible at the starting 3 digit of our 5 digit number. Clubbing all possibilities of 3 digit and 2 digit to form 5 digit number and multiplying we would arrive at our answer.

Complete step-by-step answer:
First of all, we will study the divisibility rule of 4.
Rule of divisibility of 4: A number is divisible by 4 if its last two digits are divisible by 4.
We have to make a 5 digit number, so we will consider 5 gaps as below:
\[\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\]
We are given the set as: \[\left\{ \text{1},\text{ 2},\text{ 3},\text{ 4},\text{ 5} \right\}\]
Here, we will now assume cases to see when the digit number formed is divisible by 4.
Since, a number is divisible by 4, when last two digits are divisible by 4, so we will consider all the possibilities of case two digit of \[\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\] such that they are divisible by 4.
We have possibilities of last two digits to be 12, 24, 32, 44 and 52.
So, we have total possibility as:
\[\begin{align}
  & \text{Case I: }\underline{{}}\underline{{}}\underline{{}}\underline{1}\underline{2} \\
 & \text{Case II: }\underline{{}}\underline{{}}\underline{{}}\underline{2}\underline{4} \\
 & \text{Case III: }\underline{{}}\underline{{}}\underline{{}}\underline{3}\underline{2} \\
 & \text{Case IV: }\underline{{}}\underline{{}}\underline{{}}\underline{4}\underline{4} \\
 & \text{Case V: }\underline{{}}\underline{{}}\underline{{}}\underline{5}\underline{2} \\
\end{align}\]
So, we have now fixed places of the last two digits of our 5 digit number.
We see that the fourth and fifth place of \[\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\] have all number from \[\left\{ \text{1},\text{ 2},\text{ 3},\text{ 4},\text{ 5} \right\}\] possible.
So, we have 5 possibilities at fourth place as a total of 5 cases are formed.
Now, we will consider the first 3 digits of our 5 digit number.
As we are given that, the repetition of digits is allowed, so we have 5 possibility from \[\left\{ \text{1},\text{ 2},\text{ 3},\text{ 4},\text{ 5} \right\}\] at first place of \[\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\] from left.
Similarly, we have 5 possibility from \[\left\{ \text{1},\text{ 2},\text{ 3},\text{ 4},\text{ 5} \right\}\] at second and third place of \[\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\] from left.
So, we have a total of \[5\times 5\times 5=125\] for the first three digits of the number.
So, we have from above cases, we can write the total possibilities as
\[125\times 5=625\text{ possibility}\]
Therefore, we have the total possible numbers of 5 digits number which are divisible by 4 from the set \[\left\{ \text{1},\text{ 2},\text{ 3},\text{ 4},\text{ 5} \right\}\] when repetition is allowed is 625.

Note: Students might get confused at the point where we are multiplying 5 with 5 and 5 and not adding them. Whenever one case or situation is dependent on the other case, we multiply and when they are independent then we add them.
Here we have the first term of \[\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\text{ }\underline{{}}\] is dependent on second term or on third term or so on. So, we have to multiply $5\times 5\times 5$
For example: the number 12344 is different from 12324 so we multiply.