GATE DS&AI 2024 | GA Question: 3

Question

How many $4$-digit positive integers divisible by $3$ can be formed using only the digits $\{1,3,4,6,7\}$, such that no digit appears more than once in a number?

• A. $24$
• B. $48$
• C. $72$
• D. $12$

Answer 1

Let's say the Digit $D$ is  $a_{1}a_{2}a_{3}a_{4}$ .

It is given , $a_{1} \neq a_{2} \neq a_{3} \neq a_{4}$ which is repetition not allowed . it is also given $D$ is divisible by $3$ . From the divisibiilty rule of $3$ we know that sum of digit is also divisible by $3$ .

Now , Given choice of Digits are {1,3,4,6,7} . Now sum of the digits is $1+3+4+6+7=21$ .

Now we have to $5$ digits in the set from where we have to rejects $1$ digit such that after rejecting that the sum is still divisible by $3$ . Now the question is which digits we should reject ? => The digits which are divisible by $3$ .

So , We have to keep $1$,$4$,$7$ and can choose any of the $3$ or $6$ to make the digit D.

So there are $\binom{2}{1}$ ways to choose between $3$ or $6$ and the $4$ digit can arrange in $4!$ ways .

So total number of ways is : $\binom{2}{1}$ *$4!=48$

Comments

I have a question, are you also considering 6,1,4,7 divisible by 3 in this?

Answer 2

 
Therefore, (The Final Answer is 48.)

Answer 3

Minimum number formed with 4 digit=13

Maximum number=20

Note= number divisible by 3 mean sum of digit of number divisible by 3

Find number divisible by 3 is in between minimum and maximum which can formed with digit sum is

15(1, 3,4,7)and 18(1, 4,6,7) only these number formed with sum of digit in between 13 and 20

a, b, c,d permutations =24

So 24+24=48

Answer 4

B. 48

Given 5 digits are 1,3,4,6,7 whose sum is 21 we have to make 4 digit no.(no repetition ) 

so we have to reject one digit such that their sum divisible by 3.

we have 2 cases to do 

Case 1. reject 3  

           {1,4,6,7} => 4!= 24 ways (permutations)

Case 2. reject 6 

           {1,3,4,7} => 4! = 24 ways (permutations)

so total=24+24=48 ways