JEE Main On 16 April 2018 Question 5
Question: The number of numbers between 2,000 and 5,000 that can be formed with the digits (repetition of digits 0, 1, 2, 3, 4 is not allowed) and are multiple of 3 is? [JEE Main 16-4-2018]
Options:
A) 30
B) 48
C) 24
D) 36
Show Answer
Answer:
Correct Answer: A
Solution:
There are 4 places to be filled with the given digits. The thousands place can have only 2, 3 and 4 since the number has to be greater than 2000. For the remaining 3 places, we have pick out digits such that the resultant number is divisible by 3.
The divisibility criteria for 3 states that sum of digits of the number should be divisible by 3.
Case 1: if we pick 2 for thousand place. The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are:
$ 0,1 $ ad 3 as 2 + 1 + 0 + 3 = 6 is divisible by 3. 0,3 and 4 as 2 + 3 + 0 + 4 = 9 is divisible by 3.
In both the above combination, the remaining three digits can be arranged in ways.
Total number Case 2: If we pick 3 for thousands place The remaining digits we can pick such that
sum of digits at all places is a multiple of 3 are: 0,1and as 3 + 1 + 0 + 2 = 6 is divisible by 3. 0,2 and 4 as 3 + 2 + 0 + 4 = 9 is divisible by 3.
In both the above combination, the remaining three digits can be arranged in 3! ways.
Total number $ =2\times 3!=12 $
Case 3: if we pick 4 for thousand place. The remaining digits we can pick such that sum of digits at all places is a multiple of 3 are:
0,2 and 3 as 4 + 2 + 0 + 3 = 9 is divisible by 3.
In the above combination, the remaining three digits can be arranged in 3! ways.
Total number = 3! = 6 Total number of numbers between 2000 and 5000 divisible by 3 are 12 + 12 + 6 = 30.
Option A is correct.
