How many five digit positive integers that are divisible by 4 can be formed using the digits 0,1,2,3,4 and 5 without any of the digits getting repeated in a number .
Answers
Answer:
The answer is :-
Step-by-step explanation:
216
Test of divisibility for 3
The sum of the digits of any number that is divisible by '3' is divisible by 3.
For instance, take the number 54372.
Sum of its digits is,
=5+4+3+7+2
=21
As 21 is divisible by '3', 54372 is also divisible by 3.
There are six digits viz., 0,1,2,3,4 and 5. To form 5-digit numbers we need exactly 5 digits. So we should not be using one of the digits.
The sum of all the six digits 0,1,2,3,4 and 5 is 15. We know that any number is divisible by 3 if and only if the sum of its digits is divisible by '3'.
Combining the two criteria that we use only 5 of the 6 digits and pick them in such a way that the sum is divisible by 3, we should not use either '0' or '3' while forming the five digit numbers.
Case 1
If we do not use '0', then the remaining 5 digits can be arranged in:
5 ways=120 numbers.
Case 2
If we do not use '3', then the arrangements should take into account that '0' cannot be the first digit as a 5-digit number will not start with '0'.
The first digit from the left can be any of the 4 digits 1,2,4 or 5
Then the remaining 4 digits including '0' can be arranged in the other 4 places in 4! ways.
So, there will be 4×4! numbers =4×24=96 numbers.
Combining Case 1 and Case 2, there are a total of 120+96=216, 5 digit numbers divisible by '3' that can be formed using the digits 0 to 5.
Hope that it helps...