Hi All,
Can anyone explain how this one will be solved? Answer and explanation hidden here, but I don't get the logic.
How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeating?
A. 15
B. 90
C. 216
D. 120
E. 625
Answer [spoiler]C
And this is the explanation I found:
Test of divisibility for 3 : The sum of the digits of any number that is divisible by '3' is divisible by 3.
There are six digits, 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 are 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 five digit numbers divisible by '3' that can be formed using the digits 0 to 5.[/spoiler]
Can anyone explain how this one will be solved? Answer and explanation hidden here, but I don't get the logic.
How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeating?
A. 15
B. 90
C. 216
D. 120
E. 625
Answer [spoiler]C
And this is the explanation I found:
Test of divisibility for 3 : The sum of the digits of any number that is divisible by '3' is divisible by 3.
There are six digits, 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 are 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 five digit numbers divisible by '3' that can be formed using the digits 0 to 5.[/spoiler]
