# but this question is different!!!!!!!!!!

### Asked by kvunni hj | 18th Nov, 2013, 11:41: AM

Expert Answer:

### Your **question **was as follows:
how many 5 digit nos. that are divisible by 3 can be formed using the digits 0,1,2,3,4.5 without repetion ?
**Answer:**
We know that a number is divisible by 3 if the sum of its digits 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.

Thus, there are a total of 120 + 96 = 216 five digit numbers divisible by '3' that can be formed using the digits 0 to 5.

**question**was as follows:

**Answer:**

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.

Thus, there are a total of 120 + 96 = 216 five digit numbers divisible by '3' that can be formed using the digits 0 to 5.

### Answered by | 18th Nov, 2013, 11:53: AM

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