# 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 repeated.

A. 15

B. 96

C. 216

D. 120

E. 625

**Answer: Option C **

__Solution(By Examveda Team)__

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.

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Why we should not use 0 and 3 while arranging numbers