How Many Five Digit Numbers Can be Formed Using Digits 0, 1, 2, 3, 4, 5, Which Are Divisible By 3 GMAT Problem Solving

Question: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4, 5, which are divisible by 3, without any of the digits repeating?

1. 15
2. 96
3. 120
4. 181
5. 216

“How many five digit numbers can be formed using digits 0, 1, 2, 3, 4, 5, which are divisible by 3”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:
First step:
We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.
We have six digits: 0,1,2,3,4,5. Their sum=15.
For a number to be divisible by 3 the sum of the digits must be divisible by 3.
As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be:
15-0={1, 2, 3, 4, 5} and
15-3={0, 1, 2, 4, 5}.
This means that no other 5 from given six will total the number divisible by 3.
Second step:
We have two set of numbers:
{1, 2, 3, 4, 5} and {0, 1, 2, 4, 5}.
We need to know how many 5 digit numbers can be formed using this two sets:
{1, 2, 3, 4, 5} --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.
{0, 1, 2, 4, 5} --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, total combinations 5!, minus combinations with 0 as the
first digit (combination of 4) 4! --> 5!-4!=96
120+96=216

Correct Answer: E

Approach Solution 2:

Explanation:
By the property of divisibility by 3 i.e "a no: is divisible by 3, if the sum of the digits is divisible by 3"(e.g= 12-->1+2=3)
so from 0,1,2,3,4,5 the set of 5 digit no:s that can be formed which is divisible by 3 are 0,1,2,4,5(sum=12) & 1,2,3,4,5(sum=15)
from first set(0,1,2,4,5) no:s formed are 96 i.e first digit can be formed from any 4 no: except 0. second digit from 4 no: except digit used at first place,3rd from rest 3 , 4th from rest 2
no: and in fifth remaining digit since no repetition allowed.
from second set(1,2,3,4,5) no:s formed are 120 i.e first digit can be formed from any 5 digits. second digit from 4 no: except digit used at first place,3rd from rest 3 , 4th from rest 2 no:
and in fifth remaining digit since no repetition allowed.
so total 120+96=216

Correct Answer: E

Approach Solution 3:

In this question, we are supposed to form a five digit number which will be divisible by 3, and the divisibility test of 3 is the sum of digits should be divisible by 3.
Therefore, we are only going to consider 5 digits whose sum will result in a number which will be divisible by 3.

Complete step-by-step answer:

Let us observe the digits given to us, we have 0,1,2,3,4 and 5, to make a five digit number we only need 5 digits out of the given 6 digits,
Case 1: Using digits 0,1,2,4 and 5.
The number of ways in which we can arrange these 5 digits are 

⇒4×4×3×2×1⇒4×4×3×2×1

⇒96⇒96

Case 2: Using the digits 1,2,3,4 and 5
The number of ways in which we can arrange these 5 digits are 

⇒5×4×3×2×1⇒5×4×3×2×1

⇒120⇒120

Therefore, the total number of cases =96+120=216

Correct Answer: E

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