GMAT Problem Solving - How Many Five Digit Numbers can be Formed Using the Digits 0, 1, 2, 3, 4, and 5 Which are Divisible by 3, Without Repeating the Digits?

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

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

“How many five digit numbers can be formed using the digits 0, 1, 2, 3, 4, and 5 which are divisible by 3, without repeating the digits?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide 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.

Answer:

Approach Solution 1

We should determine which 5 digits from the given 6, would form the 5-digits number divisible by3.
We have six digits: 0, 1, 2, 3, 4, 5. Their sum is 15.
For a number to be divisible by 3 the sum of the digits must be divisible by3. 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-- = {1, 2, 3, 4, 5} and 15-3 = {0, 1, 2, 3, 4, 5}. Meaning that no other 5 from the given six will total the number divisible by 3.

We have two set of numbers:
{1, 2, 3, 4, 5} and {0, 1, 2, 3, 4, 5}. How many 5-digit numbers can be formed using these two sets:
{1, 2, 3, 4, 5} This set gives 5! Numbers, as any combination of these digits would give us a 5 digit number divisible by 3. 5! = 120.
{0, 1, 2, 3, 4, 5}. Now here we cannot use 0 as the first digit, otherwise the number won’t be any more 5 digit and become 4 digit. So, desired # would be total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4!: 5! – 4! = 4! (5 – 1) = 4! * 4 = 96.

Total 120 + 96 = 216

Correct option: E

Approach Solution 2

Trick is to know which numbers are divisible by 3,
Whose sum of the digit is divisible by 3.
With 1, 2, 3, 4, and 5 we can make following numbers:
5*4*3*2*1 = 120 numbers;
Here the sum is 15 for the digits of the numbers formed;
Next are the numbers whose sum of the digits is 12.
Taking 0, 1, 2, 4, 5:
We can make 96 numbers: 4*4*3*2*1 = 96 (4 in starting as a number can’t start with zero)

Total numbers formed = 120 + 96 = 216

Correct option: E

Approach Solution 3

So 5 digit number is a requirement without repeating
There are in total 6 numbers = 0, 1, 2, 3, 4, 5
Number is divisible by 3 means sum of the digits also divisible by 3
5 + 4 + 3 + 2 + 1 = 15 is divisible by 3
So, total number of permutations = 5! = 120
Now these numbers of combinations are without taking 0 into account.
If we take 0 into account, there is only one other choice of digits which is divisible by 3
This is 0, 1, 2, 4, 5
0 + 1 + 2 + 4 + 5 = 12 is also divisible by 3
Here we cannot take 0 as the first digit, otherwise it won’t be a 5-digit number
Therefore possible permutations in this case = 4*4*3*2*1 = 96
Total number of ways possible = 120 + 96 = 216

Correct option: E

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