How many six-digit positive numbers can be formed using the digit 1 GMAT Problem Solving

Question: How many six-digit positive numbers can be formed using the digit 1, 2, 3, 4, 5, and 6 exactly once, such that the six-digit number is divisible by its unit’s digit, if the units digit is not 1?

   A. 456
   B. 492
   C. 520
   D. 528
   E. 620

Answer: D

Approach Solution (1):
Ending with 2- All 5! (120) possible numbers are divisible by 2
Ending with 3- All 5! (120) possible numbers are divisible by 3
Ending with 4- the tens digit has to be 2 or 6 for the number to be divisible by 4, so only 2^4! Numbers are divisible – 48 numbers
Ending with 5- All 5! (120) numbers are divisible by 5
Ending with 6- since all numbers are even and all the numbers are divisible by 3, all are divisible by 6 as well as 120 numbers
Overall 120*4 + 48 = 528
Correct option: D

Approach Solution (2):
Use 1, 2, 3, 4, 5, and 6 exactly once
The 6 digit numbers will be of the form 123456 or 546321 or 326154 etc. we want those numbers which are divisible by their units digit
If we fix the units digit of the number and use other 5 digits, we can arrange then in 5! Ways.
If the unit digit is 2, we get 5! such members. The numbers will be even and hence, obviously divisible by 2. Example: 134562 is divisible by 2
If the unit digit is 3, we get 5! such members. The sum of the digits is 1 + 2 + 3 + 4 + 5 + 6 = 21 which is divisible by 3
If the unit digit is 5, we get 5! such members. The number will be divisible by 5. Example 126435 is divisible by 5
If the units digit is 6, we get 5! Such members. The numbers will be even and we know that all the numbers formed will be divisible by 3 so all such members will be divisible by 6
Now, the only complication left is with 4 in unit’s digit. We will have such 5! Members. To be divisible by 4, the last two digits must be a multiple of 4.
So the last two digits are selected in 2 ways and the other 4 digits are arranged in 4! Ways to get 2*4!
Only these numbers will be divisible by 4
So 356124 will be divisible by 4 but 356214 will not be
Total members = 4*5! + 2*4! = 528
Correct option: D

Approach Solution (3):
Use 1, 2, 3, 4, 5, and 6 exactly once
Ending with 2- All 5! (120) possible numbers
Ending with 3- All 5! (120) possible numbers
Ending with 4- the tens digit has to be 2 or 6 for the number to be divisible by 4
Ending with 5- All 5! (120) numbers
Ending with 6- since all numbers are even and all the numbers are divisible by 3, all are divisible by 6 as well as 120 numbers
Overall 120*4 + 48 = 528
Correct option: D

“How many six-digit positive numbers can be formed using the digit 1, 2, 3, 4, 5, and 6 exactly once, such that the six-digit number is divisible by its unit’s digit, if the units digit is not 1?”- 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.

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