How many six-digit positive numbers can be formed using the digits 1

Question

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

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

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fall2021 · Accepted Answer

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 numbers are divisible by 3, all are divisible by 6 as well so 120 numbers

Overall 120*4+48 = 528. Option D

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KarishmaB · Answer

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 digits. 
If we fix the units digit of the number and use other 5 digits, we can arrange them in 5! ways.

If the units digit is 2, we get 5! such numbers. The numbers will be even and hence, obviously divisible by 2. e.g. 134562 is divisible by 2.

If the units digit is 3, we get 5! such numbers. The sum of digits is 1+2+3+4+5+6 = 21 which is divisible by 3. So all such numbers will be divisible by 3.

If the units digit is 5, we get 5! such numbers. The number will be divisible by 5. e.g. 126435 is divisible by 5.

If units digit is 6, we get 5! such numbers. The numbers will be even and we know that all numbers formed will be divisible by 3 so all such numbers will be divisible by 6. e.g. 423156 is even and divisible by 3 so divisible by 6.

Now, the only complication left is with 4 in units digit. We will have 5! such numbers. 
To be divisible by 4, the last two digits must be a multiple of 4. i.e. the last 2 digits must be one of 24 or 64. 
So last two digits are selected in 2 ways and 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 numbers = 4*5! + 2*4! = 528

Answer (D)

Ninokh · Answer

Also the answer is 528, in my opinion, the wording is confusing, as one can get - if the units digit is not 1 >> the rest restrictive condition. So if the units digit is 1 will work for us. So I was searching in the answers 528+120. Hopefully there was no such )

CEdward · Answer

This is a tough problem.  VeritasKarishma  I know you've written about this extensively, can you explain how to do these types of problems?

ScottTargetTestPrep · Answer

Solution:

We see that if the units digit is 2, then the number has to be divisible by 2; if the units digit is 3, then the number has to be divisible by 3, etc. Let’s look at each case individually.

Case 1 : The units digit is 2 and the number is divisible by 2

Recall that if the units digit of a number is even, then the number is divisible by 2. Since here the units digit of the number is 2 (which is an even digit), it will always be divisible by 2. In this case, we have 5! = 120 ways to rearrange the other 5 digits (i.e., the digits on the left of the units digit 2).

Case 2:  The units digit is 3 and the number is divisible by 3

Recall that if the sum of the digits of a number is divisible by 3, then the number is divisible by 3. Since here the sum of the digits of the number is always 1 + 2 + 3 + 4 + 5 + 6 = 21 (which is divisible by 3), it will always be divisible by 3. In this case, we have 5! = 120 ways to rearrange the other 5 digits (i.e., the digits on the left of the units digit 3).

Case 3:  The units digit is 4 and the number is divisible by 4

Recall that if the last two digits of a number are divisible by 4, then the number is divisible by 4. Here, the last two digits of the number must be 24 or 64 in order for it to be divisible by 4. Therefore, we have 4! = 24 ways to rearrange the other 4 digits (i.e., the digits on the left of the above-mentioned two digits). Since there are 2 possible last two-digit combinations and each has 24 choices for the remaining 4 digits, there are 24 x 2 = 48 possible 6-digit numbers in this case.

Case 4:  The units digit is 5 and the number is divisible by 5.

Recall that if the units digit of a number is divisible by 5, then the number is divisible by 5. Since here the units digit of the number is 5 (which is divisible by 5), it will always be divisible by 5. In this case, we have 5! = 120 ways to rearrange the other 5 digits (i.e., the digits on the left of the units digit 5).

Case 5 : The units digit is 6 and the number is divisible by 6.

Recall that if the number is divisible by 2 and 3, then it is divisible by 6. Since here the units digit of the number is 6 (which is even), it will be divisible by 2. Furthermore, since the sum of the digits of the number is 21, it will always be divisible by 3. In other words, when the units digit of the number is 6 and no matter how we arrange the remaining 5 digits, it will always be divisible by 6 and we have 5! = 120 ways to rearrange the remaining 5 digits.

Therefore, we have a total of 120 + 120 + 48 + 120 + 120 = 528 such numbers.

Answer: D

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How many six-digit positive numbers can be formed using the digits 1

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