Question

from the digits 1,3,5,6,7 and 9 how many 4 digits that are divisible by 3

Answer 1

Step-by-step explanation:

All integers divisible by 3 follow a simple rule; the sum of all their digits add up to a multiple of 3. (If the sum of the digits is 2-digit or greater number, you can repeat the process and add the digits of that number until you get to 3, 6 or 9)

Using the digits 1, 2, 3, 5, 6, 7, and 9, we need to find all combinations of 4 digits that add up to a multiple of 3.

Since 3, 6 and 9 are already multiples of 3, the other digits in the number must add up to a multiple of 3 if they are present.

So, we can count the amount of combinations that would be divisible by three through separating them by whether any of the individual digits are multiples of 3 (Mo3).

0 Mo3s: (1, 2, 5, 7)

1 Mo3: N/A

2 Mo3s: [(1, 2) (1, 5) (2, 7) (5, 7)] * [(3, 6) (3, 9), (6, 9)]

3 Mo3s: N/A

So, there is 1 way with no individual digits that are multiples of 3, and (4*3 = 12) ways that have 2 Multiples of 3, for a total of 13 combinations of digits that add up to a multiple of 3. Any 4-digit integer using these combinations of digits will be divisible by 3.

Now how many numbers can we make from each? If you have any 4 digits (let's say w, x, y, and z) you have a choice of 4 choices for the Ones spot, 3 remaining choices for the Tens, 2 choices left for the Hundreds, and 1 choice for the Thousands; This gives you 4*3*2*1 (or 4!) = 24 combinations.

If we multiply (13) sets of digits by (24) combinations for each set, we get the answer of 312.

I hope this answer helped!

EDIT: I solved this assuming you weren't allowed to repeat digits. If you were to repeat digits, there would be more possible combinations, but you would have to divide them by [(#of repeats of a digit)!] so as not to count the same number multiple times.

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Answer 2

Answer:

All integers divisible by 3 follow a simple rule; the sum of all their digits add up to a multiple of 3. (If the sum of the digits is 2-digit or greater number, you can repeat the process and add the digits of that number until you get to 3, 6 or 9)

Using the digits 1, 2, 3, 5, 6, 7, and 9, we need to find all combinations of 4 digits that add up to a multiple of 3.

Since 3, 6 and 9 are already multiples of 3, the other digits in the number must add up to a multiple of 3 if they are present.

So, we can count the amount of combinations that would be divisible by three through separating them by whether any of the individual digits are multiples of 3 (Mo3).

0 Mo3s: (1, 2, 5, 7)

1 Mo3: N/A

2 Mo3s: [(1, 2) (1, 5) (2, 7) (5, 7)] * [(3, 6) (3, 9), (6, 9)]

3 Mo3s: N/A

So, there is 1 way with no individual digits that are multiples of 3, and (4*3 = 12) ways that have 2 Multiples of 3, for a total of 13 combinations of digits that add up to a multiple of 3. Any 4-digit integer using these combinations of digits will be divisible by 3.

Now how many numbers can we make from each? If you have any 4 digits (let's say w, x, y, and z) you have a choice of 4 choices for the Ones spot, 3 remaining choices for the Tens, 2 choices left for the Hundreds, and 1 choice for the Thousands; This gives you 4*3*2*1 (or 4!) = 24 combinations.

If we multiply (13) sets of digits by (24) combinations for each set, we get the answer of 312.

I hope this answer helped!

EDIT: I solved this assuming you weren't allowed to repeat digits. If you were to repeat digits, there would be more possible combinations, but you would have to divide them by [(#of repeats of a digit)!] so as not to count the same number multiple times.