SOLUTION: A briefcase lock opens with the correct 4-digit code. If the digits can be any number from 0-9 with no repetition, how many 4-digit codes are possible that end in a multiple of 3?

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Question 1160695: A briefcase lock opens with the correct 4-digit code. If the digits can be any number from 0-9 with no repetition, how many 4-digit codes are possible that end in a multiple of 3?
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
A briefcase lock opens with the correct 4-digit code. If the digits can be any number from 0-9 with no repetition, how many 4-digit codes are possible that end in a multiple of 3?
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The units must be 0, 3, 6 or 9 ---> 1 of 4
The other 3 are 1 of 9,
then 1 of 8,
then 1 of 7.
---> 4*9*8*7
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If you don't want 0 to be used as a multiple of 3, then make adjustments.
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

So, there are 4 possibilities for the last digit in the 4-th position reading from left to right.

    These possibilities are 0, 3, 6 and 9.


Then there are 10-1 = 9 possibilities for the digit in  the 3-rd position;

                      8 possibilities for the digit in  the 2-nd position;

                      7 possibilities for the digit in  the 1-st position.


In all, there are  4*9*8*7 = 2016 different codes, satisfying imposed conditions.


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