Question

How many distinct 3 digit code can i create such that this code is divisible by 4.

For example these codes are rejected since they have repeating numbers/less than 3 digits:
024/100/112/996/444
​

Answer 1

The other answer is just wrong.

There are 9•9•8 = 648 distinct 3-digit codes. The first digit can be any numeral from 1-9, the next digit can be any from 0-9 minus the one used in the first position, and the last digit can be any from 0-9 minus both the numerals used in the first two positions.

But that doesn't even account for the divisibility constraint.

Let the code be
`abc`. We can expand this as

`100a + 10b + c`

In order for this to be divisible by 4, we observe that

`100a + 8b + 2b + c = 4 (25a + 2b) + (2b+c)`

so we only need
`2b+c` to be divisible by 4.

The last digit must be even, so there are only 5 choices for the last digit. I list the possibilities and outcomes below. For some integer
`k`, we need

`c=0 \implies 2b=4k \implies b=2k`

`c=2 \implies 2b+2=4k \implies b = 2k-1`

`c=4 \implies 2b+4 = 4k \implies b = 2(k-1)`

`c=6 \implies 2b+6 = 4k \implies b = 2k-3`

`c=8 \implies 2b+8=4k \implies b = 2(k-2)`

Ignoring
`a` for the moment, in the cases of
`c\in\{0,4,8\}`,
`b` is also even. This leaves 3 choices for
`c` and 2 choices for
`b`.

Likewise, in the cases of
`c\in\{2,6\}`,
`b` is odd. This leaves 2 choices for
`c` and 5 choices for
`b`.

Now taking into account the choice for
`a`, we have the following decision tree.

• If
`a\in\{2,6\}` and
`c\in\{0,4,8\}`, then
`b\in\{0,2,4,6,8\}\setminus\{a,c\}` - a total of 2•3•3 = 18 codes.

• If
`a\in\{4,8\}` and
`c\in\{0,4,8\}\setminus\{a\}`, then
`b\in\{0,2,4,6,8\}\setminus\{a,c\}` - a total of 2•2•3 = 12 codes.

• If
`a\in\{2,6\}` and
`c\in\{2,6\}\setminus\{a\}`, then
`b\in\{1,3,5,7,9\}\setminus\{a,c\}` - a total of 2•1•5 = 10 codes.

• If
`a\in\{4,8\}` and
`c \in\{2,6\}`, then
`b\in\{1,3,5,7,9\}` - a total of 2•2•5 = 20 codes.

• If
`a\in\{1,3,5,7,9\}` and
`c\in\{0,4,8\}`, then
`b\in\{0,2,4,6,8\}\setminus\{c\}` - a total of 5•3•4 = 60 codes.

• If
`a\in\{1,3,5,7,9\}` and
`c\in\{2,6\}`, then
`b\in\{1,3,5,7,9\}\setminus\{a\}` - a total of 5•2•4 = 40 codes.

Hence there are a total of 18 + 12 + 10 + 20 + 60 + 40 = 160 codes.