How many 4 digit numbers are multiples of 4 no matter how you permute them? (base 10)
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A number written in base ten is a multiple of $4$ if and only if the two-digit number formed by its last two digits is a multiple of $4$. If these two digits are $ab$ in that order, clearly $b$ must be even. It’s not hard to verify that if $b$ is a multiple of $4$ (i.e., $0,4$, or $8$), then the two-digit number $ab$ is a multiple of $4$ if and only if $a$ is even, while if $b\equiv2\pmod 4$ (i.e., $2$ or $6$), then $ab$ is a multiple of $4$ if and only if $a$ is odd.
Now suppose that you have your four-digit number $n$ whose permutations are all multiples of $4$. Clearly $n$ cannot contain an odd digit, so it also cannot contain a digit $2$ or $6$. We conclude that all four digits must be multiples of $4$, i.e., must be $0,4$, or $8$. Now just count the four-digit numbers that can be formed from these three digits, being careful to exclude leading zeroes.
Let's introduce variables for the digits. It is always useful to be able to talk about the objects that you are dealing with, and the most direct way is to give them names. So, let's say the digits you are using are $a,b,c,d$.
The number that in base $10$ is written as $abcd$ is really $1000a+100b+10c+d$. Since $1000$ and $100$ are already multiples of $4$, this number will be a multiple of $4$ precisely when $10c+d$ is a multiple of $4$. In particular, of course, $d$ is even.
But you are told that no matter how you permute the digits, the result is again a multiple of $4$. So, in particular, $abdc$ is also a multiple of $4$. (Why did I look at this specific number? Well, I already know that $10c+d$ is a multiple of $4$, I may as well try to learn something more about $c$ and $d$ before bringing the other two variables into the mix.)
Ok, so $10d+c$ is a multiple of $4$. Subtracting, we see that $9(c-d)$ is a multiple of $4$, so $c-d$ is a multiple of $4$.
Now, since the same happens with all the other permutations, we see that also $c-a,c-b,a-b$ are multiples of $4$. If one of them is $2$ or $6$ (remember, they must be even), the others must also be among $2$ and $6$. But $62$ is not a multiple of $4$, so this cannot be.
So the only remaining options are that $a,b,c,d$ are among $0,4,8$. Now we see all the possible numbers formed this way work, and counting should be easy. (You may want to double check whether something like $0484$ is a valid number, since it is not a four digit integer. If not, $0$ cannot be used, unless $a=b=c=d=0$.)
I could be wrong; but a number is divisible by 4 if and only if the last 2 numbers are. There are of course 25 = 100/4 of these combinations of 2 numbers. Then the same applies to the first two which might be permuted with the last two and we have 25^2 = 625 of these numbers. If we do not accept the leading 0, it would be (88/4)*25 = 550 numbers or not far from it. More or less. Permutations between digits 1 and 2 and between 3 and 4 are not counted.