How many 4-digit numbers with $3$, $4$, $6$ and $7$ are divisible by $44$?

Question

Consider all four-digit numbers where each of the digits $3$, $4$, $6$ and $7$ occurs exactly once. How many of these numbers are divisible by $44$?

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My attack:

There are $24$ possible four digit numbers where $3$, $4$, $6$ and $7$ occur exactly once. I thought of writing them all down and checking divisibility, but isn't there a better way to do this?

Also, how do I check divisibility by $44$ easily? I read on the internet there was a trick* to determine if a number is divisible by $11$, but a number which is divisible by $11$ doesn't have to be divisible by $44$, does it?

*For example $3729$, you write down $(7+9)-(3+2)=11$, which is divisible by $11$, so $3729$ is divisible by $11$.

I'm only looking for $\large{\textbf{a hint}}$.

Answer 1

For a number to be divisible by $4$, the last 2 digits have to form a 2-digit number that is divisible by $4$. This should simplify things a lot.

The trick for $11$: you already know.

And if $ABCD$ is divisible by both $4$ and $11$, it is divisible by $44$.

Answer 2

Start with the 11 trick.

It means you must have 3,7 in odd and 4,6 in even positions or vice versa. It is even, so you must have 3476 or one of the other three formed by transposing the 3 and the 7 or the 4 and the 6. That is only four to check. The number cannot end in 4, because 34 and 74 are not multiples of 4. So that leaves 2 to check.