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

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 $44$?

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 $\large{\textbf{a hint}}$.

• No but a number that is divisible by 11 AND divisible by 4 must be divisible by 44. A simple check for divisibility by 4 is if the last two digits of the number are divisible by 4 then the whole number is divisible by 4 – KBusc Sep 8 '14 at 18:03

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 $44$.
• So for $3467$ I should apply the trick for $11$ and for $4$ check if $67$ is divisible by $4$, is that correct? – rae306 Sep 8 '14 at 18:10
• Because of divisibility by $4$, you already know that your number has to end in an even number. So $3467$ doesn't need to be considered at all. Say your number ends in $6$. Then the 10-th digit must be 3 or 7 because 36 and 76 are divisible by $4$ but $46$ is not. – Kim Jong Un Sep 8 '14 at 18:12