A8. Consider all four-digit numbers in which each of the digits $3,4,6$ and 7 appears exactly once. How many of these numbers are divisible by 44?
(A) 2
(B) 4
(C) 6
(D) 8
(E) 12

B-questions
A8. Consider all four-digit numbers in which each of the digits 3, 4, 6, and 7 appears exactly once. How many of these numbers are divisible by 44? (A) 2

Solution Let the number $n$ consisting of the digits $a, b, c$, and $d (n=1000a+100b+10c+d)$ satisfy the condition, so it is divisible by 44. It must then certainly be divisible by 11. Since $m=1001a+99b+11c=11(91a+9b+c)$ is also a multiple of 11, $m-n$ must also be a multiple of 11, so $m-n=a-b+c-d$ must be a multiple of 11. But this is at most the two highest minus the two lowest (so $13-7=6$) and at least -6, so it must be 0. Therefore, $a+c=b+d$. And since the digits sum to 20, it follows that $a+c=b+d=10$.

Suppose $d=4$, then $b=6$ and we get 3674 and 7634, but neither of these satisfy the condition (since 74 and 34 are not divisible by 4). Suppose $d=6$, then $b=4$ and we get 3476 and 7436, both of which do satisfy the condition. There are no other options for $d$, as $d$ must be even. This gives a total of 2 solutions.

Alternative solution Check the 24 possibilities, or the 12 even possibilities, or the 6 multiples of 4.