AMC10 2016 B

Answer (D): Let $k$ be the common difference for the arithmetic sequence. If $b=c$ or $c=d$, then $k=bc-ab=cd-bc$ must be a multiple of $10$, so $b=c=d$. However, the two-digit integers $bc$ and $cd$ are then equal, a contradiction. Therefore either $(b,c,d)$ or $(b,c,d+10)$ is an increasing arithmetic sequence. Case 1: $(b,c,d)$ is an increasing arithmetic sequence. In this case the additions of $k$ to $ab$ and $bc$ do not involve any carries, so $(a,b,c)$ also forms an increasing arithmetic sequence, as does $(a,b,c,d)$. Let $n=b-a$. If $n=1$, the possible values of $a$ are $1,2,3,4,5,$ and $6$. If $n=2$, the possible values of $a$ are $1,2,$ and $3$. There are no possibilities with $n\ge 3$. Thus in this case there are $9$ integers that have the required property: $1234,2345,3456,4567,5678,6789,1357,2468,$ and $3579$. Case 2: $(b,c,d+10)$ is an increasing arithmetic sequence. In this case the addition of $k$ to $bc$ involves a carry, so $(a,b,c-1)$ forms a nondecreasing arithmetic sequence, as does $(b,c-1,(d+10)-2)=(b,c-1,d+8)$. Hence $(a,b,c-1,d+8)$ is a nondecreasing arithmetic sequence. Again letting $n=b-a$, note that $0\le c=d+(9-n)\le 9$ and $1\le a=d+(8-3n)\le 9$. The only integers with the required properties are $8890$ with $n=0$; $5680$ and $6791$ with $n=1$; $2470,3581,$ and $4692$ with $n=2$; and $1482$ and $2593$ with $n=3$. Thus in this case there are $8$ integers that have the required property. The total number of integers with the required property is $9+8=17$.