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Difference between revisions of "2016 AMC 10B Problems/Problem 24"

Problem

How many four-digit integers $abcd$, with $a \neq 0$, have the property that the three two-digit integers $ab form an increasing arithmetic sequence? One such number is $4692$, where $a=4$, $b=6$, $c=9$, and $d=2$. $\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20$

Solution 1

The numbers are $10a+b, 10b+c,$ and $10c+d$. Note that only $d$ can be zero, the numbers $ab$, $bc$, and $cd$ cannot start with a zero, and $a\le b\le c$.

To form the sequence, we need $(10c+d)-(10b+c)=(10b+c)-(10a+b)$. This can be rearranged as $10(c-2b+a)=2c-b-d$. Notice that since the left-hand side is a multiple of $10$, the right-hand side can only be $0$ or $10$. (A value of $-10$ would contradict $a\le b\le c$.) Therefore we have two cases: $a+c-2b=1$ and $a+c-2b=0$.

Case 1

If $c=9$, then $b+d=8,\ 2b-a=8$, so $5\le b\le 8$. This gives $2593, 4692, 6791, 8890$. If $c=8$, then $b+d=6,\ 2b-a=7$, so $4\le b\le 6$. This gives $1482, 3581, 5680$. If $c=7$, then $b+d=4,\ 2b-a=6$, so $b=4$, giving $2470$. There is no solution for $c=6$. Added together, this gives us $8$ answers for Case 1.

Case 2

This means that the digits themselves are in an arithmetic sequence. This gives us $9$ answers, $$1234, 1357, 2345, 2468, 3456, 3579, 4567, 5678, 6789.$$ Adding the two cases together, we find the answer to be $8+9=$ $\boxed{\textbf{(D) }17}$.

Solution 2 (Brute Force, when you have lots of time)

Looking at the answer options, all the numbers are pretty small so it is easy to make a list. $12|23|34 \rightarrow 1234$ $13|35|57 \rightarrow 1357$ $14|48|82 \rightarrow 1482$ $23|34|45 \rightarrow 2345$ $24|46|68 \rightarrow 2468$ $24|47|70 \rightarrow 2470$ $25|59|93 \rightarrow 2593$ $34|45|56 \rightarrow 3456$ $35|57|79 \rightarrow 3579$ $35|58|81 \rightarrow 3581$ $45|56|67 \rightarrow 4567$ $46|69|92 \rightarrow 4692$ $56|67|78 \rightarrow 5678$ $56|68|80 \rightarrow 5680$ $67|78|89 \rightarrow 6789$ $67|79|91 \rightarrow 6791$ $88|89|90 \rightarrow 8890$

Counting all the cases we get our answer of $17$ which is $\boxed{D}$ -srisainandan6

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 