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

m (Case 1)
m (Case 2)
Line 20: Line 20:
 
===Case 2===
 
===Case 2===
 
This means that the digits themselves are in arithmetic sequence. This gives us <math>9</math> answers, <math>1234, 1357, 2345, 2468, 3456, 3579, 4567, 5678, 6789</math>.
 
This means that the digits themselves are in arithmetic sequence. This gives us <math>9</math> answers, <math>1234, 1357, 2345, 2468, 3456, 3579, 4567, 5678, 6789</math>.
 
+
<cmath></cmath>
 
Adding the two cases together, we find the answer to be <math>\boxed{\textbf{(D) }17}</math>.
 
Adding the two cases together, we find the answer to be <math>\boxed{\textbf{(D) }17}</math>.
  

Revision as of 17:25, 4 February 2017

Problem

How many four-digit integers $abcd$, with $a \neq 0$, have the property that the three two-digit integers $ab<bc<cd$ 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

The numbers are $10a+b, 10b+c,$ and $10c+d$. Note that only $d$ can be zero, and that $a\le b\le c$.

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 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 $\boxed{\textbf{(D) }17}$.

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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

Invalid username
Login to AoPS