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

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If <math>c=7</math>, then <math>b+d=4,\ 2b-a=6</math>, so <math>b=4</math>, giving <math>2470</math>.
 
If <math>c=7</math>, then <math>b+d=4,\ 2b-a=6</math>, so <math>b=4</math>, giving <math>2470</math>.
 
There is no solution for <math>c=6</math>.
 
There is no solution for <math>c=6</math>.
−
<cmath></cmath>
 
 
Added together, this gives us <math>8</math> answers for Case 1.
 
Added together, this gives us <math>8</math> answers for Case 1.
  

Revision as of 17:27, 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 (Problems • Answer Key • Resources)
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

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