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

Line 17: Line 17:
 
Added together, this gives us <math>8</math> answers for Case 1.
 
Added together, this gives us <math>8</math> answers for Case 1.
  
Solution 2(Answer Choices)
+
 
  
 
===Case 2===
 
===Case 2===
Line 27: Line 27:
 
{{AMC10 box|year=2016|ab=B|num-b=23|num-a=25}}
 
{{AMC10 box|year=2016|ab=B|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
Solution 2(Answer Choices)

Revision as of 15:47, 24 December 2020

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 for the numbers ab, bc, and cd cannot start with a 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 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 $8+9=$ $\boxed{\textbf{(D) }17}$.

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

Solution 2(Answer Choices)

Invalid username
Login to AoPS