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

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==Solution==
 
==Solution==
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Answer is (D) for those of you wondering (verified by coding). Still need solution.
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The numbers are <math>10a+b, 10b+c,</math> and <math>10c+d</math>. Note that only <math>d</math> can be zero, and that <math>a\le b\le c</math>.
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abcd that satisfy problem conditions:
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−
1234 1357 1482 2345 2468 2470 2593 3456 3579 3581 4567 4692 5678 5680 6789 6791 8890
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To form the sequence, we need <math>(10c+d)-(10b+c)=(10b+c)-(10a+b)</math>. This can be rearranged as <math>10(c-2b+a)=2c-b-d</math>. Notice that since the left-hand side is a multiple of <math>10</math>, the right-hand side can only be <math>0</math> or <math>10</math>. (A value of <math>-10</math> would contradict <math>a\le b\le c</math>.) Therefore we have two cases: <math>a+c-2b=1</math> and <math>a+c-2b=0</math>.
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===Case 1===
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If <math>c=9</math>, then <math>b+d=8,\ 2b-a=8</math>, so <math>5\le b\le 8</math>. This gives <math>2593, 4692, 6791, 8890</math>.
 +
If <math>c=8</math>, then <math>b+d=6,\ 2b-a=7</math>, so <math>4\le b\le 6</math>. This gives <math>1482, 3581, 5680</math>.
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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>.
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There is no solution for <math>c=6</math>.
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===Case 2===
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This means that the digits themselves are in arithmetic sequence. This gives <math>1234, 1357, 2345, 2468, 3456, 3579, 4567, 5678, 6789</math>.
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 +
Counting the solutions, the answer is <math>\textbf{(D) }17</math>.
  
 
==See Also==
 
==See Also==
 
{{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}}

Revision as of 16:38, 21 February 2016

Problem

How many four-digit integers $abcd$, with $a \not\equiv 0$, 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 $c=6$.

Case 2

This means that the digits themselves are in arithmetic sequence. This gives $1234, 1357, 2345, 2468, 3456, 3579, 4567, 5678, 6789$.

Counting the solutions, the answer is $\textbf{(D) }17$.

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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