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

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<math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20</math>
 
<math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20</math>
  
==Solution==
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==Solution 1==
 
The numbers are <math>10a+b, 10b+c,</math> and <math>10c+d</math>. Note that only <math>d</math> can be zero, the numbers <math>ab</math>, <math>bc</math>, and <math>cd</math> cannot start with a zero, and <math>a\le b\le c</math>.
 
The numbers are <math>10a+b, 10b+c,</math> and <math>10c+d</math>. Note that only <math>d</math> can be zero, the numbers <math>ab</math>, <math>bc</math>, and <math>cd</math> cannot start with a zero, and <math>a\le b\le c</math>.
  
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Adding the two cases together, we find the answer to be <math>8+9=</math> <math>\boxed{\textbf{(D) }17}</math>.
 
Adding the two cases together, we find the answer to be <math>8+9=</math> <math>\boxed{\textbf{(D) }17}</math>.
  
==Solution 2==
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==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.  
 
Looking at the answer options, all the numbers are pretty small so it is easy to make a list.  
  

Latest revision as of 13:03, 27 January 2021

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 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

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

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