2016 AMC 10B Problems/Problem 24: Difference between revisions

Revision as of 12:54, 27 January 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 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 $8+9=$ $\boxed{\textbf{(D) }17}$ .

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

@@ Line 6: / Line 6: @@
 ==Solution==
-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>.
+The numbers are <math>10a+b, 10b+c,</math> and <math>10c+d</math>. Note that only <math>d</math> can be zero for the numbers ab, bc, and cd cannot start with a zero and that <math>a\le b\le c</math>.
 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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