Difference between revisions of "2016 AMC 10B Problems/Problem 24"
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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>. | + | 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>. | 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>. |
Revision as of 12:54, 27 January 2020
Contents
Problem
How many four-digit integers , with , have the property that the three two-digit integers form an increasing arithmetic sequence? One such number is , where , , , and .
Solution
The numbers are and . Note that only can be zero for the numbers ab, bc, and cd cannot start with a zero and that .
To form the sequence, we need . This can be rearranged as . Notice that since the left-hand side is a multiple of , the right-hand side can only be or . (A value of would contradict .) Therefore we have two cases: and .
Case 1
If , then , so . This gives . If , then , so . This gives . If , then , so , giving . There is no solution for . Added together, this gives us answers for Case 1.
Case 2
This means that the digits themselves are in arithmetic sequence. This gives us answers, . Adding the two cases together, we find the answer to be .
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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