Difference between revisions of "2016 AMC 10B Problems/Problem 24"
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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>. | If <math>c=7</math>, then <math>b+d=4,\ 2b-a=6</math>, so <math>b=4</math>, giving <math>2470</math>. | ||
There is no solution for <math>c=6</math>. | There is no solution for <math>c=6</math>. | ||
− | + | <cmath></cmath> | |
+ | Added together, this gives us <math>8</math> answers for Case 1. | ||
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===Case 2=== | ===Case 2=== | ||
This means that the digits themselves are in arithmetic sequence. This gives us <math>9</math> answers, <math>1234, 1357, 2345, 2468, 3456, 3579, 4567, 5678, 6789</math>. | This means that the digits themselves are in arithmetic sequence. This gives us <math>9</math> answers, <math>1234, 1357, 2345, 2468, 3456, 3579, 4567, 5678, 6789</math>. |
Revision as of 18:24, 4 February 2017
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, 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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