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== | + | ==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>. | ||
Revision as of 23:53, 22 January 2021
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 1
The numbers are and . Note that only can be zero, the numbers , , and cannot start with a zero, and .
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 an arithmetic sequence. This gives us answers, Adding the two cases together, we find the answer to be .
Solution 2
Looking at the answer options, all the numbers are pretty small so it is easy to make a list.
Counting all the cases we get our answer of which is -srisainandan6
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.