Difference between revisions of "2016 AMC 10B Problems/Problem 24"
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<cmath>10a + 8c + d = 19b.</cmath> | <cmath>10a + 8c + d = 19b.</cmath> | ||
− | Notice that <math>b > c</math>. Because the values of <math>a</math> and <math>d</math> are irrelevant compared to the other numbers, we can just find pairs of <math>(b, c)</math> that | + | Notice that <math>b > c</math>. Because the values of <math>a</math> and <math>d</math> are irrelevant compared to the other numbers, we can just find pairs of <math>(b, c)</math> such that <math>a > 0</math>. Trying out each value of <math>b</math> from <math>2</math> to <math>9</math> and summing the number of pairs yields <math>1 + 2 + 4 + 4 + 3 + 2 + 1 + 0 = \boxed{\textbf{(D) }17}</math> |
- cappucher | - cappucher |
Revision as of 23:18, 27 April 2024
Contents
[hide]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 (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.
Counting all the cases we get our answer of which is -srisainandan6
Solution 3
Let be the difference between the numbers , , and . We then have and
Subtracting the second equation from the first and then simplifying, we are left with:
Notice that . Because the values of and are irrelevant compared to the other numbers, we can just find pairs of such that . Trying out each value of from to and summing the number of pairs yields
- cappucher
Video Solution
https://www.youtube.com/watch?v=UhPxvZ6V4Zs
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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