Difference between revisions of "2016 AMC 10B Problems/Problem 24"
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Counting all the cases we get our answer of <math>17</math> which is <math>\boxed{D}</math> | Counting all the cases we get our answer of <math>17</math> which is <math>\boxed{D}</math> | ||
-srisainandan6 | -srisainandan6 | ||
+ | |||
+ | ==Solution 3== | ||
+ | Let <math>x</math> be the difference between the numbers <math>ab</math>, <math>bc</math>, and <math>cd</math>. We then have | ||
+ | <cmath>10a + b = 10b + c - x</cmath> and | ||
+ | <cmath>10b + c = 10c + d - x.</cmath> | ||
+ | |||
+ | Subtracting the second equation from the first and then simplifying, we are left with: | ||
+ | <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 work. 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 = \boxed{\textbf{(D) }17}</math> | ||
==Video Solution== | ==Video Solution== |
Revision as of 23:16, 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 that work. Trying out each value of from to and summing the number of pairs yields
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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