Difference between revisions of "2023 AMC 8 Problems/Problem 25"
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==Problem== | ==Problem== | ||
− | Fifteen integers <math>a_1, a_2, a_3, \dots, a_{15}</math> are arranged in order on a number line. The integers are equally spaced and have | + | Fifteen integers <math>a_1, a_2, a_3, \dots, a_{15}</math> are arranged in order on a number line. The integers are equally spaced and have the property that |
<cmath>1 \le a_1 \le 10, \thickspace 13 \le a_2 \le 20, \thickspace \text{ and } \thickspace 241 \le a_{15}\le 250.</cmath> | <cmath>1 \le a_1 \le 10, \thickspace 13 \le a_2 \le 20, \thickspace \text{ and } \thickspace 241 \le a_{15}\le 250.</cmath> | ||
What is the sum of digits of <math>a_{14}?</math> | What is the sum of digits of <math>a_{14}?</math> |
Revision as of 20:24, 30 August 2024
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2 (most intuitive solution)
- 4 Video Solution by Math-X (First understand the problem!!!)
- 5 Video Solution (Solve under 60 seconds!!!)
- 6 Video Solution
- 7 Video Solution(🚀Just 3 min!🚀)
- 8 Video Solution 1 by OmegaLearn (Divisibility makes diophantine equation trivial)
- 9 Video Solution by SpreadTheMathLove Using Arithmetic Sequence
- 10 Animated Video Solution
- 11 Video Solution by Magic Square
- 12 Video Solution by Interstigation
- 13 Video Solution by WhyMath
- 14 Video Solution by harungurcan
- 15 See Also
Problem
Fifteen integers are arranged in order on a number line. The integers are equally spaced and have the property that What is the sum of digits of
Solution 1
We can find the possible values of the common difference by finding the numbers which satisfy the conditions. To do this, we find the minimum of the last two: , and the maximum–. There is a difference of between them, so only and work, as , so satisfies . The number is similarly found. , however, is too much.
Now, we check with the first and last equations using the same method. We know . Therefore, . We test both values we just got, and we can realize that is too large to satisfy this inequality. On the other hand, we can now find that the difference will be , which satisfies this inequality.
The last step is to find the first term. We know that the first term can only be from to since any larger value would render the second inequality invalid. Testing these three, we find that only will satisfy all the inequalities. Therefore, . The sum of the digits is therefore .
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
Solution 2 (most intuitive solution)
Let the common difference between consecutive be . Since , we find from the first and last inequalities that . As must be an integer, this means . Substituting this into all of the given inequalities so we may extract information about gives The second inequality tells us that while the last inequality tells us , so we must have . Finally, to solve for , we simply have , so our answer is .
~eibc (edited by CHECKMATE2021)
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/Ku_c1YHnLt0?si=HaykiiKOmQl2ugA_&t=6010 ~Math-X
Video Solution (Solve under 60 seconds!!!)
https://youtu.be/6O5UXi-Jwv4?si=DXihmbcAl8cHISp3&t=1174
~hsnacademy
Video Solution
~please like and subscribe
Video Solution(🚀Just 3 min!🚀)
~Education, the Study of Everything
Video Solution 1 by OmegaLearn (Divisibility makes diophantine equation trivial)
Video Solution by SpreadTheMathLove Using Arithmetic Sequence
https://www.youtube.com/watch?v=EC3gx7rQlfI
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=1047
Video Solution by Interstigation
https://youtu.be/DBqko2xATxs&t=3550
Video Solution by WhyMath
~savannahsolver
Video Solution by harungurcan
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1864s
~harungurcan
See Also
2023 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.