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 the property that <cmath>1 \le a_1 \le 10, \thickspace 13 \le a_2 \le 20, \thickspace 241 \le a_{15}\le 250.</cmath> | + | 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 |
− | What is the sum of digits of <math>a_{14}</math> | + | <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> | ||
− | <math>\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } | + | <math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12</math> |
==Solution 1== | ==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: <math>241-20=221</math>, and the maximum–<math>250-13=237</math>. There is a difference of <math>13</math> between them, so only <math>17</math> and <math>18</math> work, as <math>17\cdot13=221</math>, so <math>17</math> satisfies <math>221\leq 13x\leq237</math>. The number <math>18</math> is similarly found. <math>19</math>, however, is too much. |
− | < | + | Now, we check with the first and last equations using the same method. We know <math>241-10\leq 14x\leq250-1</math>. Therefore, <math>231\leq 14x\leq249</math>. We test both values we just got, and we can realize that <math>18</math> is too large to satisfy this inequality. On the other hand, we can now find that the difference will be <math>17</math>, which satisfies this inequality. |
− | < | + | The last step is to find the first term. We know that the first term can only be from <math>1</math> to <math>3</math> since any larger value would render the second inequality invalid. Testing these three, we find that only <math>a_1=3</math> will satisfy all the inequalities. Therefore, <math>a_{14}=13\cdot17+3=224</math>. The sum of the digits is therefore <math>\boxed{\textbf{(A)}\ 8}</math>. |
+ | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat | ||
+ | ==Solution 2 (most intuitive solution)== | ||
+ | Let the common difference between consecutive <math>a_i</math> be <math>d</math>. | ||
+ | Since <math>a_{15} - a_1 = 14d</math>, we find from the first and last inequalities that <math>231 \le 14d \le 249</math>. As <math>d</math> must be an integer, this means <math>d = 17</math>. Substituting this into all of the given inequalities so we may extract information about <math>a_1</math> gives | ||
+ | <cmath>1 \le a_1 \le 10, \thickspace 13 \le a_1 + 17 \le 20, \thickspace 241 \le a_1 + 238 \le 250.</cmath> | ||
+ | The second inequality tells us that <math>1 \le a_1 \le 3</math> while the last inequality tells us <math>3 \le a_1 \le 12</math>, so we must have <math>a_1 = 3</math>. Finally, to solve for <math>a_{14}</math>, we simply have <math>a_{14} = a_1 + 13d = 3 + 13(17) = 3 + 221 = 224</math>, so our answer is <math>2 + 2 + 4 = \boxed{\textbf{(A)}\ 8}</math>. | ||
− | + | ~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== | |
+ | https://youtu.be/wYjg-sE-QWs | ||
+ | ~please like and subscribe | ||
− | + | ==Video Solution(🚀Just 3 min!🚀)== | |
+ | https://youtu.be/X95x9iseAB8 | ||
− | + | <i>~Education, the Study of Everything </i> | |
− | |||
− | |||
− | |||
− | |||
− | |||
==Video Solution 1 by OmegaLearn (Divisibility makes diophantine equation trivial)== | ==Video Solution 1 by OmegaLearn (Divisibility makes diophantine equation trivial)== | ||
https://youtu.be/5LLl26VI-7Y | https://youtu.be/5LLl26VI-7Y | ||
− | ==Video Solution by SpreadTheMathLove Using | + | ==Video Solution by SpreadTheMathLove Using Arithmetic Sequence== |
https://www.youtube.com/watch?v=EC3gx7rQlfI | https://www.youtube.com/watch?v=EC3gx7rQlfI | ||
Line 48: | Line 57: | ||
==Video Solution by Magic Square== | ==Video Solution by Magic Square== | ||
https://youtu.be/-N46BeEKaCQ?t=1047 | https://youtu.be/-N46BeEKaCQ?t=1047 | ||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/DBqko2xATxs&t=3550 | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/iP1ous_RW3M | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by harungurcan== | ||
+ | https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1864s | ||
+ | |||
+ | ~harungurcan | ||
+ | |||
+ | ==Video Solution by Dr. David== | ||
+ | https://youtu.be/j8b6cHHHb0c | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2023|num-b=24|after=Last Problem}} | {{AMC8 box|year=2023|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:38, 10 November 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 Video Solution by Dr. David
- 16 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
Video Solution by Dr. David
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.