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Difference between revisions of "2023 AMC 8 Problems/Problem 25"

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(→‎Solution 4 (Simple, intuitive, and fast))
 
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<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12</math>
 
<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12</math>
  
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==Official Video Solution(Using Sequence properties for simple multiplication!)==
 
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https://youtu.be/fhqSU0qhsrE?feature=shared
 
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~TheMathGeek (Plz sub)
 
  
  
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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
 
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
  
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==Solution 2 (most intuitive solution)==
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==Solution 2 (complicated but fast)==
 
Let the common difference between consecutive <math>a_i</math> be <math>d</math>.  
 
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  
 
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>
 
<cmath>1 \le a_1 \le 10, \thickspace 13 \le a_1 + 17 \le 20, \thickspace 241 \le a_1 + 238 \le 250.</cmath>
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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>.
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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>.
  
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~eibc (edited by CHECKMATE2021)
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~eibc (edited by CHECKMATE2021 & PRAYER914)
  
 
==Solution 3(simplified)==
 
==Solution 3(simplified)==
 
We are given 15 equally spaced integers, with \( a_1 \) between 1 and 10, \( a_2 \) between 13 and 20, and \( a_{15} \) between 241 and 250. The integers form an arithmetic sequence, so we use the formula \( a_n = a_1 + (n-1) \cdot d \), where \( d \) is the common difference. From the given information, we calculate the range for \( d \) and find that \( d = 17 \) works. Substituting \( a_1 = 3 \) and \( d = 17 \) into the formula, we find \( a_{14} = 3 + 13 \cdot 17 = 224 \). The problem asks for the sum of the digits of \( a_{14} \), so we sum the digits of 224, which are 2, 2, and 4, giving \( 2 + 2 + 4 = 8 \), and the final answer is \( \boxed{8} \).
 
We are given 15 equally spaced integers, with \( a_1 \) between 1 and 10, \( a_2 \) between 13 and 20, and \( a_{15} \) between 241 and 250. The integers form an arithmetic sequence, so we use the formula \( a_n = a_1 + (n-1) \cdot d \), where \( d \) is the common difference. From the given information, we calculate the range for \( d \) and find that \( d = 17 \) works. Substituting \( a_1 = 3 \) and \( d = 17 \) into the formula, we find \( a_{14} = 3 + 13 \cdot 17 = 224 \). The problem asks for the sum of the digits of \( a_{14} \), so we sum the digits of 224, which are 2, 2, and 4, giving \( 2 + 2 + 4 = 8 \), and the final answer is \( \boxed{8} \).
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==Solution 4 (Simple, intuitive, and fast)==
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This solution is similar to solution 1, but useful if you are not familiar with inequalities. The distance from a1 to a15 should be 14d. a15 would be 14d + a1. 14d should be the multiple of 14 the closest to but not exceeding (a1 cannot be negative) 241. We find that 14*17 =238 is that number. Since d=17, a1 has to be less than or equal to 3. If a1 were to be 4 or greater, a2 would be greater than 20. Let’s return to the equation 14d + a1 equals a15. 238 + a1 = a15, and since a15 has to be at least 241, a1 can only be 3. That makes a15=241. Subtract 17 from that and we get 224. Then we add all the digits together and get <math>\boxed{\textbf{(A)}\ 8}</math>.
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-ChamzC19
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==Video Solution by Math-X==
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https://youtu.be/Ku_c1YHnLt0?si=HaykiiKOmQl2ugA_&t=6010
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~Math-X
  
 
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 
https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=3376
 
https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=3376
 
~hsnacademy
 
~hsnacademy
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==Video Solution by Math-X (First understand the problem!!!)==
 
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https://youtu.be/Ku_c1YHnLt0?si=HaykiiKOmQl2ugA_&t=6010
 
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~Math-X
 
  
 
==Video Solution==
 
==Video Solution==
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{{AMC8 box|year=2023|num-b=24|after=Last Problem}}
 
{{AMC8 box|year=2023|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Number Theory Problems]]

Latest revision as of 00:31, 28 September 2025

Problem

Fifteen integers $a_1, a_2, a_3, \dots, a_{15}$ are arranged in order on a number line. The integers are equally spaced and have the property that \[1 \le a_1 \le 10, \thickspace 13 \le a_2 \le 20, \thickspace \text{ and } \thickspace 241 \le a_{15}\le 250.\] What is the sum of digits of $a_{14}?$

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12$



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: $241-20=221$, and the maximum–$250-13=237$. There is a difference of $13$ between them, so only $17$ and $18$ work, as $17\cdot13=221$, so $17$ satisfies $221\leq 13x\leq237$. The number $18$ is similarly found. $19$, however, is too much.

Now, we check with the first and last equations using the same method. We know $241-10\leq 14x\leq250-1$. Therefore, $231\leq 14x\leq249$. We test both values we just got, and we can realize that $18$ is too large to satisfy this inequality. On the other hand, we can now find that the difference will be $17$, which satisfies this inequality.

The last step is to find the first term. We know that the first term can only be from $1$ to $3$ since any larger value would render the second inequality invalid. Testing these three, we find that only $a_1=3$ will satisfy all the inequalities. Therefore, $a_{14}=13\cdot17+3=224$. The sum of the digits is therefore $\boxed{\textbf{(A)}\ 8}$.

~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat

Solution 2 (complicated but fast)

Let the common difference between consecutive $a_i$ be $d$. Since $a_{15} - a_1 = 14d$, we find from the first and last inequalities that $231 \le 14d \le 249$. As $d$ must be an integer, this means $d = 17$. Substituting this into all of the given inequalities so we may extract information about $a_1$ gives \[1 \le a_1 \le 10, \thickspace 13 \le a_1 + 17 \le 20, \thickspace 241 \le a_1 + 238 \le 250.\] The second inequality tells us that $1 \le a_1 \le 3$, while the last inequality tells us $3 \le a_1 \le 12$, so we must have $a_1 = 3$. Finally, to solve for $a_{14}$, we simply have $a_{14} = a_1 + 13d = 3 + 13(17) = 3 + 221 = 224$, so our answer is $2 + 2 + 4 = \boxed{\textbf{(A)}\ 8}$.

~eibc (edited by CHECKMATE2021 & PRAYER914)

Solution 3(simplified)

We are given 15 equally spaced integers, with \( a_1 \) between 1 and 10, \( a_2 \) between 13 and 20, and \( a_{15} \) between 241 and 250. The integers form an arithmetic sequence, so we use the formula \( a_n = a_1 + (n-1) \cdot d \), where \( d \) is the common difference. From the given information, we calculate the range for \( d \) and find that \( d = 17 \) works. Substituting \( a_1 = 3 \) and \( d = 17 \) into the formula, we find \( a_{14} = 3 + 13 \cdot 17 = 224 \). The problem asks for the sum of the digits of \( a_{14} \), so we sum the digits of 224, which are 2, 2, and 4, giving \( 2 + 2 + 4 = 8 \), and the final answer is \( \boxed{8} \).

Solution 4 (Simple, intuitive, and fast)

This solution is similar to solution 1, but useful if you are not familiar with inequalities. The distance from a1 to a15 should be 14d. a15 would be 14d + a1. 14d should be the multiple of 14 the closest to but not exceeding (a1 cannot be negative) 241. We find that 14*17 =238 is that number. Since d=17, a1 has to be less than or equal to 3. If a1 were to be 4 or greater, a2 would be greater than 20. Let’s return to the equation 14d + a1 equals a15. 238 + a1 = a15, and since a15 has to be at least 241, a1 can only be 3. That makes a15=241. Subtract 17 from that and we get 224. Then we add all the digits together and get $\boxed{\textbf{(A)}\ 8}$. -ChamzC19

Video Solution by Math-X

https://youtu.be/Ku_c1YHnLt0?si=HaykiiKOmQl2ugA_&t=6010

~Math-X

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=3376 ~hsnacademy

Video Solution

https://youtu.be/wYjg-sE-QWs

~please like and subscribe

Video Solution(🚀Just 3 min!🚀)

https://youtu.be/X95x9iseAB8

~Education, the Study of Everything

Video Solution 1 by OmegaLearn (Divisibility makes diophantine equation trivial)

https://youtu.be/5LLl26VI-7Y

Video Solution by SpreadTheMathLove Using Arithmetic Sequence

https://www.youtube.com/watch?v=EC3gx7rQlfI

Animated Video Solution

https://youtu.be/itDH7AgxYFo

~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

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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