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

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What is the sum of digits of <math>a_{14}?</math>
 
What is the sum of digits of <math>a_{14}?</math>
  
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$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12
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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>
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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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==Solution 1==
 
==Solution 1==
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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.
 
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.
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~eibc (edited by CHECKMATE2021)
 
~eibc (edited by CHECKMATE2021)
  
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==Solution 3(simplified)==
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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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-SharpWhiz17
  
 
==Video Solution by Math-X (First understand the problem!!!)==
 
==Video Solution by Math-X (First understand the problem!!!)==

Latest revision as of 14:07, 2 January 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 $a_{14}?$

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



Official Video Solution(Using Sequence properties for simple multiplication!)

https://youtu.be/fhqSU0qhsrE?feature=shared ~TheMathGeek (Plz sub)


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 (most intuitive solution)

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)

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 \( \boxed{8} \).

-SharpWhiz17

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

~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
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All AJHSME/AMC 8 Problems and Solutions

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