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>
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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>?
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<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>
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What is the sum of digits of <math>a_{14}?</math>
  
<cmath>\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } ~12</cmath>
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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>
  
==Video Solution 1 by OmegaLearn (Divisibility makes diophantine equation trivial)==
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==Solution 1==
https://youtu.be/5LLl26VI-7Y
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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.
 +
 
 +
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>.
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 +
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
  
==Animated Video Solution==
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==Solution 2 (most intuitive solution)==
https://youtu.be/itDH7AgxYFo
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Let the common difference between consecutive <math>a_i</math> be <math>d</math>.  
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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>
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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>.
  
~Star League (https://starleague.us)
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~eibc (edited by CHECKMATE2021)
  
==Solution 1==
 
  
<cmath>1\leq a_1\leq10</cmath>
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==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/Ku_c1YHnLt0?si=HaykiiKOmQl2ugA_&t=6010
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~Math-X
  
<cmath>13\leq a_2\leq20</cmath>
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==Video Solution (Solve under 60 seconds!!!)==
 +
https://youtu.be/6O5UXi-Jwv4?si=DXihmbcAl8cHISp3&t=1174
  
<cmath>241\leq a_{15}\leq250</cmath>
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~hsnacademy
  
 +
==Video Solution==
 +
https://youtu.be/wYjg-sE-QWs
  
 +
~please like and subscribe
  
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 13 between them, so only <math>17</math> and <math>18</math> work, as <math>17*13=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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==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)==
 +
https://youtu.be/5LLl26VI-7Y
  
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.
+
==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)
  
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{\text{(A)} \hspace{0.1 in} 8}</math>
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==Video Solution by Magic Square==
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https://youtu.be/-N46BeEKaCQ?t=1047
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==Video Solution by Interstigation==
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https://youtu.be/DBqko2xATxs&t=3550
  
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==Video Solution by WhyMath==
 +
https://youtu.be/iP1ous_RW3M
  
~ apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
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~savannahsolver
  
==Solution 2==
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==Video Solution by harungurcan==
Let the common difference between consecutive <math>a_i</math> be <math>d</math>. Then, since <math>a_{15} - a_1 = 14d</math>, we find from the first and last inequalities that <math>231 \le d \le 249</math>. As <math>d</math> must be an integer, this means <math>d = 17</math>. Plugging this into all of the given inequalities so we may extract information about <math>a_1</math> gives
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https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1864s
<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>a_1 \le 3</math>, while the last inequality tells us <math>3 \le a_1</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 + 221 = 224</math>, so our answer is <math>\boxed{\textbf{(A)}\ 8}</math>.
 
  
~eibc
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~harungurcan
  
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==Video Solution by Dr. David==
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https://youtu.be/j8b6cHHHb0c
  
 
==See Also==  
 
==See Also==  
{{AMC8 box|year=2023|num-b=24}}
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{{AMC8 box|year=2023|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:38, 10 November 2024

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 (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)


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 (ProblemsAnswer KeyResources)
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

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