Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 8 Problems/Problem 25"
 
(60 intermediate revisions by 29 users not shown)
Line 1: Line 1:
−
<math>1\leq a_1\leq10</math>,
+
==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 \text{ and } \thickspace 241 \le a_{15}\le 250.</cmath>
 +
What is the sum of digits of <math>a_{14}?</math>
  
−
<math>13\leq a_2\leq20</math>,
+
<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12</math>
  
−
<math>241\leq a_15\leq250</math>,
+
==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 are <math>13</math> 7 differences between them, so only <math>17</math> and <math>18</math> work, as <math>17*13=221</math>, so <math>17</math> satisfied <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.
  
 
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.
 
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{\text{(D)}11</math>
+
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
+
 
 +
~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)==
 +
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==
 +
{{AMC8 box|year=2023|num-b=24|after=Last Problem}}
 +
{{MAA Notice}}

Latest 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 (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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_8_Problems/Problem_25&oldid=233070"