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

(→‎Solution 2)
(→‎Solution 2)
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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
 
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
  
−
==Solution 2==
+
==Solution 2 (most intuitive)==
 
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  
Line 22: Line 22:
 
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>.
 
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
+
~eibc (edited by CHECKMATE2021)
  
 
==Video Solution by Math-X (First understand the problem!!!)==
 
==Video Solution by Math-X (First understand the problem!!!)==

Revision as of 22:59, 19 January 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)

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(🚀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

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

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