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2023 AMC 8 Problems/Problem 25: Difference between revisions

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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>


==Solution 1==
==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 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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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat


==Solution 2 (most intuitive solution)==
==Solution 2 (complicated but fast)==Step 1: Express everything in terms of a_1 and the common difference d
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 the 15 integers are equally spaced, they form an arithmetic sequence:
<cmath>1 \le a_1 \le 10, \thickspace 13 \le a_1 + 17 \le 20, \thickspace 241 \le a_1 + 238 \le 250.</cmath>
a_1, a_1 + d, a_1 + 2d, \dots, a_1 + 14d.
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>.
 
So:
a_2 = a_1 + d, \quad a_{15} = a_1 + 14d.
 
⸻
 
Step 2: Translate the given inequalities
 
We are told:
1 \le a_1 \le 10, \quad 13 \le a_2 \le 20, \quad 241 \le a_{15} \le 250.
 
⸻


~eibc (edited by CHECKMATE2021)
Step 3: Find the possible range for a_{15} - a_2


We have:
a_{15} - a_2 = (a_1 + 14d) - (a_1 + d) = 13d.


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


-SHARPWHIZ17
So the smallest possible value of a_{15} - a_2 happens when a_{15} is smallest and a_2 is largest:
241 - 20 = 221.
The largest possible value of a_{15} - a_2 happens when a_{15} is largest and a_2 is smallest:
250 - 13 = 237.


Hence,
221 \le a_{15} - a_2 \le 237.


==Video Solution by Math-X (First understand the problem!!!)==
⸻
 
Step 4: Solve for d
 
Since a_{15} - a_2 = 13d,
221 \le 13d \le 237.
Divide through by 13:
17 \le d \le 18.23.
Because d is an integer, d can only be 17 or 18.
 
⸻
 
Step 5: Check each possible d
 
Case 1: d = 17
Then
a_2 = a_1 + 17 \implies 13 \le a_1 + 17 \le 20 \Rightarrow -4 \le a_1 \le 3.
But since 1 \le a_1 \le 10, the overlap gives 1 \le a_1 \le 3.
 
Now check a_{15} = a_1 + 14(17) = a_1 + 238.
We need 241 \le a_{15} \le 250:
241 \le a_1 + 238 \le 250 \Rightarrow 3 \le a_1 \le 12.
The overlap of 1 \le a_1 \le 3 and 3 \le a_1 \le 12 is a_1 = 3.
 
So a_1 = 3, d = 17.
 
⸻
 
Case 2: d = 18
Then
a_2 = a_1 + 18 \implies 13 \le a_1 + 18 \le 20 \Rightarrow -5 \le a_1 \le 2.
Within 1 \le a_1 \le 10, this gives a_1 = 1 or 2.
 
But a_{15} = a_1 + 14(18) = a_1 + 252,
so
a_{15} \ge 1 + 252 = 253 > 250,
which doesn’t satisfy the given range.
Hence d = 18 is impossible.
 
⸻
 
Step 6: Compute a_{14}
 
With a_1 = 3 and d = 17,
a_{14} = a_1 + 13d = 3 + 13(17) = 3 + 221 = 224.
 
⸻
 
Step 7: Find the sum of digits
 
2 + 2 + 4 = 8.
 
⸻
 
✅ Final Answer:
\boxed{8}
 
==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 <math>\boxed{\textbf{(A)}\ 8}</math>.
 
-ChamzC19
 
==Video Solution by Math-X==
https://youtu.be/Ku_c1YHnLt0?si=HaykiiKOmQl2ugA_&t=6010  
https://youtu.be/Ku_c1YHnLt0?si=HaykiiKOmQl2ugA_&t=6010  
~Math-X
~Math-X


==Video Solution (Solve under 60 seconds!!!)==
==Video Solution (A Clever Explanation You’ll Get Instantly)==
https://youtu.be/6O5UXi-Jwv4?si=DXihmbcAl8cHISp3&t=1174
https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=3376
 
~hsnacademy
~hsnacademy


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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}}
[[Category:Introductory Number Theory Problems]]

Latest revision as of 14:22, 10 November 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)==Step 1: Express everything in terms of a_1 and the common difference d

Since the 15 integers are equally spaced, they form an arithmetic sequence: a_1, a_1 + d, a_1 + 2d, \dots, a_1 + 14d.

So: a_2 = a_1 + d, \quad a_{15} = a_1 + 14d.

⸻

Step 2: Translate the given inequalities

We are told: 1 \le a_1 \le 10, \quad 13 \le a_2 \le 20, \quad 241 \le a_{15} \le 250.

⸻

Step 3: Find the possible range for a_{15} - a_2

We have: a_{15} - a_2 = (a_1 + 14d) - (a_1 + d) = 13d.

Now, based on the ranges given: a_{15} \text{ can be between } 241 \text{ and } 250, a_2 \text{ can be between } 13 \text{ and } 20.

So the smallest possible value of a_{15} - a_2 happens when a_{15} is smallest and a_2 is largest: 241 - 20 = 221. The largest possible value of a_{15} - a_2 happens when a_{15} is largest and a_2 is smallest: 250 - 13 = 237.

Hence, 221 \le a_{15} - a_2 \le 237.

⸻

Step 4: Solve for d

Since a_{15} - a_2 = 13d, 221 \le 13d \le 237. Divide through by 13: 17 \le d \le 18.23. Because d is an integer, d can only be 17 or 18.

⸻

Step 5: Check each possible d

Case 1: d = 17 Then a_2 = a_1 + 17 \implies 13 \le a_1 + 17 \le 20 \Rightarrow -4 \le a_1 \le 3. But since 1 \le a_1 \le 10, the overlap gives 1 \le a_1 \le 3.

Now check a_{15} = a_1 + 14(17) = a_1 + 238. We need 241 \le a_{15} \le 250: 241 \le a_1 + 238 \le 250 \Rightarrow 3 \le a_1 \le 12. The overlap of 1 \le a_1 \le 3 and 3 \le a_1 \le 12 is a_1 = 3.

So a_1 = 3, d = 17.

⸻

Case 2: d = 18 Then a_2 = a_1 + 18 \implies 13 \le a_1 + 18 \le 20 \Rightarrow -5 \le a_1 \le 2. Within 1 \le a_1 \le 10, this gives a_1 = 1 or 2.

But a_{15} = a_1 + 14(18) = a_1 + 252, so a_{15} \ge 1 + 252 = 253 > 250, which doesn’t satisfy the given range. Hence d = 18 is impossible.

⸻

Step 6: Compute a_{14}

With a_1 = 3 and d = 17, a_{14} = a_1 + 13d = 3 + 13(17) = 3 + 221 = 224.

⸻

Step 7: Find the sum of digits

2 + 2 + 4 = 8.

⸻

✅ Final Answer: \boxed{8}

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)
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