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 10B Problems/Problem 23"
(→‎Problem)
 
(8 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
An arithmetic sequence of positive integers has <math>\text{n} \ge 3</math> terms, initial term <math>a</math>, and common difference <math>d > 1</math>. Carl wrote down all the terms in this sequence correctly except for one term, which was off by <math>1</math>. The sum of the terms he wrote was <math>222</math>. What is <math>a + d + n</math>?
+
An arithmetic sequence of positive integers has <math>n \ge 3</math> terms, initial term <math>a</math>, and common difference <math>d > 1</math>. Carl wrote down all the terms in this sequence correctly except for one term, which was off by <math>1</math>. The sum of the terms he wrote was <math>222</math>. What is <math>a + d + n</math>?
  
 
<math>\textbf{(A) } 24 \qquad \textbf{(B) } 20 \qquad \textbf{(C) } 22 \qquad \textbf{(D) } 28 \qquad \textbf{(E) } 26</math>
 
<math>\textbf{(A) } 24 \qquad \textbf{(B) } 20 \qquad \textbf{(C) } 22 \qquad \textbf{(D) } 28 \qquad \textbf{(E) } 26</math>
Line 8: Line 8:
 
Since one of the terms was either <math>1</math>  more or <math>1</math> less than it should have been, the sum should have been <math>222-1=221</math> or <math>222+1=223.</math>  
 
Since one of the terms was either <math>1</math>  more or <math>1</math> less than it should have been, the sum should have been <math>222-1=221</math> or <math>222+1=223.</math>  
  
−
The formula for an arithmetic series is <math>an+d\left(\dfrac{(n-1)n}2\right)=\dfrac n2\left(a+d(n-1)\right).</math> This can quickly be rederived by noticing that the sequence goes <math>a,a+d,a+2d,a+3d,\dots,a+(n-1)d</math>, and grouping terms.  
+
The formula for an arithmetic series is <math>an+d\left(\dfrac{(n-1)n}2\right)=\dfrac n2\left(2a+d(n-1)\right).</math> This can quickly be rederived by noticing that the sequence goes <math>a,a+d,a+2d,a+3d,\dots,a+(n-1)d</math>, and grouping terms.  
  
 
We know that <math>\dfrac n2(2a+d(n-1))=221</math> or <math>223</math>. Let us now show that <math>223</math> is not possible.  
 
We know that <math>\dfrac n2(2a+d(n-1))=221</math> or <math>223</math>. Let us now show that <math>223</math> is not possible.  
  
−
If <math>\dfrac n2(2a+d(n-1))=223</math>, we can simplify this to be <math>n(a+d(n-1))=223\cdot2.</math> Since every expression in here should be an integer, we know that either <math>n=2</math> and <math>a+d(n-1)=223</math> or <math>n=223</math> and <math>a+d(n-1)=2.</math> The latter is not possible, since <math>n\ge3,d>1,</math> and <math>a>0.</math> The former is also impossible, as <math>n\ge3.</math> Thus, <math>\dfrac n2(2a+d(n-1))\neq223\implies\dfrac n2(2a+d(n-1))=221</math>.  
+
If <math>\dfrac n2(2a+d(n-1))=223</math>, we can simplify this to be <math>n(2a+d(n-1))=223\cdot2.</math> Since every expression in here should be an integer, we know that either <math>n=2</math> and <math>2a+d(n-1)=223</math> or <math>n=223</math> and <math>2a+d(n-1)=2.</math> The latter is not possible, since <math>n\ge3,d>1,</math> and <math>a>0.</math> The former is also impossible, as <math>n\ge3.</math> Thus, <math>\dfrac n2(2a+d(n-1))\neq223\implies\dfrac n2(2a+d(n-1))=221</math>.  
  
 
We can factor <math>221</math> as <math>13\cdot17</math>. Using similar reasoning, we see that <math>221\cdot2</math> can not be paired as <math>2</math> and <math>221</math>, but rather must be paired as <math>13</math> and <math>17</math> with a factor of <math>2</math> somewhere.  
 
We can factor <math>221</math> as <math>13\cdot17</math>. Using similar reasoning, we see that <math>221\cdot2</math> can not be paired as <math>2</math> and <math>221</math>, but rather must be paired as <math>13</math> and <math>17</math> with a factor of <math>2</math> somewhere.  
Line 28: Line 28:
  
  
−
The summation of the set is <math>222 \pm 1 = 221,223</math>. First, <math>221</math>: its only possible factors are <math>1,13,17,221</math>, and as said by the problem, <math>n\ge3</math>, so <math>n</math> must be <math>13,17,</math>or <math>221</math>. Let's start with <math>n=13</math>. Then, <math>a+6d=17</math>, and this means <math>a=5</math>,<math>d=2</math>. Summing gives <math>13+5+2=20</math>. We don't need to test any more cases, since the problem writes that all <math>a+d+n</math> are the same, so we don't need to test other cases.
+
The summation of the set is <math>222 \pm 1 = 221,223</math>. First, <math>221</math>: its only possible factors are <math>1,13,17,221</math>, and as said by the problem, <math>n\ge3</math>, so <math>n</math> must be <math>13,17,</math>or <math>221</math>. Let's start with <math>n=13</math>. Then, <math>a+6d=17</math>, and this means <math>a=5</math>,<math>d=2</math>. Summing gives <math>13+5+2=\boxed{\textbf{(B) }20}</math>. We don't need to test any more cases, since the problem writes that all <math>a+d+n</math> are the same.
  
 
-HIA2020
 
-HIA2020
Line 70: Line 70:
  
 
Case 2: <math>n=13</math>
 
Case 2: <math>n=13</math>
−
Substituting for <math>n</math> gives us <math>a+6d=17</math>. Using the same logic from case 1, we get <math>3>d>1</math>, so <math>d=2</math>. Solving for <math>a</math>, we get <math>a=5</math>. Therefore, <math>a+d+n=5+2+12=20</math>, so the answer is <math>\boxed{\textbf{(B) }20.}</math> .
+
Substituting for <math>n</math> gives us <math>a+6d=17</math>. Using the same logic from case 1, we get <math>3>d>1</math>, so <math>d=2</math>. Solving for <math>a</math>, we get <math>a=5</math>. Therefore, <math>a+d+n=5+2+13=20</math>, so the answer is <math>\boxed{\textbf{(B) }20.}</math> .
 
~azc1027
 
~azc1027
  
 
==Video Solution 1 by SpreadTheMathLove==
 
==Video Solution 1 by SpreadTheMathLove==
 
https://www.youtube.com/watch?v=F30LJeoaNWo
 
https://www.youtube.com/watch?v=F30LJeoaNWo
 +
 +
==Video Solution 2==
 +
 +
https://youtu.be/IDYnZmNh1io
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2023|ab=B|num-b=22|num-a=24}}
 
{{AMC10 box|year=2023|ab=B|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:55, 23 October 2024

Problem

An arithmetic sequence of positive integers has $n \ge 3$ terms, initial term $a$, and common difference $d > 1$. Carl wrote down all the terms in this sequence correctly except for one term, which was off by $1$. The sum of the terms he wrote was $222$. What is $a + d + n$?

$\textbf{(A) } 24 \qquad \textbf{(B) } 20 \qquad \textbf{(C) } 22 \qquad \textbf{(D) } 28 \qquad \textbf{(E) } 26$

Solution 1

Since one of the terms was either $1$ more or $1$ less than it should have been, the sum should have been $222-1=221$ or $222+1=223.$

The formula for an arithmetic series is $an+d\left(\dfrac{(n-1)n}2\right)=\dfrac n2\left(2a+d(n-1)\right).$ This can quickly be rederived by noticing that the sequence goes $a,a+d,a+2d,a+3d,\dots,a+(n-1)d$, and grouping terms.

We know that $\dfrac n2(2a+d(n-1))=221$ or $223$. Let us now show that $223$ is not possible.

If $\dfrac n2(2a+d(n-1))=223$, we can simplify this to be $n(2a+d(n-1))=223\cdot2.$ Since every expression in here should be an integer, we know that either $n=2$ and $2a+d(n-1)=223$ or $n=223$ and $2a+d(n-1)=2.$ The latter is not possible, since $n\ge3,d>1,$ and $a>0.$ The former is also impossible, as $n\ge3.$ Thus, $\dfrac n2(2a+d(n-1))\neq223\implies\dfrac n2(2a+d(n-1))=221$.

We can factor $221$ as $13\cdot17$. Using similar reasoning, we see that $221\cdot2$ can not be paired as $2$ and $221$, but rather must be paired as $13$ and $17$ with a factor of $2$ somewhere.

Let us first try $n=13.$ Our equation simplifies to $2a+12d=34\implies a+6d=17.$ We know that $d>1,$ so we try the smallest possible value: $d=2.$ This would give us $a=17-2\cdot6=17-12=5.$ (Indeed, this is the only possible $d$.)

There is nothing wrong with the values we have achieved, so it is reasonable to assume that this is the only valid solution (or all solutions sum to the same thing), so we answer $a+d+n=5+2+13=\boxed{\textbf{(B) }20.}$

For the sake of completeness, we can explore $n=17.$ It turns out that we reach a contradiction in this case, so we are done.

~Technodoggo

Solution 2

There are $n$ terms, the $x$th term is $a+(x-1)d$, summation is $an+dn(n-1)/2=n(a+\frac{d(n-1)}{2})$.


The summation of the set is $222 \pm 1 = 221,223$. First, $221$: its only possible factors are $1,13,17,221$, and as said by the problem, $n\ge3$, so $n$ must be $13,17,$or $221$. Let's start with $n=13$. Then, $a+6d=17$, and this means $a=5$,$d=2$. Summing gives $13+5+2=\boxed{\textbf{(B) }20}$. We don't need to test any more cases, since the problem writes that all $a+d+n$ are the same.

-HIA2020

Solution 3

We must have the sum of terms of the arithmetic sequence is $222\pm 1$, which is $221$ or $223$.

Since we have $223$ is prime, it cannot be the sum of the arithmetic sequence.

We see that $221$ is just $13\times 17$.

We can write any arithmetic sequence with an odd amount of terms like this: $b-md,\cdots ,b-2d,b-d,b,b+d,b+2d,\cdots b+md$ where b is the middle term and d is the common difference.

By the sum of an arithmetic sequence, we have $13b=221$ and $17b=221$ and therefore $(b,n)$ $=$ $(13,17)$ or $(17,13)$.

Then $a+d+n=b-md+d+n=30-(m-1)d$.

We must have that m is either $\frac{17-1}{2}$ or $\frac{13-1}{2}$, so m is either $6$ or $8$.

So $a+d+n=30-5d$ or $30-7d$.

Taking$\mod 7$, we have no answer choices that give $2$, and then taking$\mod 5$ gives the only answer that works is $20$.

Therefore we have $a+d+n=\boxed{\textbf{(B) }20.}$

~ESAOPS

Solution 4

The formula for the sum of an arithmetic sequence is $n(\frac{a_1+a_n}{2})$, where $a$ is the first term, $a_n$ is the last term, and $n$ is the number of terms. Let $a$ be the first term, $d$ be the common difference, and $n$ be the number of terms of Carl's sequence. Since the sum the sequence is $1$ less or $1$ more than $222$, we have \[n(\frac{a+d(n-1)}{2})\pm{1} = 222\] \[n(\frac{a+d(n-1)}{2}) = 222\pm{1}.\] The right-hand side is either $221$ or $223$. We know that it has to be divisible by $n$ so we can find the factors of $221$ and $223$. Checking all the primes less than $15$, we find that $223$ is prime and $221=13\cdot17$.

Because $n\ge3$, the sum must be $221$ and the only possible values of $n$ are $13$ and $17$. We can test both cases.


Case 1: $n=17$ Substituting for $n$ gives us $a+8d=17$. Since the sequence consists of only positive integers, $d$ is an integer. We know that $d>1$ but if $d\ge2$, then $8d>13$. Hence, this case is not possible.


Case 2: $n=13$ Substituting for $n$ gives us $a+6d=17$. Using the same logic from case 1, we get $3>d>1$, so $d=2$. Solving for $a$, we get $a=5$. Therefore, $a+d+n=5+2+13=20$, so the answer is $\boxed{\textbf{(B) }20.}$ . ~azc1027

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=F30LJeoaNWo

Video Solution 2

https://youtu.be/IDYnZmNh1io

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 AMC 10 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_10B_Problems/Problem_23&oldid=230493"