Difference between revisions of "2023 AMC 10B Problems/Problem 23"
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+ | ==Problem== | ||
An arithmetic sequence of positive integers has <math>\text{n} \ge 3</math> terms, initial term a, and common difference <math>\text{d} > 1</math>. 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? | An arithmetic sequence of positive integers has <math>\text{n} \ge 3</math> terms, initial term a, and common difference <math>\text{d} > 1</math>. 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? | ||
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+ | <math>\textbf{(A) } 24 \qquad \textbf{(B) } 20 \qquad \textbf{(C) } 22 \qquad \textbf{(D) } 28 \qquad \textbf{(E) } 26</math> | ||
==Solution 1== | ==Solution 1== | ||
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==Video Solution 1 by SpreadTheMathLove== | ==Video Solution 1 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=F30LJeoaNWo | https://www.youtube.com/watch?v=F30LJeoaNWo | ||
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+ | ==See also== | ||
+ | {{AMC10 box|year=2023|ab=B|num-b=22|num-a=24}} | ||
+ | {{MAA Notice}} |
Revision as of 20:51, 15 November 2023
Contents
[hide]Problem
An arithmetic sequence of positive integers has terms, initial term a, and common difference . 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?
Solution 1
Since one of the terms was either more or less than it should have been, the sum should have been or
The formula for an arithmetic series is This can quickly be rederived by noticing that the sequence goes , and grouping terms.
We know that or . Let us now show that is not possible.
If , we can simplify this to be Since every expression in here should be an integer, we know that either and or and The latter is not possible, since and The former is also impossible, as Thus, .
We can factor as . Using similar reasoning, we see that can not be paired as and , but rather must be paired as and with a factor of somewhere.
Let us first try Our equation simplifies to We know that so we try the smallest possible value: This would give us (Indeed, this is the only possible .)
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
For the sake of completeness, we can explore It turns out that we reach a contradiction in this case, so we are done.
~Technodoggo
Solution 2
There are terms, the th term is , summation is .
The summation of the set is . First, : its only possible factors are , and as said by the problem, , so must be or. Let's start with . Then, , and this means ,. Summing gives . We don't need to test any more cases, since the problem writes that all are the same, so we don't need to test other cases.
-HIA2020
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=F30LJeoaNWo
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.