2023 AMC 10B Problems/Problem 23
Contents
Problem
An arithmetic sequence of positive integers has terms, initial term , and common difference . Carl wrote down all the terms in this sequence correctly except for one term, which was off by . The sum of the terms he wrote was . What is ?
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.
-HIA2020
Solution 3
We must have the sum of terms of the arithmetic sequence is , which is or .
Since we have is prime, it cannot be the sum of the arithmetic sequence.
We see that is just .
We can write any arithmetic sequence with an odd amount of terms like this: where b is the middle term and d is the common difference.
By the sum of an arithmetic sequence, we have and and therefore or .
Then .
We must have that m is either or , so m is either or .
So or .
Taking, we have no answer choices that give , and then taking gives the only answer that works is .
Therefore we have
~ESAOPS
Solution 4
The formula for the sum of an arithmetic sequence is , where is the first term, is the last term, and is the number of terms. Let be the first term, be the common difference, and be the number of terms of Carl's sequence. Since the sum the sequence is less or more than , we have The right-hand side is either or . We know that it has to be divisible by so we can find the factors of and . Checking all the primes less than , we find that is prime and .
Because , the sum must be and the only possible values of are and . We can test both cases.
Case 1:
Substituting for gives us . Since the sequence consists of only positive integers, is an integer. We know that but if , then . Hence, this case is not possible.
Case 2:
Substituting for gives us . Using the same logic from case 1, we get , so . Solving for , we get . Therefore, , so the answer is .
~azc1027
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=F30LJeoaNWo
Video Solution 2
~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 |
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