2023 AMC 10 B Problem 23

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In This Page:

To help you master 2023 AMC 10 B Problem 23, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

Key ideas, useful formulas, and takeaways you can apply to similar problems

Related problems from past AMC exams, plus new practice problems that build the same skills

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2023 AMC10 B Problem 23

Solution

Solution 1

Algebraic Factoring and Bounding Method

By setting the arithmetic sum formula equal to the possible true sums, we create a Diophantine equation. We can severely restrict the number of terms by establishing a lower bound based on the minimum possible sequence.

Sum of an Arithmetic Sequence: S = $\frac{n}{2}$ (2a + (n-1)d)

Step 1: Determine the possible true sums

Carl wrote one term off by 1, which means his total was off by exactly 1. Therefore, the true sum of the arithmetic sequence must be either 1 less or 1 more than his recorded total.

S = 222 - 1 = 221

S = 222 + 1 = 223

Step 2: Set up the factoring equation

We apply the standard sum formula for an arithmetic sequence. By setting this equal to our possible true sums and multiplying by 2 to clear the fraction, we express twice the sum as a product of two integer factors.

S = \frac{n}{2} (2 a + (n - 1) d)

n (2 a + (n - 1) d) = 442 or 446

Step 3: Establish a lower bound for the sum

In AMC sequence problems, unless specified otherwise, it is standard to test positive integer terms first. This means the first term is at least 1. We are given that the common difference is an integer strictly greater than 1, so it must be at least 2. Substituting these minimums into the sum formula gives a lower bound in terms of the number of terms.

a \geq 1

d \geq 2

S \geq \frac{n}{2} (2 (1) + (n - 1) 2)

S \geq \frac{n}{2} (2 n) = n^{2}

Step 4: Bound the number of terms

Because the true sum is at most 223, we can use our inequality to find the maximum possible number of terms. Taking the square root restricts the number of terms to a narrow window.

n^{2} \leq S \leq 223

n \leq 223 \approx 14.93

n \leq 14

Step 5: Identify valid factors

We are given that the number of terms must be at least 3. We check the prime factorizations of 442 and 446 to find any factors between 3 and 14. The number 446 has no factors in this range, while 442 has exactly one.

3 \leq n \leq 14

446 = 2 \cdot 223 ⟹ Factors: 1, 2, 223, 446

442 = 2 \cdot 13 \cdot 17 ⟹ Factors: 1, 2, 13, 17, 26, 34, 221, 442

n = 13

Step 6: Solve for the sequence parameters

Substitute the valid number of terms back into the factored equation for 442. Simplify the resulting linear equation. Since the common difference must be at least 2, we test values to find a positive integer for the first term.

13 (2 a + 12 d) = 442

2 a + 12 d = 34

a + 6 d = 17

If d = 2, then a = 17 - 12 = 5

If d \geq 3, then a \leq 17 - 18 = - 1 (not positive)

Step 7: Calculate the final requested sum

Add the first term, common difference, and number of terms together to find the final answer.

a + d + n = 5 + 2 + 13 = 20

Correct answer: B. $20$
20 (Option B). Check: A sequence starting at 5 with a difference of 2 and 13 terms ends at 29. The sum is 13 * (5 + 29) / 2 = 221. If Carl recorded one term 1 too high, his sum would be 222, satisfying all conditions.

Solution 2

Median and Average Term Method

In an arithmetic sequence, the average of the terms equals the median. By analyzing the median, we can quickly determine the number of terms and the middle term without heavy algebra.

Average equals Median: In an arithmetic sequence, the average of all terms is exactly the median (middle) value.

Step 1: Relate the average term to the median

In any arithmetic sequence, the average of all terms is equal to the median term. For a sequence of integers, the median must be an integer if the number of terms is odd, or a half-integer if the number of terms is even. Therefore, the average term must be a multiple of one-half.

Average term = \frac{S}{n}

Average term = Median term

\frac{S}{n} = \frac{k}{2} for some integer k

Step 2: Determine the divisibility condition

Rearranging the relationship shows that the number of terms must perfectly divide twice the true sum.

n divides 2 S

2 S \in {442, 446}

Step 3: Test factors of 446

Consider the factors of 446 that are at least 3. If the number of terms were 223, the median would be extremely small. A sequence of 223 distinct positive integers cannot have a median of 1, so we reject 446.

446 = 2 \cdot 223 ⟹ n \in {223, 446}

If n = 223, Median = \frac{223}{223} = 1

Step 4: Test factors of 442

Consider the factors of 442. We test 13, which gives a reasonable median. Since 13 is odd, the median is exactly the middle term, which is the 7th term.

442 = 2 \cdot 13 \cdot 17

Test n = 13

Median = \frac{221}{13} = 17

a_{7} = 17

Step 5: Find the first term and common difference

Express the 7th term using the arithmetic sequence formula. We need the first term to be positive and the common difference to be at least 2. Testing the minimum common difference yields a valid positive first term.

a_{7} = a + 6 d = 17

If d = 2, then a = 5

If d = 3, then a = - 1 (rejected)

Step 6: Verify the next factor

Briefly check the next factor, 17. The median would be the 9th term. However, forcing the common difference to be at least 2 makes the first term negative, so this factor is rejected for positive sequences.

Test n = 17

Median = \frac{221}{17} = 13

a_{9} = a + 8 d = 13

If d \geq 2, then 8 d \geq 16 ⟹ a < 0

Step 7: Calculate the final sum

Using the unique positive integer parameters, compute the requested sum.

a + d + n = 5 + 2 + 13 = 20

Correct answer: B. $20$
20 (Option B). The statistical view confirms our algebraic result perfectly, validating the number of terms through the size of the median.

Solution 3

Exhaustive Case Analysis (Multiple-Choice Shortcut)

If we do not want to assume the sequence is entirely positive, we can test all valid integer factors and eliminate cases using the given multiple-choice options.

Multiple-Choice Filtering: When equations yield multiple valid integer families, test cases systematically and use the provided answer choices to identify the intended domain.

Step 1: List all possible numbers of terms

Without assuming the sequence contains only positive integers, the number of terms can be any factor of 442 or 446 that is at least 3.

n \in {13, 17, 26, 34, 221, 223, 442, 446}

Step 2: Evaluate the case for 13 terms

We previously found the relationship between the first term and common difference for 13 terms. If the common difference is 2, we get a sum that matches an answer choice. If the common difference is 3, we get a mathematically valid sequence, but its sum is not among the options.

n = 13 ⟹ a + 6 d = 17

If d = 2 ⟹ a = 5 ⟹ a + d + n = 20 (Option B)

If d = 3 ⟹ a = - 1 ⟹ a + d + n = 15 (Not an option)

Step 3: Evaluate the case for 17 terms

Test the next factor. The resulting requested sums are too small to match any answer choice.

n = 17 ⟹ 2 a + 16 d = 26 ⟹ a + 8 d = 13

If d = 2 ⟹ a = - 3 ⟹ a + d + n = 16

If d = 3 ⟹ a = - 11 ⟹ a + d + n = 9

Step 4: Observe the trend for larger factors

As the number of terms increases, the first term must become increasingly negative to compensate for the large number of terms and the common difference. This causes the requested sum to drop rapidly, moving further away from the answer choices.

n = 26 ⟹ 2 a + 25 d = 17

If d = 3 ⟹ 2 a = - 58 ⟹ a = - 29

a + d + n = - 29 + 3 + 26 = 0

Step 5: Conclude the unique matching solution

Since all other valid integer sequences produce sums well below the minimum answer choice, the only solution that fits the multiple-choice options is the one we found first.

a + d + n = 20

Correct answer: B. $20$
20 (Option B). Verifying that negative sequences generate sums outside the option range proves that assuming positive terms is a safe and intended shortcut.

Adapted Problems to Practice

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What Formulas Do You Need for This Problem?

Sum of an Arithmetic Sequence:

S = \frac{n}{2} (2 a + (n - 1) d)

S is the sum, n is the number of terms, a is the first term, d is the common difference. Used to relate the given sum to the unknown parameters of the sequence.

Median of an Arithmetic Sequence:

In an arithmetic sequence, the average of all terms is exactly equal to the median term.

$Average$ = $\frac{S}{n}$ = $Median$ Allows us to bypass the full algebraic sum formula and use divisibility rules on the average term to find the number of terms.

Key Takeaway: Remember This Strategy for Next Time

A problem asks for integer parameters of a sequence given its sum.

Try: Write the sum formula and clear the denominator to create a factoring problem (Diophantine equation) where the number of terms must divide twice the sum.

Why: Integer constraints severely limit the possible values for the number of terms to the integer factors of 2S.

An arithmetic sequence problem does not explicitly state the terms are positive, but it is a multiple-choice competition problem.

Try: Assume the terms are positive integers to establish a quick bound (like S ≥ n²), but understand that the answer choices can be used to filter out unintended negative cases.

Why: Competition problems often imply positive integers for sequences unless negative terms are necessary. If negative terms are valid, they usually produce requested quantities outside the range of the answer choices.

You need to relate the sum, number of terms, and the middle of an arithmetic sequence.

Try: Use the property that the average term equals the median term (S/n = median).

Why: It bypasses heavy algebra and allows you to use logic about the size of the middle term to quickly eliminate impossible values for the number of terms.

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