AMC 10 · 2023 · #23

Grade 8 arithmetic

sequences-arithmeticdivisibility-rulesbound-inequality-then-enumeratesystematic-enumerationlinear-diophantine easier-related-problemsystematic-enumerationcaseworkbound-inequality-then-enumerate ↑ Prerequisites: sequences-arithmeticdivisibility-rules

📏 Long solution 💡 3 insights

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 $a + d + n$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: An arithmetic sequence of positive integers has first term $a$, common difference $d > 1$, and $n \ge 3$ terms. Carl wrote the sum but made one term off by $1$, giving total $222$. Find $a + d + n$.

Givens: Arithmetic sequence: first term $a \ge 1$ (positive integer), common difference $d \ge 2$ (integer, since $d > 1$ and terms are positive integers); Number of terms $n \ge 3$; Carl's recorded sum $= 222$; exactly one term was off by $\pm 1$, so the true sum $S = 221$ or $S = 223$; Sum formula: $S = \tfrac{n}{2}(2a + (n-1)d)$, i.e. $2S = n \cdot \big(2a + (n-1)d\big)$; Answer choices: (A) $24$, (B) $20$, (C) $22$, (D) $28$, (E) $26$

Unknowns: $a + d + n$

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #2 Make a Systematic List, #6 Guess and Check, #3 Eliminate Possibilities

Tool #9 (Easier Problem): the unknown number $n$ of terms is the toughest part, so doubling the sum makes $2S = n(2a + (n-1)d)$ a clean product of two integers. This turns the problem into 'factor $442$ or $446$, then check'. Tool #2 (Systematic List) sweeps each factor of $442 = 2 \cdot 13 \cdot 17$ and $446 = 2 \cdot 223$ as a candidate for $n$. Tool #6 (Guess and Check) plugs each candidate $n$ into the formula and solves for $(a, d)$. Tool #3 (Eliminate) drops candidates with $a < 1$ or $d < 2$ or non-integer.

Execute — Answer: B

#9 Solve an Easier Related Problem 8.F.B.4 Step 1

Set up the true sum.
Carl's $222$ is off by $\pm 1$, so the true sum is $S = 221$ or $S = 223$.
Using the standard arithmetic-sum formula $S = \tfrac{n}{2}(2a + (n-1)d)$, multiply by $2$: $2S = n \cdot K$ where $K = 2a + (n-1)d$ is a positive integer.
So $n$ divides $2S$.

$$2S = n \cdot \big(2a + (n-1)d\big),\quad 2S \in \{442,\;446\}$$

💡 Grade 8 'construct a function for a linear relationship' — the sum of a linear sequence is captured by one product formula.

#3 Eliminate Possibilities 6.NS.B.4 Step 2

Try $S = 223$.
Then $2S = 446 = 2 \cdot 223$ (and $223$ is prime).
Divisors are $1, 2, 223, 446$, and $n \ge 3$ leaves only $n = 223$ or $446$.
For $n = 223$: $K = 446/223 = 2$, so $2a + 222 d = 2$; but $a \ge 1$ and $d \ge 2$ gives left side $\ge 2 + 444 = 446$, impossible.
$n = 446$ is even worse.
So $S = 223$ is ruled out.

$$S=223:\;446 = 2 \cdot 223,\;n \in \{223, 446\}\;\text{both fail}\;\text{✗}$$

💡 Grade 6 'find factor pairs' — $223$ is prime, so divisor pool is tiny and the leftover term $K$ is too small for $a \ge 1, d \ge 2$.

#2 Make a Systematic List 6.NS.B.4 Step 3

So $S = 221 = 13 \cdot 17$.
Then $2S = 442 = 2 \cdot 13 \cdot 17$.
Divisors of $442$ are $1, 2, 13, 14, 17, 26, 34, 221, 442$.
Keep those with $n \ge 3$: candidate list is $n \in \{13, 14, 17, 26, 34, 221, 442\}$.

$$2S = 442 = 2 \cdot 13 \cdot 17,\quad n \in \{13, 14, 17, 26, 34, 221, 442\}$$

💡 Grade 6 'GCF and factor pairs' — list every divisor of $442$; $n$ must be one of them.

#9 Solve an Easier Related Problem 8.EE.A.2 Step 4

Bound $n$ further.
With $a \ge 1$ and $d \ge 2$, the smallest possible $K = 2a + (n-1)d$ is at least $2 + 2(n - 1) = 2n$.
So $nK \ge n \cdot 2n = 2n^{2}$, giving $2n^{2} \le 442$ so $n^{2} \le 221 < 225 = 15^{2}$.
Therefore $n \le 14$.
The only candidates left are $n = 13$ and $n = 14$.

$$2n^{2} \le 2S = 442 \;\Rightarrow\; n \le 14$$

💡 Grade 8 'square roots' — $n^{2} \le 221$ caps $n$ at $14$.

#6 Guess and Check 4.OA.B.4 Step 5

Test $n = 14$.
Then $K = 442/14 = 31.57\ldots$, not an integer.
Reject.

$$n=14:\;K = 442/14 \notin \mathbb{Z}\;\text{✗}$$

💡 Grade 4 'factor pairs' — $14$ does not divide $442$, so $K$ is not whole.

#6 Guess and Check 8.EE.C.7 Step 6

Test $n = 13$.
Then $K = 442/13 = 34$, so $2a + 12d = 34$, i.e.
$a + 6d = 17$.
With $d \ge 2$: if $d = 2$, $a = 17 - 12 = 5 \ge 1$ ✓.
If $d = 3$, $a = 17 - 18 = -1 < 0$ ✗.
So the unique solution is $a = 5, d = 2, n = 13$.

$$n=13:\;a + 6d = 17 \;\Rightarrow\; (a, d) = (5, 2)\;\checkmark$$

💡 Grade 8 'solve a linear equation' — one equation in two integers with tight positivity is forced.

#3 Eliminate Possibilities 3.OA.D.8 Step 7

Verify.
Sequence: $5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29$ ($13$ terms, $d = 2$).
Sum $= \tfrac{13}{2}(5 + 29) = \tfrac{13 \cdot 34}{2} = 13 \cdot 17 = 221$.
Carl's $222$ is $221 + 1$, so he wrote one term too high by $1$ — consistent.
Compute $a + d + n = 5 + 2 + 13 = 20$.
Choice (B).

$$a + d + n = 5 + 2 + 13 = 20 \;\Rightarrow\; \textbf{(B)}$$

💡 Grade 3 'two-step word problem' — add the three integers to finish.

[1] #9 8.F.B.4 Set up the true sum. Carl's $222$ is off by $\pm 1$, so the true sum is $S = 221

[2] #3 6.NS.B.4 Try $S = 223$. Then $2S = 446 = 2 \cdot 223$ (and $223$ is prime). Divisors are

[3] #2 6.NS.B.4 So $S = 221 = 13 \cdot 17$. Then $2S = 442 = 2 \cdot 13 \cdot 17$. Divisors of $

[4] #9 8.EE.A.2 Bound $n$ further. With $a \ge 1$ and $d \ge 2$, the smallest possible $K = 2a +

[5] #6 4.OA.B.4 Test $n = 14$. Then $K = 442/14 = 31.57\ldots$, not an integer. Reject.

[6] #6 8.EE.C.7 Test $n = 13$. Then $K = 442/13 = 34$, so $2a + 12d = 34$, i.e. $a + 6d = 17$. W

[7] #3 3.OA.D.8 Verify. Sequence: $5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29$ ($13$ terms,

Review

Reasonableness: Sequence check passes: $5 + 7 + 9 + \cdots + 29$ has $13$ terms with first $5$ and last $29$, sum $\tfrac{13(5 + 29)}{2} = 221$ ✓. All terms are positive integers, $d = 2 > 1$, $n = 13 \ge 3$ — every constraint satisfied. Carl's $222 = 221 + 1$ matches the 'off by $1$' clue (he could have written $14$ instead of $13$ for the term that should be $7$? Actually any term $t$ written as $t + 1$ gives $+1$ — many valid mistakes). The competing case $S = 223$ was ruled out cleanly. Final answer $20$ matches choice (B).

Alternative: Tool #1 (Draw a Diagram): write the sequence as a row of dots labeled $a, a+d, a+2d, \ldots, a + (n-1)d$. The sum equals the number of dots times the average (first plus last over two). Pair the dots from outside in (Gauss's trick), getting $n \cdot \tfrac{a + (a+(n-1)d)}{2}$. The same formula appears without algebra, and the factor-pair logic for $2 \cdot 221 = 442$ proceeds identically.

CCSS standards used (min grade 8)

3.OA.D.8 Solve two-step word problems using the four operations (Final addition $5 + 2 + 13$ to combine $a + d + n$.)
4.OA.B.4 Find all factor pairs for a whole number; recognize multiples (Checking that $14$ does not divide $442$ (so $n = 14$ is rejected).)
6.NS.B.4 Find the greatest common factor and least common multiple (Listing the divisors of $442 = 2 \cdot 13 \cdot 17$ and of $446 = 2 \cdot 223$ to enumerate candidate values of $n$.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Bounding $n \le \sqrt{221} < 15$ so $n \le 14$, dramatically shrinking the candidate list.)
8.EE.C.7 Solve linear equations in one variable (Solving $a + 6d = 17$ for integer $a \ge 1, d \ge 2$ to find the unique pair $(5, 2)$.)
8.F.B.4 Construct a function to model a linear relationship between two quantities (Stating the arithmetic-sum formula $S = \tfrac{n}{2}(2a + (n-1)d)$ as the linear-sequence partial-sum function.)

⭐ This AMC 10 problem only needs Grade 8 algebra you already know — the off-by-1 clue means the true sum is $221$ or $223$, so doubling gives $n \cdot K = 442$ or $446$. Factor pairs plus the positivity bound $n \le 14$ leave only $n = 13$, which forces $a = 5, d = 2$. Add up: (B) $20$.