AMC 8 · 2023 · #25
Grade 6 arithmeticProblem
Fifteen integers a1,a2,a3,…,a15 are arranged in order on a number line. The integers are equally spaced and have the property that
1≤a1≤10,13≤a2≤20, and 241≤a15≤250.
What is the sum of digits of a14?
Pick an answer.
AMC 8 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.
Toolkit + CCSS Solution
Understand
Restated: Fifteen integers $a_1, a_2, \dots, a_{15}$ sit on a number line in equally spaced order. We know the first term satisfies $1 \le a_1 \le 10$, the second term satisfies $13 \le a_2 \le 20$, and the last term satisfies $241 \le a_{15} \le 250$. Find the sum of the digits of $a_{14}$.
Givens: $15$ integers $a_1, a_2, \dots, a_{15}$ are equally spaced — i.e., they form an arithmetic sequence with a common difference $d$; All terms are integers, so the common difference $d$ is an integer; $1 \le a_1 \le 10$; $13 \le a_2 \le 20$; $241 \le a_{15} \le 250$; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $11$, (E) $12$
Unknowns: The value of $a_{14}$, then the sum of its digits
Understand
Restated: Fifteen integers $a_1, a_2, \dots, a_{15}$ sit on a number line in equally spaced order. We know the first term satisfies $1 \le a_1 \le 10$, the second term satisfies $13 \le a_2 \le 20$, and the last term satisfies $241 \le a_{15} \le 250$. Find the sum of the digits of $a_{14}$.
Givens: $15$ integers $a_1, a_2, \dots, a_{15}$ are equally spaced — i.e., they form an arithmetic sequence with a common difference $d$; All terms are integers, so the common difference $d$ is an integer; $1 \le a_1 \le 10$; $13 \le a_2 \le 20$; $241 \le a_{15} \le 250$; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $11$, (E) $12$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
Equally spaced integers immediately signal Tool #5 (Look for a Pattern) — the arithmetic-sequence pattern $a_n = a_1 + (n-1)d$ lets us express every term using just $a_1$ and $d$. From there, Tool #7 (Identify Subproblems) cleanly splits the work into two smaller hunts: first pin down $d$, then pin down $a_1$. Differences like $a_{15} - a_2 = 13d$ erase $a_1$ from the picture and squeeze $d$ between two bounds. Tool #3 (Eliminate Possibilities) finishes the job — only integer values inside both ranges survive, and only one $a_1$ makes all three given ranges true.
Execute — Answer: A
4.OA.C.5 Step 1 - Use the arithmetic-sequence pattern.
- Because the $15$ integers are equally spaced, each term is the previous plus a fixed integer step $d$.
- The $n$-th term obeys $a_n = a_1 + (n-1)d$, so $a_2 = a_1 + d$, $a_{14} = a_1 + 13d$, and $a_{15} = a_1 + 14d$.
💡 Spotting the "add the same amount each time" rule is exactly the Grade 4 pattern standard.
6.EE.B.8 Step 2 - Subproblem 1: bound $d$.
- Subtract $a_2$ from $a_{15}$ to eliminate $a_1$ — the result is $13d$.
- Using the smallest possible $a_{15}$ with the largest $a_2$, and vice versa, gives a tight range for $13d$.
💡 Subtracting two inequalities to bound an unknown is the Grade 6 inequality-solving skill.
6.EE.B.8 Step 3 - Narrow $d$ to a single integer.
- Dividing the bounds by $13$ gives $17 \le d \le 18.23\ldots$, so $d \in \{17, 18\}$.
- Test $d = 18$: then $a_2 = a_1 + 18 \ge 19$ forces $a_1 \le 2$, and $a_{15} = a_1 + 252 \ge 253$, which violates $a_{15} \le 250$.
- So $d = 17$ is the only survivor — this is the key uniqueness step.
💡 Listing the integer candidates inside an inequality and crossing out the impossible ones is Tool #3 in action — still Grade 6 inequality reasoning.
6.EE.B.8 Step 4 - Subproblem 2: pin down $a_1$ now that $d = 17$.
- The three given ranges become three ranges for $a_1$: $1 \le a_1 \le 10$, $13 \le a_1 + 17 \le 20$ (so $-4 \le a_1 \le 3$), and $241 \le a_1 + 238 \le 250$ (so $3 \le a_1 \le 12$).
- The only integer in all three is $a_1 = 3$.
💡 Intersecting three integer ranges into a single value uses Grade 6 inequality logic.
4.NBT.B.4 Step 5 - Compute $a_{14}$ and add its digits.
- With $a_1 = 3$ and $d = 17$, $a_{14} = 3 + 13 \times 17 = 3 + 221 = 224$.
- The digits $2, 2, 4$ sum to $8$, matching choice (A).
💡 Adding the three digits of a multi-digit whole number is the Grade 4 multi-digit arithmetic standard.
4.OA.C.5 Use the arithmetic-sequence pattern. Because the $15$ integers are equally space 6.EE.B.8 Subproblem 1: bound $d$. Subtract $a_2$ from $a_{15}$ to eliminate $a_1$ — the r 6.EE.B.8 Narrow $d$ to a single integer. Dividing the bounds by $13$ gives $17 \le d \le 6.EE.B.8 Subproblem 2: pin down $a_1$ now that $d = 17$. The three given ranges become th 4.NBT.B.4 Compute $a_{14}$ and add its digits. With $a_1 = 3$ and $d = 17$, $a_{14} = 3 + Review
Reasonableness: Sanity check: with $a_1 = 3$ and $d = 17$, $a_2 = 20$ (inside $[13, 20]$ ✓), and $a_{15} = 3 + 14 \cdot 17 = 3 + 238 = 241$ (inside $[241, 250]$ ✓). Notice both endpoints are pressed against the boundary — the constraints really do force the unique solution. The arithmetic step $d = 17$ is steady (no jumps), all $15$ terms are integers between $3$ and $241$, and $a_{14} = a_{15} - d = 241 - 17 = 224$ confirms the calculation. Digit sum $2+2+4 = 8$ is reasonable for a $3$-digit number near $200$.
Alternative: Tool #6 (Guess and Check) on $d$: only a handful of integer step sizes can possibly stretch from the low teens at $a_2$ to roughly $245$ at $a_{15}$ across $13$ steps. Estimating $\tfrac{245 - 16}{13} \approx 17.6$ suggests $d = 17$ or $18$; quickly checking $d = 18$ overshoots $a_{15}$, so $d = 17$. Then $a_{15} = 241$ forces $a_1 = 3$, and $a_{14} = a_{15} - d = 241 - 17 = 224$ — same answer with less inequality manipulation.
CCSS standards used (min grade 6)
4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing the equally-spaced integers as an arithmetic pattern with rule $a_n = a_1 + (n-1)d$ and writing $a_2$, $a_{14}$, $a_{15}$ accordingly.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Computing $3 + 221 = 224$ and the digit sum $2 + 2 + 4 = 8$ in the final step.)6.EE.B.8Write an inequality of the form x > c or x < c and graph on a number line (Combining the three given range conditions, subtracting them to bound $13d$, narrowing $d$ to a single integer, and intersecting three ranges for $a_1$ down to one value.)
⭐ This AMC 8 problem only needs Grade 6 inequality reasoning — combining range conditions to pin down one whole-number answer — that you already know!
⭐ This AMC 8 problem only needs Grade 6 inequality reasoning — combining range conditions to pin down one whole-number answer — that you already know!
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