AMC 8 · 2023 · #25

Grade 6 arithmetic

Archetype Small-Domain Constrained Search

sequences-arithmeticlinear-equations-two-vardigit-sum bound-inequality-then-enumeratesystematic-enumerationpattern-recognition ↑ Prerequisites: sequences-arithmeticmulti-digit-arithmetic

📏 Medium solution 💡 4 insights

Problem

Fifteen integers $a_1, a_2, a_3, \dots, a_{15}$ are arranged in order on a number line. The integers are equally spaced and have the property that
$1 \le a_1 \le 10, \thickspace 13 \le a_2 \le 20, \thickspace \text{ and } \thickspace 241 \le a_{15}\le 250.$
What is the sum of digits of $a_{14}?$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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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

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

#5 Look for a Pattern 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 $a_{15} = a_1 + 14d$.

$$a_n = a_1 + (n-1)d \;\Rightarrow\; a_2 = a_1 + d,\; a_{14} = a_1 + 13d,\; a_{15} = a_1 + 14d$$

💡 Spotting the "add the same amount each time" rule is exactly the Grade 4 pattern standard.

#7 Identify Subproblems 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 $13d$.

$$a_{15} - a_2 = 13d,\;\; 241 - 20 \le 13d \le 250 - 13,\;\; 221 \le 13d \le 237$$

💡 Subtracting two inequalities to bound an unknown is the Grade 6 inequality-solving skill.

#3 Eliminate Possibilities 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.

$$\tfrac{221}{13} \le d \le \tfrac{237}{13}\;\Rightarrow\; d \in \{17, 18\};\;\; d = 18 \Rightarrow a_{15} \ge 253 > 250\;\Rightarrow\; d = 17$$

💡 Listing the integer candidates inside an inequality and crossing out the impossible ones is Tool #3 in action — still Grade 6 inequality reasoning.

#7 Identify Subproblems 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 $a_1 = 3$.

$$1 \le a_1 \le 10,\;\; -4 \le a_1 \le 3,\;\; 3 \le a_1 \le 12\;\Rightarrow\; a_1 = 3$$

💡 Intersecting three integer ranges into a single value uses Grade 6 inequality logic.

#5 Look for a Pattern 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).

$$a_{14} = 3 + 13 \cdot 17 = 224,\;\; 2 + 2 + 4 = 8 \;\Rightarrow\; \textbf{(A)}$$

💡 Adding the three digits of a multi-digit whole number is the Grade 4 multi-digit arithmetic standard.

[1] #5 4.OA.C.5 Use the arithmetic-sequence pattern. Because the $15$ integers are equally space

[2] #7 6.EE.B.8 Subproblem 1: bound $d$. Subtract $a_2$ from $a_{15}$ to eliminate $a_1$ — the r

[3] #3 6.EE.B.8 Narrow $d$ to a single integer. Dividing the bounds by $13$ gives $17 \le d \le

[4] #7 6.EE.B.8 Subproblem 2: pin down $a_1$ now that $d = 17$. The three given ranges become th

[5] #5 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.5 Generate 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.4 Fluently 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.8 Write 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

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