Competition · AMC preparation · step 4 of 4
AMC 8 · 2023 · #25
Grade 6 arithmeticPick an answer.
AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Equally spaced integers immediately signal Tool #5 (Look for a Pattern) — the arithmetic-sequence pattern a_n = a₁ + (n-1)d lets us express every term using just a₁ and d. From there, Tool #7 (Identify Subproblems) cleanly splits the work into two smaller hunts: first pin down d, then pin down a₁. Differences like a₁₅ - a₂ = 13d erase a₁ 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 a₁ makes all three given ranges true.
Write the sequence formula
Equally spaced integers form an arithmetic sequence, so every term follows the rule a_n = a₁ + (n-1)d.
Spotting the "add the same amount each time" rule is exactly the Grade 4 pattern standard.
4.OA.C.5Look For A PatternBound the common difference
Subtract a₂ from a₁₅ to cancel a₁, leaving 221 ≤ 13d ≤ 237.
Subtracting two inequalities to bound an unknown is the Grade 6 inequality-solving skill.
6.EE.B.8Identify SubproblemsNarrow the difference to one value
Dividing by 13 gives d ∈ {17, 18}; d = 18 forces a₁₅ ≥ 253, too big, so d = 17 is the only survivor.
Listing the integer candidates inside an inequality and crossing out the impossible ones is Tool #3 in action — still Grade 6 inequality reasoning.
The common difference between consecutive terms can only be 17.
▸ Why?
Only two whole-number steps survive the range facts — 17 and 18 — and a step of 18 pushes the 15th term past its allowed ceiling, so 17 is the only value left.
▸ Why?
The climb from the 2nd term to the 15th term is 13 equal steps, so its size must land between 221 and 237, and split among 13 steps that forces each step to sit between 17 and about 18.2 — leaving only the whole numbers 17 and 18.
▸ Why?
From the 2nd term to the 15th term there are 13 equal steps, so the total climb is 13 copies of the step size stacked up.
▸ Why?
The climb is the leftover part of the 15th term once the 2nd term is removed, and it is widest when the 15th term is as high as allowed and the 2nd term as low as allowed, giving at most 250 - 13 = 237 and at least 241 - 20 = 221.
▸ Why?
Sharing that 221-to-237 climb equally among the 13 steps undoes the stacking, leaving each single step between 17 and just over 18.
▸ Why?
A step of 18 would make the 15th term at least 253, but the 15th term is allowed to reach only 250, so a step of 18 is impossible.
▸ Why?
The 15th term is the first term with 14 equal steps added on, so a step of 18 adds a climb of 14 copies of 18, which is 252.
▸ Why?
Since the first term is at least 1, the 15th term is at least 1 + 252 = 253, which is more than the largest value, 250, the 15th term is allowed to take.
Pin down the first term
With d = 17 the three given ranges collapse onto a single integer: a₁ = 3.
Intersecting three integer ranges into a single value uses Grade 6 inequality logic.
6.EE.B.8Identify SubproblemsFind the 14th term and add digits
So a₁₄ = 3 + 13 × 17 = 224, whose digits give 2 + 2 + 4 = 8, choice (A).
Adding the three digits of a multi-digit whole number is the Grade 4 multi-digit arithmetic standard.
4.NBT.B.4Look For A PatternThis AMC 8 problem only needs Grade 6 inequality reasoning — combining range conditions to pin down one whole-number answer — that you already know!
- Write the sequence formula
- Bound the common difference
- Narrow the difference to one value
- Pin down the first term
- Find the 14th term and add digits
A parent dashboard for the family lives at sensimlab.com.