AMC 8 · 2009 · #5
Easy mode Grade 4Problem
Imagine a list of numbers. The first three numbers are 1, 2, and 3.
From the fourth number on, here is the rule: each new number is the sum of the three numbers right before it. So the fourth number is 1+2+3=6.
Keep using that same rule to find the next numbers. What is the eighth number in the list?
Pick an answer.
AMC 8 2009 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: A sequence begins $1, 2, 3$. From the fourth term on, every term equals the sum of the three terms right before it. Find the eighth term.
Givens: First three terms: $a_1 = 1$, $a_2 = 2$, $a_3 = 3$; Rule: $a_n = a_{n-1} + a_{n-2} + a_{n-3}$ for $n \ge 4$; Worked example: $a_4 = 1 + 2 + 3 = 6$; Answer choices: (A) $11$, (B) $20$, (C) $37$, (D) $68$, (E) $99$
Unknowns: The eighth term of the sequence, $a_8$
Understand
Restated: A sequence begins $1, 2, 3$. From the fourth term on, every term equals the sum of the three terms right before it. Find the eighth term.
Givens: First three terms: $a_1 = 1$, $a_2 = 2$, $a_3 = 3$; Rule: $a_n = a_{n-1} + a_{n-2} + a_{n-3}$ for $n \ge 4$; Worked example: $a_4 = 1 + 2 + 3 = 6$; Answer choices: (A) $11$, (B) $20$, (C) $37$, (D) $68$, (E) $99$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #2 Make a Systematic List
The sequence is defined by a clear repeating rule: each term is the sum of the previous three. Tool #5 (Look for a Pattern) fits exactly — we follow the rule that generates the pattern, term by term. Tool #2 (Make a Systematic List) keeps the work organized: write the terms in a single row, in order, so the "previous three" needed at each step are always the last three numbers on the list. With only $a_4$ through $a_8$ to find, this beats setting up algebra (tool #13).
Execute — Answer: D
4.OA.C.5 Step 1 - Write the three starting terms in order.
- This is the seed of the list.
💡 Putting the known values on the list first means the "previous three" we need next is just the last three numbers visible.
3.NBT.A.2 Step 2 Apply the rule to get $a_4$ by summing the last three terms on the list.
💡 The problem hands us this one as the worked example, so it doubles as a check that we read the rule correctly.
3.NBT.A.2 Step 3 - Now the list is $1, 2, 3, 6$.
- Sum the last three to get $a_5$.
💡 Slide the "window of three" one step to the right and add.
3.NBT.A.2 Step 4 - List: $1, 2, 3, 6, 11$.
- Sum the last three for $a_6$.
💡 Same window-slide move — the rule never changes, only the three numbers we add.
3.NBT.A.2 Step 5 - List: $1, 2, 3, 6, 11, 20$.
- Sum the last three for $a_7$.
💡 By now the terms are growing fast — sanity-check each sum because a single addition error spreads forward.
3.NBT.A.2 Step 6 - List: $1, 2, 3, 6, 11, 20, 37$.
- Sum the last three for $a_8$, which is the answer.
💡 One final window-slide gives the target term. The completed list is $1, 2, 3, 6, 11, 20, 37, 68$.
4.OA.C.5 Write the three starting terms in order. This is the seed of the list. 3.NBT.A.2 Apply the rule to get $a_4$ by summing the last three terms on the list. 3.NBT.A.2 Now the list is $1, 2, 3, 6$. Sum the last three to get $a_5$. 3.NBT.A.2 List: $1, 2, 3, 6, 11$. Sum the last three for $a_6$. 3.NBT.A.2 List: $1, 2, 3, 6, 11, 20$. Sum the last three for $a_7$. 3.NBT.A.2 List: $1, 2, 3, 6, 11, 20, 37$. Sum the last three for $a_8$, which is the answe Review
Reasonableness: The terms grow but not crazily: each new term is roughly double the previous one (because it adds two smaller terms to the largest). From $a_7 = 37$, doubling gives about $74$ — and $68$ is right under that, which fits. Among the choices, $11, 20, 37$ are earlier terms in the sequence (classic distractors) and $99$ is too large to be a sum of $11 + 20 + 37$. Only $68$ fits, confirming (D).
Alternative: Tool #3 (Eliminate Possibilities) on the choices works too. The eighth term must equal the sum of the three terms before it, so it must be at least $a_5 + a_6 + a_7 \ge 11 + 20 + 37 = 68$ and exactly equal that. Choices $11, 20, 37$ are themselves earlier terms — too small — and $99$ exceeds the only valid sum. Only (D) $68$ survives.
CCSS standards used (min grade 4)
4.OA.C.5Generate a number or shape pattern that follows a given rule (Reading the recurrence rule $a_n = a_{n-1} + a_{n-2} + a_{n-3}$ and using it to extend the pattern term by term from $a_3$ to $a_8$.)3.NBT.A.2Fluently add and subtract within 1000 (Computing each new term as a sum of three earlier terms: $1+2+3$, $2+3+6$, $3+6+11$, $6+11+20$, and $11+20+37$.)
⭐ This AMC 8 problem only needs a Grade 4 skill — follow a given pattern rule — plus plain addition you've known since Grade 3!
⭐ This AMC 8 problem only needs a Grade 4 skill — follow a given pattern rule — plus plain addition you've known since Grade 3!
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