AMC 8 · 2009 · #5

Easy mode Grade 4

Problem

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.

(A)

(B)

(C)

(D)

(E)

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.

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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.

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

#2 Make a Systematic List 4.OA.C.5 Step 1

Write the first three terms

Write the three starting terms in order.
This is the seed of the list.

$$a_1, a_2, a_3 = 1, 2, 3$$

💡 Putting the known values on the list first means the "previous three" we need next is just the last three numbers visible.

#5 Look for a Pattern 3.NBT.A.2 Step 2

Find the fourth term

Apply the rule to get $a_4$ by summing the last three terms on the list.

$$a_4 = 1 + 2 + 3 = 6$$

💡 The problem hands us this one as the worked example, so it doubles as a check that we read the rule correctly.

#5 Look for a Pattern 3.NBT.A.2 Step 3

Find the fifth term

Now the list is $1, 2, 3, 6$.
Sum the last three to get $a_5$.

$$a_5 = 2 + 3 + 6 = 11$$

💡 Slide the "window of three" one step to the right and add.

#5 Look for a Pattern 3.NBT.A.2 Step 4

Find the sixth term

List: $1, 2, 3, 6, 11$.
Sum the last three for $a_6$.

$$a_6 = 3 + 6 + 11 = 20$$

💡 Same window-slide move — the rule never changes, only the three numbers we add.

#5 Look for a Pattern 3.NBT.A.2 Step 5

Find the seventh term

List: $1, 2, 3, 6, 11, 20$.
Sum the last three for $a_7$.

$$a_7 = 6 + 11 + 20 = 37$$

💡 By now the terms are growing fast — sanity-check each sum because a single addition error spreads forward.

#5 Look for a Pattern 3.NBT.A.2 Step 6

Find the eighth term

List: $1, 2, 3, 6, 11, 20, 37$.
Sum the last three for $a_8$, which is the answer.

$$a_8 = 11 + 20 + 37 = 68 \;\Rightarrow\; \textbf{(D)}$$

💡 One final window-slide gives the target term. The completed list is $1, 2, 3, 6, 11, 20, 37, 68$.

[1] #2 4.OA.C.5 Write the three starting terms in order. This is the seed of the list.

[2] #5 3.NBT.A.2 Apply the rule to get $a_4$ by summing the last three terms on the list.

[3] #5 3.NBT.A.2 Now the list is $1, 2, 3, 6$. Sum the last three to get $a_5$.

[4] #5 3.NBT.A.2 List: $1, 2, 3, 6, 11$. Sum the last three for $a_6$.

[5] #5 3.NBT.A.2 List: $1, 2, 3, 6, 11, 20$. Sum the last three for $a_7$.

[6] #5 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.5 Generate 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.2 Fluently 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!

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Write the first three terms

Find the fourth term

Find the fifth term

Find the sixth term

Find the seventh term

Find the eighth term

Review

CCSS standards used (min grade 4)

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