AMC 10 · 2006 · #18

Grade 6 arithmetic

Archetype Modular Cycle Frequency Count └ Sub-archetype Position-in-Cycle Lookup

recursive-sequencepattern-recognitionmodular-arithmetic identify-subproblems ↑ Prerequisites: recursive-sequence

📏 Medium solution 💡 2 insights

Problem

Let $a_1 , a_2 , ...$ be a sequence for which $a_1=2$ , $a_2=3$ , and $a_n=\frac{a_{n-1}}{a_{n-2}}$ for each positive integer $n \ge 3$ . What is $a_{2006}$ ?

Pick an answer.

(A)

$\frac{1}{2}$

(B)

$\frac{2}{3}$

(C)

$\frac{3}{2}$

(D)

(E)

AMC 10 2006 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 starts with a_1 = 2 and a_2 = 3. From the third term on, each term equals the previous term divided by the term before that: a_n = a_{n-1} / a_{n-2}. Find the value of the 2006th term.

Givens: a_1 = 2; a_2 = 3; a_n = a_{n-1} / a_{n-2} for every n >= 3

Unknowns: The value of a_2006

Understand

Givens: a_1 = 2; a_2 = 3; a_n = a_{n-1} / a_{n-2} for every n >= 3

Plan

Primary tool: #5 Look for a Pattern

Secondary: #2 Make a Systematic List, #9 Solve an Easier Related Problem

The term number 2006 is far too large to reach by computing one term at a time. But the rule only looks back two steps, so once a pair of consecutive terms repeats, the whole sequence must repeat from there. The plan is to list the first several terms, watch for the repeat, and then use the repeat length to jump straight to term 2006.

Execute — Answer: E

#2 Make a Systematic List 6.NS.A.1 Step 1

List the first several terms

Apply the rule term by term, dividing the previous term by the one before it.
Dividing by a fraction means multiplying by its reciprocal.

$$a_3 = \frac{a_2}{a_1} = \frac{3}{2}, \quad a_4 = \frac{a_3}{a_2} = \frac{3/2}{3} = \frac{1}{2}, \quad a_5 = \frac{a_4}{a_3} = \frac{1/2}{3/2} = \frac{1}{3}, \quad a_6 = \frac{a_5}{a_4} = \frac{1/3}{1/2} = \frac{2}{3}$$

💡 Just follow the rule a few times and write down what you get.

#5 Look for a Pattern 4.OA.C.5 Step 2

Spot the repeat

Keep going two more terms.
a_7 comes back to 2 and a_8 comes back to 3 — the same pair we started with (a_1, a_2).
Since the rule only depends on the two most recent terms, once that starting pair reappears the whole sequence must cycle.
So the sequence repeats every 6 terms.

$$a_7 = \frac{a_6}{a_5} = \frac{2/3}{1/3} = 2 = a_1, \quad a_8 = \frac{a_7}{a_6} = \frac{2}{2/3} = 3 = a_2$$

💡 When the starting pair shows up again, the sequence has no choice but to loop.

#9 Solve an Easier Related Problem 6.NS.B.2 Step 3

Jump to term 2006 with the cycle

Terms with the same position within a block of 6 are equal: a_n depends only on the remainder of n when divided by 6.
Divide 2006 by 6: it goes 334 times with 2 left over, so 2006 leaves the same remainder as 2.
That means a_2006 equals a_2, which is 3.
The matching choice is (E).

$$2006 = 6 \times 334 + 2 \;\Rightarrow\; a_{2006} = a_2 = 3$$

💡 Only where a number lands inside the repeating block matters, and 2006 lands where 2 does.

[1] #2 6.NS.A.1 Apply the rule term by term, dividing the previous term by the one before it. Di

[2] #5 4.OA.C.5 Keep going two more terms. a_7 comes back to 2 and a_8 comes back to 3 — the sam

[3] #9 6.NS.B.2 Terms with the same position within a block of 6 are equal: a_n depends only on

Review

Reasonableness: Check the cycle claim directly: the six listed values 2, 3, 3/2, 1/2, 1/3, 2/3 do return to 2, 3 at positions 7 and 8, so the period of 6 is real. The remainder step is easy to test on small term numbers: a_8 should equal a_2 because 8 = 6 + 2, and indeed a_8 = 3. The same logic gives a_2006 = a_2 = 3, so the answer sits inside the given choices as expected.

Alternative: Instead of matching remainders, you could keep listing positions that equal a_2: those are terms 2, 8, 14, 20, ... — every number that is 2 more than a multiple of 6. Since 2006 - 2 = 2004 = 6 x 334 is a multiple of 6, term 2006 is on that list and equals a_2 = 3.

CCSS standards used (min grade 6)

6.NS.A.1 Interpret and compute quotients of fractions and solve word problems (Dividing one term by another when those terms are fractions, e.g. (1/2) / (3/2) = 1/3.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Recognizing that the listed terms form a repeating cycle of length 6.)
6.NS.B.2 Fluently divide multi-digit numbers using the standard algorithm (Dividing 2006 by 6 to find the remainder 2 and locate term 2006 within its cycle.)

⭐ When a sequence starts repeating, find how long the loop is, then use the remainder to see where a far-away term lands in the loop.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

List the first several terms

Spot the repeat

Jump to term 2006 with the cycle

Review

CCSS standards used (min grade 6)

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