AMC 10 · 2011 · #23

Grade 6 arithmetic

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

base-conversionsequences-geometric pattern-recognitionconvert-to-algebra ↑ Prerequisites: pattern-recognition

📏 Medium solution 💡 3 insights

Problem

Seven students count from 1 to 1000 as follows:

Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.

Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.

Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.

Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.

Finally, George says the only number that no one else says.

What number does George say?

Pick an answer.

(A)

(B)

242

(C)

365

(D)

728

(E)

998

AMC 10 2011 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: Seven students, Alice through George in alphabetical order, take turns dealing with the integers from $1$ to $1000$. On her turn, a student considers every number no earlier student has said, groups those numbers in consecutive threes, and skips the middle number of each group while saying the rest. George ends up saying exactly one number that nobody else says. Which number is it?

Givens: The numbers run from $1$ to $1000$.; The students act in alphabetical order: Alice, Barbara, Candice, Debbie, Eliza, Fatima, George.; Each student looks only at the numbers no earlier student said, groups them in consecutive threes, and skips the middle of each group.; Alice's spoken list begins $1, 3, 4, 6, 7, 9, \ldots$, so she skips $2, 5, 8, \ldots$; Answer choices: (A) $37$, (B) $242$, (C) $365$, (D) $728$, (E) $998$.

Unknowns: The single number George says.

Understand

Plan

Primary tool: #5 Look for a Pattern

Secondary: #4 Introduce a Variable, #7 Identify Subproblems, #3 Eliminate Possibilities

Chasing 1000 numbers through seven students by hand is hopeless, so the move is Tool #5 (Look for a Pattern): watch how one student's leftovers turn into the next student's list. Tool #4 (Introduce a Variable) captures that hand-off cleanly — a list with first term $a$ and common difference $d$ passes on a list with first term $a+d$ and difference $3d$. Tool #7 (Identify Subproblems) splits the work into two easy tracks: how the difference grows (powers of three) and how the starting number grows (one addition per student). Tool #3 (Eliminate Possibilities) finishes: George's very next number already passes $1000$, so only one survives.

Execute — Answer: C

#5 Look for a Pattern 4.OA.C.5 Step 1

What Alice leaves behind

Alice looks at $1, 2, 3, \ldots, 1000$ and skips the middle of each triple $(1,2,3), (4,5,6), (7,8,9), \ldots$, i.e.
she skips $2, 5, 8, \ldots, 998$.
Those skipped numbers are exactly the list Barbara must work with, and they form an arithmetic sequence: first term $2$, common difference $3$.

$2,\, 5,\, 8,\, \ldots,\, 998$ — first term $2$, common difference $3$

💡 The numbers one student throws away are precisely the list the next student picks up.

#4 Introduce a Variable 6.EE.B.6 Step 2

The hand-off rule, with variables

Suppose a student is looking at an arithmetic sequence with first term $a$ and common difference $d$: $a,\, a+d,\, a+2d,\, a+3d,\, \ldots$ She keeps the 1st, 3rd, 4th, 6th, $\ldots$ terms and skips the 2nd, 5th, 8th, $\ldots$ The skipped terms are $a+d,\, a+4d,\, a+7d,\, \ldots$ — again an arithmetic sequence, now with first term $a+d$ and common difference $3d$.

$$(a,\, d) \;\longrightarrow\; (a+d,\; 3d)$$

💡 Skipping every third number starting from the second triples the gap and bumps the start up by one step.

#5 Look for a Pattern 6.EE.A.1 Step 3

The gap triples each round

Alice's list has common difference $d = 1$.
By the rule, the difference is multiplied by $3$ every time the list is handed on, so student number $n$ works with common difference $d_n = 3^{\,n-1}$.
For Alice through George these are $1, 3, 9, 27, 81, 243, 729$.

$$d_1, \ldots, d_7 = 3^0, 3^1, \ldots, 3^6 = 1,\,3,\,9,\,27,\,81,\,243,\,729$$

💡 Each hand-off multiplies the spacing by three, so the gaps are simply the powers of three.

#7 Identify Subproblems 4.NBT.B.4 Step 4

Track each starting number

The starting number updates by $a_{n+1} = a_n + d_n$ (the first skipped term is the old start plus the old gap).
Beginning from Alice's $a_1 = 1$: Barbara $1+1 = 2$, Candice $2+3 = 5$, Debbie $5+9 = 14$, Eliza $14+27 = 41$, Fatima $41+81 = 122$, George $122+243 = 365$.

$$a_1,\ldots,a_7 = 1,\, 2,\, 5,\, 14,\, 41,\, 122,\, 365$$

💡 Each new start is the old start plus the old gap — one addition per student.

#3 Eliminate Possibilities 4.NBT.A.2 Step 5

George is left with one number

George works with first term $365$ and common difference $729$, so his numbers would be $365,\, 365 + 729 = 1094,\, \ldots$ But $1094 > 1000$, so only $365$ lies in range.
George therefore says exactly one number, $365$, which is choice (C).

$365,\; 1094,\; \ldots$; only $365 \le 1000 \;\Rightarrow\; \textbf{(C)}\ 365$

💡 The next number already overshoots $1000$, so George is left with a single value.

[1] #5 4.OA.C.5 Alice looks at $1, 2, 3, \ldots, 1000$ and skips the middle of each triple $(1,2

[2] #4 6.EE.B.6 Suppose a student is looking at an arithmetic sequence with first term $a$ and c

[3] #5 6.EE.A.1 Alice's list has common difference $d = 1$. By the rule, the difference is multi

[4] #7 4.NBT.B.4 The starting number updates by $a_{n+1} = a_n + d_n$ (the first skipped term is

[5] #3 4.NBT.A.2 George works with first term $365$ and common difference $729$, so his numbers w

Review

Reasonableness: A closed form confirms the last start: $a_7 = 1 + (1 + 3 + 9 + 27 + 81 + 243) = 1 + 364 = 365$, equivalently $1 + \frac{3^6 - 1}{2} = 1 + 364 = 365$. It also makes sense that only George gets a lone number: the earlier gaps $1, 3, 9, 27, 81, 243$ all fit inside $1000$ many times, so those students say lots of numbers, but George's gap of $729$ leaves room for just one value below $1000$. Finally, $365$ is exactly answer (C), while the distractors $37$, $242$, $728$, $998$ never appear as a starting number in the sequence.

Alternative: Skip the table with a closed form. The starts satisfy $a_{n+1} = a_n + 3^{\,n-1}$ with $a_1 = 1$, so $a_n = 1 + \frac{3^{\,n-1} - 1}{2}$. For George ($n = 7$): $a_7 = 1 + \frac{729 - 1}{2} = 1 + 364 = 365$. A base-three view gives the same answer: George's number is the smallest one whose position forces six straight "middle" skips.

CCSS standards used (min grade 6)

4.OA.C.5 Generate a number or shape pattern that follows a given rule and identify apparent features of the pattern (Seeing that the numbers each student skips form a new arithmetic sequence handed to the next student.)
6.EE.B.6 Use variables to represent numbers and write expressions when solving a real-world or mathematical problem (Abstracting the hand-off rule as $(a, d) \to (a+d,\, 3d)$ for any first term $a$ and difference $d$.)
6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Expressing the seven common differences as powers of three, $d_n = 3^{\,n-1}$.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm (Computing each starting number $a_{n+1} = a_n + d_n$ up through $122 + 243 = 365$.)
4.NBT.A.2 Compare two multi-digit numbers based on meanings of the digits in each place (Checking that George's next number $1094$ exceeds $1000$, so only $365$ survives.)

⭐ Whatever numbers one student skips become the next student's list — and each hand-off triples the gap (the powers of three) while nudging the start up by one gap. Follow that for seven students and George is left with just $365$.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

What Alice leaves behind

The hand-off rule, with variables

The gap triples each round

Track each starting number

George is left with one number

Review

CCSS standards used (min grade 6)

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