AMC 10 · 2020 · #15

Grade 6 arithmetic

modular-arithmeticpattern-recognitionlcm easier-related-problempattern-recognitionsystematic-enumeration ↑ Prerequisites: modular-arithmeticlcm

📏 Medium solution 💡 2 insights

Problem

Steve wrote the digits $1$ , $2$ , $3$ , $4$ , and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$ rd, $6$ th, $9$ th, $\ldots$ digits from the left), then erased every fourth digit from the resulting list (that is, the $4$ th, $8$ th, $12$ th, $\ldots$ digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions $2019, 2020, 2021$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Steve writes $123451234512345\ldots$ — the digits $1,2,3,4,5$ cycling, for $10{,}000$ digits. He then performs three erasure passes in order: erase every $3$rd digit, then every $4$th digit of what's left, then every $5$th digit of what's still left. What is the sum of the digits now sitting in positions $2019, 2020, 2021$?

Givens: Original list: $1,2,3,4,5,1,2,3,4,5,\ldots$ for $10{,}000$ digits (cycle length $5$); Pass 1: delete positions $3, 6, 9, \ldots$ from the original; Pass 2: delete positions $4, 8, 12, \ldots$ from the list that survives pass 1; Pass 3: delete positions $5, 10, 15, \ldots$ from the list that survives pass 2; Answer choices: $7, 9, 10, 11, 12$

Unknowns: Sum of the three digits at positions $2019, 2020, 2021$ in the final list

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #5 Look for a Pattern, #2 Make a Systematic List, #3 Eliminate Possibilities

Tool #9 (Easier Problem): replace the daunting list of $10{,}000$ digits with one repeating block whose length is divisible by every relevant period. Tool #5 (Pattern): each erasure pass preserves a periodic structure, so the final list is also periodic — find its period. Tool #2 (Systematic List): write the block out explicitly, perform the three passes on it, and read off positions $2019, 2020, 2021$ via modular arithmetic. Tool #3 (Eliminate): match the digit sum to the five answer choices.

Execute — Answer: D

#9 Solve an Easier Related Problem 6.NS.B.4 Step 1

Pick a working block long enough for all three passes to behave cleanly.
Original cycle is $5$ digits; pass 1 deletes every $3$rd, so a block of $\mathrm{lcm}(5, 3) = 15$ original digits gives a clean repeat after pass 1.
Use that block: $123451234512345$.

$$\text{block} = 123\,451\,234\,512\,345 \quad (15 \text{ digits})$$

💡 Grade 6 LCM: pick a block size that the original cycle and the deletion period both divide.

#2 Make a Systematic List 5.OA.B.3 Step 2

Pass 1: delete positions $3, 6, 9, 12, 15$ (the bold/underlined digits below).
Block becomes $10$ digits long.

$$12\underline{3}45\underline{1}23\underline{4}51\underline{2}34\underline{5} \;\to\; 1{,}2{,}4{,}5{,}2{,}3{,}5{,}1{,}3{,}4 = 1245235134$$

💡 Grade 5 systematic: just keep what isn't a multiple of $3$ in the block.

#2 Make a Systematic List 5.OA.B.3 Step 3

Pass 2 deletes every $4$th from the surviving list.
The current block has length $10$; double it to length $20 = \mathrm{lcm}(10, 4)$ so the cycle structure stays clean.
Doubled: $12452351341245235134$.
Delete positions $4, 8, 12, 16, 20$.

$$124\underline{5}235\underline{1}341\underline{2}452\underline{3}513\underline{4} \;\to\; 124\,235\,341\,452\,513 = 124235341452513$$

💡 Grade 5 systematic: double to fit the new period, then drop multiples of $4$.

#2 Make a Systematic List 5.OA.B.3 Step 4

Pass 3 deletes every $5$th from the new list.
The current block has length $15$, already divisible by $5$.
Delete positions $5, 10, 15$.

$$1242\underline{3}5341\underline{4}5251\underline{3} \;\to\; 1242\,5341\,5251 = 124253415251$$

💡 Grade 5 systematic: drop the $5$th, $10$th, $15$th — three deletions in a $15$-block.

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

After all three passes the final list is the $12$-digit cycle $\mathbf{124253415251}$ repeating forever.
Sanity check: each original $15$-digit block contributed $15 \cdot \tfrac{2}{3} \cdot \tfrac{3}{4} \cdot \tfrac{4}{5} = 8$ surviving digits — so two original blocks ($30$ digits) become $16$ survivors per outer round… but our doubled block tracking accounts for that and gives a $12$-digit final period.

$$\text{final cycle} = 124253415251 \quad (\text{length } 12)$$

💡 Grade 4 pattern: the final list is just one $12$-digit pattern repeated.

#5 Look for a Pattern 6.NS.B.2 Step 6

Find positions $2019, 2020, 2021$ in the $12$-cycle.
$2019 = 12 \cdot 168 + 3$, so position $2019$ is character $3$ of the cycle; similarly positions $2020, 2021$ are characters $4, 5$.
Reading $124253415251$: character $3$ is $4$, character $4$ is $2$, character $5$ is $5$.

$$2019 \equiv 3,\; 2020 \equiv 4,\; 2021 \equiv 5 \pmod{12} \;\Rightarrow\; \text{digits } 4, 2, 5$$

💡 Grade 6 division with remainder: $2019 \div 12$ has remainder $3$.

#3 Eliminate Possibilities 2.NBT.B.5 Step 7

Sum the three digits and match to the choices.

$$4 + 2 + 5 = 11 \;\Rightarrow\; \textbf{(D)}$$

💡 Grade 2 addition within $100$: $4 + 2 + 5 = 11$.

[1] #9 6.NS.B.4 Pick a working block long enough for all three passes to behave cleanly. Origina

[2] #2 5.OA.B.3 Pass 1: delete positions $3, 6, 9, 12, 15$ (the bold/underlined digits below). B

[3] #2 5.OA.B.3 Pass 2 deletes every $4$th from the surviving list. The current block has length

[4] #2 5.OA.B.3 Pass 3 deletes every $5$th from the new list. The current block has length $15$,

[5] #5 4.OA.C.5 After all three passes the final list is the $12$-digit cycle $\mathbf{124253415

[6] #5 6.NS.B.2 Find positions $2019, 2020, 2021$ in the $12$-cycle. $2019 = 12 \cdot 168 + 3$,

[7] #3 2.NBT.B.5 Sum the three digits and match to the choices.

Review

Reasonableness: Each pass leaves a fraction of the digits: $\tfrac{2}{3} \cdot \tfrac{3}{4} \cdot \tfrac{4}{5} = \tfrac{2}{5}$ overall. So the final list has roughly $10{,}000 \cdot \tfrac{2}{5} = 4{,}000$ digits — well above position $2021$, confirming the question is even answerable. The three digits $4, 2, 5$ are all in $\{1, 2, 3, 4, 5\}$ (the only digits Steve ever wrote), so their sum $11$ lies in $[3, 15]$, comfortably matching choice (D).

Alternative: Tool #6 (Guess and Check) by simulation: write a short program that builds the $10{,}000$-digit string, then performs the three deletions and reads positions $2019, 2020, 2021$. Running the simulation reproduces the digits $4, 2, 5$ and the sum $11$, confirming (D) without trusting the LCM bookkeeping.

CCSS standards used (min grade 6)

2.NBT.B.5 Fluently add and subtract within 100 (Computing $4 + 2 + 5 = 11$.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Observing that the final list is a $12$-digit repeating cycle.)
5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships (Carrying out the three deletion passes on the $15$-digit block systematically.)
6.NS.B.2 Fluently divide multi-digit numbers using the standard algorithm (Computing $2019 \div 12$ with remainder $3$ to locate position $2019$.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Choosing block sizes $\mathrm{lcm}(5, 3) = 15$ and $\mathrm{lcm}(10, 4) = 20$ so that each pass acts cleanly.)

⭐ This AMC 10 problem only needs Grade 6 LCM and remainders you already know! Pick a $15$-digit block $123451234512345$ — the LCM of cycle and period works for pass 1. After the three deletions ($\tfrac{2}{3} \cdot \tfrac{3}{4} \cdot \tfrac{4}{5} = \tfrac{2}{5}$ surviving) the final list is a $12$-digit cycle $\mathbf{124253415251}$. Positions $2019, 2020, 2021$ correspond to cycle slots $3, 4, 5$ (since $2019 \equiv 3 \pmod{12}$), giving digits $4, 2, 5$ — sum $\mathbf{11}$, answer (D).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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