Competition · AMC preparation · step 4 of 4

AMC 10 · 2020B · #15

Grade 6 arithmetic

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

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

📏 Medium solution 💡 2 insights

Problem

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

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

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.

1STEP 1

Pick a long enough block

Original cycle is 5, pass 1 deletes every 3rd, so use a block of lcm(5,3)=15 digits: 123451234512345.

block = 123 451 234 512 345 (15 digits)

2STEP 2

Run the first pass

Pass 1: delete positions 3, 6, 9, 12, 15 from the block, leaving 10 digits: 1245235134.

123451234512345 → 1,2,4,5,2,3,5,1,3,4 = 1245235134

3STEP 3

Run the second pass

Pass 2: double the 10-block to length 20 = lcm(10,4), then delete every 4th, leaving 15 digits: 124235341452513.

12452351341245235134 → 124 235 341 452 513 = 124235341452513

4STEP 4

Run the third pass

Pass 3: length 15 is already divisible by 5, so delete positions 5, 10, 15, leaving 12 digits: 124253415251.

124235341452513 → 1242 5341 5251 = 124253415251

5STEP 5

Read off the final cycle

After all three passes the final list is the 12-digit cycle 124253415251 repeating forever.

final cycle = 124253415251 (length 12)

6STEP 6

Locate the three positions

2019 = 12·168+3, so positions 2019, 2020, 2021 are cycle slots 3, 4, 5 of 124253415251: digits 4, 2, 5.

2019 ≡ 3, 2020 ≡ 4, 2021 ≡ 5 (mod 12) → digits 4, 2, 5

7STEP 7

Add the three digits

Sum the three digits: 4 + 2 + 5 = 11, matching choice (D).

4 + 2 + 5 = 11 → (D)

Answer

Each pass leaves a fraction of the digits:

\frac{2}{3}

\frac{3}{4}

\frac{4}{5}

\frac{2}{5}

overall. So the final list has roughly 10,000 ·

\frac{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).

💡Key takeaway

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 ( $\frac{2}{3}$ · $\frac{3}{4}$ · $\frac{4}{5}$ = $\frac{2}{5}$ surviving) the final list is a 12-digit cycle 124253415251. Positions 2019, 2020, 2021 correspond to cycle slots 3, 4, 5 (since 2019 ≡ 3 (mod 12)), giving digits 4, 2, 5 — sum 11, answer (D).

Pick a long enough block
Run the first pass
Run the second pass
Run the third pass
Read off the final cycle
Locate the three positions
Add the three digits

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