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2020 AMC 10B Problem 15

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2020 AMC 10B Problem 15AMC 10 Step by Step
Name:                             Date: September 13, 2026
Modular Arithmetic · difficulty 3 of 5 · about 6 min · position P11-15

Steve wrote the digits 11 , 22 , 33 , 44 , and 55 in order repeatedly from left to right, forming a list of 10,00010,000 digits, beginning 123451234512….123451234512\ldots. He then erased every third digit from his list (that is, the 33 rd, 66 th, 99 th, …\ldots digits from the left), then erased every fourth digit from the resulting list (that is, the 44 th, 88 th, 1212 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 2019,2020,20212019, 2020, 2021 ?

  1. A)

    77

  2. B)

    99

  3. C)

    1010

  4. D)

    1111

  5. E)

    1212

Answer

D

Key insight

Each erasure keeps the list periodic: 30 original digits become 20, then 15, then 12, so the final list repeats every 12 digits.

Solution

Track the period. The original list repeats every 55 digits. Erasing every third digit is periodic with period lcm⁡(5,3)=15\operatorname{lcm}(5,3) = 15, leaving 1010 digits per 1515; equivalently, per 3030 original digits, 2020 remain. Erasing every fourth of those 2020 leaves 1515 per block, and erasing every fifth of those 1515 leaves 1212. Each block length is divisible by the erasure step, so the erasures line up with block boundaries and the final list has period 1212.

Work out one block of 3030 original digits, 123451234512345123451234512345123451234512345123451234512345:

  • Remove every 3rd: 1 2 4 5 2 3 5 1 3 4 1 2 4 5 2 3 5 1 3 41\,2\,4\,5\,2\,3\,5\,1\,3\,4\,1\,2\,4\,5\,2\,3\,5\,1\,3\,4 (20 digits).
  • Remove every 4th: 1 2 4 2 3 5 3 4 1 4 5 2 5 1 31\,2\,4\,2\,3\,5\,3\,4\,1\,4\,5\,2\,5\,1\,3 (15 digits).
  • Remove every 5th: 1 2 4 2 5 3 4 1 5 2 5 11\,2\,4\,2\,5\,3\,4\,1\,5\,2\,5\,1 (12 digits).

Now 2019=12⋅168+32019 = 12 \cdot 168 + 3, so positions 2019,2020,20212019, 2020, 2021 correspond to positions 3,4,53, 4, 5 of the block: digits 4,2,54, 2, 5. Their sum is 1111.

(The list is long enough: 10000→6667→5001→400110000 \to 6667 \to 5001 \to 4001 digits remain, more than 20212021.)

The answer is (D) 11\boxed{\textbf{(D)}\ 11}.

Why this works

Deleting every kkth term of a periodic list produces another periodic list, with period lcm⁡(period,k)\operatorname{lcm}(\text{period}, k) shrunk by the factor k−1k\frac{k-1}{k}. Chaining three erasures just chains the lcm's. Once the final period is known, a single index reduction mod 1212 replaces any attempt to simulate thousands of digits.

The trap

Applying the three erasures to the original position numbers simultaneously instead of to the shrinking list, or using the wrong period.

Common mistakes

  • Applying the three erasures to the original position numbers simultaneously instead of to the shrinking list, or using the wrong period.
  • Reducing 2019 mod 122019 \bmod 12 as 00 or off by one; write 2019=2016+32019 = 2016 + 3 explicitly.

Techniques

Organized listing / direct enumeration · Compute small cases, spot the pattern, generalize