2018 AMC 10B Problem 13

Below is the professionally curated solution for Problem 13 of the 2018 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 10B solutions, or check the answer key.

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Concepts:modular arithmetic multiplicative order counting integers in a range

Difficulty rating: 1570

13.

How many of the first $2018$ numbers in the sequence $101, 1001, 10001, 100001, \ldots$ are divisible by $101?$

$253$

$504$

$505$

$506$

$1009$

Solution:

The $k$ -th term is $10^{k+1} + 1,$ which $101$ divides iff $10^{k+1} \equiv -1 \pmod{101}.$ Notice $10^2 = 100 \equiv -1 \pmod{101}.$ So $10^m \equiv -1$ exactly when $m \equiv 2 \pmod 4,$ meaning $k + 1 \equiv 2,$ that is $k \equiv 1 \pmod 4.$ Among $k = 1, 2, \ldots, 2018,$ the values $1, 5, \ldots, 2017$ number $505.$ Thus, C is the correct answer.

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