2018 AMC 10B Problem 13

Attempt Problem 13 of the 2018 AMC 10B below, then check your answer against the professionally curated solution 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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13.

How many of the first 20182018 numbers in the sequence 101,1001,10001,100001,101, 1001, 10001, 100001, \ldots are divisible by 101?101?

253253

504504

505505

506506

10091009

Answer: C
Concepts:modular arithmeticmultiplicative ordercounting integers in a range
Difficulty rating: 1570
Solution:

The kk-th term is 10k+1+1,10^{k+1} + 1, which 101101 divides iff 10k+11(mod101).10^{k+1} \equiv -1 \pmod{101}. Notice 102=1001(mod101).10^2 = 100 \equiv -1 \pmod{101}. So 10m110^m \equiv -1 exactly when m2(mod4),m \equiv 2 \pmod 4, meaning k+12,k + 1 \equiv 2, that is k1(mod4).k \equiv 1 \pmod 4. Among k=1,2,,2018,k = 1, 2, \ldots, 2018, the values 1,5,,20171, 5, \ldots, 2017 number 505.505. Thus, C is the correct answer.

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