2023 AMC 10A Problem 14

Attempt Problem 14 of the 2023 AMC 10A 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 2023 AMC 10A solutions, or check the answer key.

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14.

A number is chosen at random from among the first 100100 positive integers, and a positive integer divisor of that number is then chosen at random. What is the probability that the chosen divisor is divisible by 11?11?

4100\dfrac{4}{100}

9200\dfrac{9}{200}

120\dfrac{1}{20}

11200\dfrac{11}{200}

350\dfrac{3}{50}

Answer: B
Concepts:factor countingbasic probability
Difficulty rating: 1630
Solution:

A number n100n \le 100 can only have a divisor divisible by 1111 when 11n,11 \mid n, so n{11,22,,99}.n \in \{11, 22, \ldots, 99\}. Write n=11mn = 11m with m9.m \le 9. Here 11m,11 \nmid m, so d(11m)=2d(m),d(11m) = 2\,d(m), and the divisors that are multiples of 1111 are exactly the d(m)d(m) numbers 11d.11d. That makes the chance d(m)2d(m)=12\frac{d(m)}{2\,d(m)} = \frac12 for each such n.n. Averaging over all 100100 starting numbers, the probability is 1100m=1912=9200.\frac{1}{100}\sum_{m=1}^{9}\frac12 = \frac{9}{200}. Therefore, the answer is B.

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