2023 AMC 10A Problem 14

A divisor divisible by $11$ can only come from a number that is itself divisible by $11$. Among $1, \ldots, 100$ these are $11, 22, \ldots, 99$: nine numbers, each of the form $11k$ with $1 \le k \le 9$. Since $k < 11$, the factor $11$ appears exactly once in each.

For such a number $n = 11k$, pair each divisor $d$ of $k$ with $11d$. The divisors of $n$ not divisible by $11$ are exactly the divisors of $k$, and the divisors divisible by $11$ are exactly the numbers $11d$. The pairing is a perfect match, so exactly half of the divisors of $n$ are multiples of $11$.

Therefore the probability is
$$\frac{9}{100} \cdot \frac12 = \frac{9}{200}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{9}{200}}$.

Two-stage random choices multiply: probability of a good first choice times the probability of success given it. The second factor looks like it depends on $n$, but the pairing $d \leftrightarrow 11d$ shows it is $\frac12$ for every $n$ with exactly one factor of $11$. Whenever a prime $p$ divides $n$ exactly once, half of $n$'s divisors contain $p$.

• Answering 9/100 by forgetting the second random step, or computing the proportion of divisors as 1/11 rather than 1/2.
• Listing the divisors of all nine multiples individually and making an arithmetic slip; the pairing argument makes this unnecessary.