Question

A number is chosen at random from among the first $100$ 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 $11?$

$\dfrac{4}{100}$

$\dfrac{9}{200}$

$\dfrac{1}{20}$

$\dfrac{11}{200}$

$\dfrac{3}{50}$

Solution:

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

2023 AMC 10A Problem 14

Solution:

Problem 14 in Other Years