2023 AMC 10A Problem 14

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}

Answer: B

Concepts:factor counting basic probability

Difficulty rating: 1630

Small hint:

Only multiples of

11

up to

100

can have a divisor divisible by

11

Big hint:

For

n = 11m

with

m \le 9,

exactly half of

n

’s divisors are multiples of

11,

since

m

is not a multiple of

11

Solution:

A number

n \le 100

can only have a divisor divisible by

11

when

n

is divisible by

11,

so

n \in \{11, 22, \ldots, 99\}.

Write

n = 11m

with

m \le 9.

Here

m

is not a multiple of

11,

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.