Library / /1162 of 1394

, 2020

Number theory Difficulty 5.7 AIME, harder Prove it United States

Problem:

Ana and Banana are rolling a standard six-sided die. Ana rolls the die twice, obtaining $a_{1}$ and $a_{2}$ , then Banana rolls the die twice, obtaining $b_{1}$ and $b_{2}$ . After Ana's two rolls but before Banana's two rolls, they compute the probability $p$ that $a_{1} b_{1} + a_{2} b_{2}$ will be a multiple of $6$ . What is the probability that $p = \frac{1}{6}$ ?

Proposed by: James Lin

Solution

Solution:

If either $a_{1}$ or $a_{2}$ is relatively prime to $6$ , then $p = \frac{1}{6}$ . If one of them is a multiple of $2$ but not $6$ , while the other is a multiple of $3$ but not $6$ , we also have $p = \frac{1}{6}$ . In other words, $p = \frac{1}{6}$ if $\operatorname{gcd}(a_{1}, a_{2})$ is coprime to $6$ , and otherwise $p \neq \frac{1}{6}$ . The probability that $p = \frac{1}{6}$ is $\frac{(3^{2}-1)(2^{2}-1)}{6^{2}} = \frac{2}{3}$ where $\frac{q^{2}-1}{q^{2}}$ corresponds to the probability that at least one of $a_{1}$ and $a_{2}$ is not divisible by $q$ for $q=2,3$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.