Solution to Exercise 3 First, we notice that $2 \times 2019$ is Japanese, since the integers 1, 2, and 2019 are among its divisors.

Let $n$ be the minimal Japanese integer: it is the LCM of the integers $d_{1}, d_{2}$, and $d_{3}$. We then set $a_{k}=n / d_{k}$, so that $1 \leqslant a_{3}2$. We deduce that $a_{1}=672 / 3=224$, and then that $d_{1}=2 \times 3=6$, so $n=6 \times 224=1344$. In particular, 1344 is a multiple of 1 and 2, so it is a Japanese integer, with $d_{1}=6, d_{2}=672$, and $d_{3}=1344$.

We are therefore looking for a potential Japanese integer $n\frac{2022}{1344}>1+\frac{1}{2}=1+\frac{1}{3}+\frac{1}{6}
$$

so $a_{1} \leqslant 5$.

We are left to study the triplets $\left(a_{1}, a_{2}, a_{3}\right)=(4,3,1),(5,3,1),(5,4,1),(4,3,2),(5,3,2)$, $(5,4,2)$, and $(5,4,3)$, which lead us respectively to the fractions $p / q=19 / 12,23 / 15,29 / 20$, $13 / 12,31 / 30,19 / 20$, and $47 / 60$. The numerator $p$ never divides $2 \times 3 \times 337$, so none of these cases is possible.

In conclusion, the smallest Japanese integer is $n=1344$.
Exercise 3. An integer $n \geqslant 1$ is said to be Nippon if it admits three divisors $d_{1}, d_{2}$ and $d_{3}$ such that $1 \leqslant d_{1}<d_{2}<d_{3}$ and $d_{1}+d_{2}+d_{3}=2022$.

What is the smallest Nippon integer?