Japan Junior Mathematical Olympiad First Round

Let us first prove the following Theorem:

Theorem
Suppose that a positive integer $n$ has the prime factorization $n = p_1^{e_1} p_2^{e_2} \cdots p_m^{e_m}$. (Here, $p_1, p_2, \cdots, p_m$ are distinct prime numbers and $e_1, e_2, \cdots, e_m$ are positive integers.) Then the total number of factors of $n$ equals $(e_1 + 1)(e_2 + 1) \cdots (e_m + 1)$.

Proof: Any factor of $n$ can be represented in the form $p_1^{f_1} p_2^{f_2} \cdots p_m^{f_m}$ with $f_j$ being an integer satisfying $0 \le f_j \le e_j$ for each $j = 1, 2, \cdots, m$. It is clear that the different $m$-tuples of integers $(f_1, f_2, \cdots, f_m)$ correspond to different factors of $n$. Therefore, the total number of factors of $n$ coincides with the number of $m$-tuples $(f_1, f_2, \cdots, f_m)$ satisfying the conditions above. Since there are exactly $e_j + 1$ choices for $f_j$ for each $j = 1, 2, \cdots, m$ there are $(e_1 + 1)(e_2 + 1) \cdots (e_m + 1)$ such $m$-tuples, which proves the claim of the Theorem.

Let us now show that the desired answer for the problem is 30. Suppose $n$ is an integer satisfying the condition of the problem. If $n$ has a prime factor besides 2, 3, 5, 7, then since $14 = 2 \cdot 7$, $16 = 2^4$, $18 = 2 \cdot 3^2$, $20 = 2^2 \cdot 5$, the integer obtained by dividing $n$ by that prime factor is smaller than $n$ and satisfies the condition of the problem as well. So, we may assume that $n$ has only $2, 3, 5, 7$ as its prime factors. Therefore, we can write $n = 2^a \cdot 3^b \cdot 5^c \cdot 7^d$, where $a, b, c, d$ are non-negative integers. If we now assume that the number of factors of $14n, 16n, 18n, 20n$ are $p, q, r, s$, respectively, then by the Theorem above, we have
$$p = (a + 2)(b + 1)(c + 1)(d + 2), \quad q = (a + 5)(b + 1)(c + 1)(d + 1), \\ r = (a + 2)(b + 3)(c + 1)(d + 1), \quad s = (a + 3)(b + 1)(c + 2)(d + 1).$$
It then follows that we have
$$\begin{align*} q &= p \iff (a+5)(d+1) = (a+2)(d+2) &\iff a-1 &= 3d, \\ q &= r \iff (a+5)(b+1) = (a+2)(b+3) &\iff 2a+1 &= 3b, \\ q &= s \iff (a+5)(c+1) = (a+3)(c+2) &\iff a+1 &= 2c. \end{align*}$$
If we make the value of $a$ bigger, $b, c, d$ also get bigger, so if $n$ is the smallest possible integer satisfying the requirement, then $a$ also has to be the smallest non-negative integer satisfying the conditions above. By substituting $a = 0, 1, \dots$, and checking to see when all of $b, c, d$ become non-negative integers, we see that $a = 1$ is the smallest value for $a$ for which all of $b, c, d$ are non-negative integers and their values are $b = 1, c = 1, d = 0$. Thus the answer to the problem is $n = 2^1 \cdot 3^1 \cdot 5^1 \cdot 7^0 = 30$.