B2 For each positive integer $n$, define $\varphi(n)$ to be the number of positive divisors of $n$. For example, $\varphi(10)=4$, since 10 has 4 positive divisors, namely $\{1,2,5,10\}$.
Suppose $n$ is a positive integer such that $\varphi(2 n)=6$. Determine the minimum possible value of $\varphi(6 n)$.
Solution: The answer is 8 .

Solution 1: Recall that if a positive integer $m$ has prime factorization $p_{1}^{e_{1}} p_{2}^{e_{2}} \ldots p_{t}^{e_{t}}$, where $p_{1}, \ldots, p_{t}$ are distinct primes, then the number of positive divisors of $m$ is $\varphi(m)=\left(e_{1}+\right.$ 1) $\left(e_{2}+1\right) \ldots\left(e_{t}+1\right)\left({ }^{*}\right)$. Note that each term in this product is at least 2 .

Since $2 n$ is an even positive integer with 6 positive divisors, $2 n=2^{5}, 2^{2} \cdot p$ or $2 \cdot p^{2}$, where $p$ is some odd prime number. Therefore, $n=2^{4}=16,2 p$ or $p^{2}$. Therefore, $6 n=6 \times 16=96,12 p$ or $6 p^{2}$.

Note that $\varphi(96)=\varphi\left(2^{5} \times 3^{1}\right)=6 \times 2=12$.
If $p=3$, then $\varphi(12 p)=\varphi(36)=\varphi\left(2^{2} \times 3^{2}\right)=3 \times 3=9$ and $\varphi\left(6 p^{2}\right)=\varphi(54)=\varphi\left(2^{1} \times 3^{3}\right)=$ $2 \times 4=8$.

It remains to show the case when $p>3$. So far the minimum value obtained for $\varphi(6 n)=8$. If $p>3$, then $6 n$ contains at least 3 different prime divisors. Then by $\left({ }^{*}\right)$, the number of positive divisors of $6 n$ is at least $2 \times 2 \times 2=8$. Therefore, $\varphi(6 n) \geq 8$ for all positive integers $n$. As we have shown, $n=9$ yields $\varphi(6 n)=8$. Therefore, the answer is 8 .

Solution 2: Note that four positive divisors of $2 n$ are $1,2, n$ and $2 n$. Note also that $n=2$ does not satisfy $\varphi(2 n)=6$. Therefore, $n \geq 2$ and consequently, $1,2, n$ and $2 n$ are all distinct.

Since $2 n$ has 6 positive divisors, there are two other positive divisors $a, b$ of $2, n$, with $a, b>2$. Then the set of positive divisors of $2 n$ is $\{1,2, a, b, n, 2 n\}$.

Now consider the positive divisors of $6 n$. Note that the set of positive divisors of $6 n$ contains those of $2 n$. Further note that $3 n$ and $6 n$ are positive divisors of $6 n$, which are not positive divisors of $2 n$. Hence, the set of positive divisors of $6 n$ contains $\{1,2, a, b, n, 2 n, 3 n, 6 n\}$. Therefore, $\varphi(6 n) \geq 8$.

We will show that this minimum can be obtained. Since $\{1,2, a, b, n, 2 n, 3 n, 6 n\}$ are the positive divisors of $6 n$ and appear in increasing order, $a \cdot 2 n=6 n$ and $b n=6 n$. Multiplying both equations and dividing both sides by $2 n^{2}$ yield $a b=18$. But since $\{1,2, a, b, n, 2 n\}$ are the positive divisors of $2 n, a b=2 n$. Therefore, $2 n=18$, from which we can conclude that $n=9$ is a candidate which yields $\varphi(2 n)=6$ and $\varphi(6 n)=8$.

This can be easily verified, since the positive divisors of 18 are $\{1,2,3,6,9,18\}$. Since the positive divisors of 54 are $\{1,2,3,6,9,18,27,54\}, \varphi(54)=8$.