5. For a positive integer $n$, denote by $\varphi(n)$ the number of positive integers $k \leq n$ relatively prime to $n$. How many positive integers $n$ less than or equal to 100 are divisible by $\varphi(n)$ ?
Answer: 16
Solution: We claim that any such integer $n$ must be either equal to 1 or of the form $2^{a} 3^{b}$, where $a \geq 1$ and $b \geq 0$.

First, we note that if $n>1$, it must be even. This is because if $n$ admits a prime factorization $n=\prod_{i=1}^{k} p_{i}^{r_{i}}$ over distinct primes $p_{i}$, then $\varphi(n)=\prod_{i=1}^{k}\left(p_{i}-1\right) p_{i}^{r_{i}-1} 1$ in particular, if an odd prime $p_{i}$ divides $n$ then $p_{i}-1$ divides $\varphi(n)$, and consequently $\varphi(n)$ is even.
Then, we note that at most one odd prime divides $n$. Suppose otherwise, i.e., $2^{a} p_{1}^{r_{1}} p_{2}^{r_{2}}$ is a part of the prime factorization of $n$ for odd primes $p_{1}, p_{2}$. Then by the multiplicativity of $\varphi$, we know that $\varphi\left(2^{a} p_{1}^{r_{1}} p_{2}^{r_{2}}\right)=2^{a-1}\left(p_{1}-1\right) p_{1}^{r_{1}-1}\left(p_{2}-1\right) p_{2}^{r_{2}-1}$ must divide $\varphi(n)$. Note that the largest power of 2 dividing $n$ is $2^{a}$, but since $p_{1}-1$ and $p_{2}-1$ are both even, $\left.2^{(} a-1\right) 2^{2}=2^{a+1}$ divides $\varphi(n)$. This is a contradiction.

Finally, we show that the said odd prime $p$ dividing $n$ must in fact be equal to 3 . Indeed, if $n=2^{a} p^{r}$ where $a$ and $r$ are positive integers, we have $\frac{n}{\varphi(n)}=\frac{2 p}{p-1}=2+\frac{2}{p-1}$. For this to be an integer, $p=3$. Hence, we count all such integers of the form $n=2^{a} 3^{r}$ by summing over all possible values of $r$ (because there are fewer):
If $r=0$, then $0 \leq a \leq 6$. (This is the only case in which $a=0$ is allowed.)
If $r=1$, then $1 \leq a \leq 5$.
If $r=2$, then $1 \leq a \leq 3$.
If $r=3$, then $a=1$.
This gives us a total of $7+5+3+1=16$ values of $n$.