Exactly 5 Factors

Suppose the integer we are looking for can be prime factorised as $p_1^{k_1}p_2^{k_2}...$ where $p_1, p_2...$ are the prime factors while $k_1, k_2...$ are their respective powers (nonnegative integers).

First for the integer to have exactly five factors, we have:

$(k_1+1)(k_2+1)...=5$

Now note that 5 is actually a prime number which means it has 2 and only 2 factors. To be exact, the only way of factorising 5 is $5\times1$ . This means we have:

$k_1+1=5$ and $k_2+1=1$ or we can exchange the role of the two unknown powers.

In other words, $k_1=4$ and $k_2=0$

This means the integer is $p_1^{4}p_2^{0}=p_1^{4}$ since $p_2^{0}=1$ regardless the value of $p_2$ , as long as it is non-zero.

Considering that the integer is even, the only possibility is $p_1=2$ as $p_1$ is prime and the only even prime number is 2.

Therefore, $2^{4}=\boxed{16}$

Relevant wiki: Factors of natural numbers

We know that any natural number $N>1$ is of form $N={p_1}^{k_1}\cdot{p_2}^{k_2}\cdot{p_3}^{k_3}\cdot\cdots{p_n}^{k_n}$ , where $p_i$ and $k_i$ denote the prime factors of $N$ and their powers respectively and $i=1,2,3,\cdots,n$ .

Now, number of divisors is expressed as $D(N)=({k_1}+1)({k_2}+1)({k_3}+1)\cdots({k_n}+1)$ where $D(N)$ denotes number of divisors.

Since each of $k_i$ is also positive natural number, so for $D(N)=5$ we have only one pair, i.e., $1\times5$ .

So, $({k_1}+1)({k_2}+1)=1\times5 ~~ \Longrightarrow {k_1}=0~~ \text{and} ~~{k_2}=4$

Hence, the number $N$ must be of form $N={p_1}^{0}\cdot{p_2}^{4}$ .

Also, the number must be even, so at least one of its prime divisors must be $2$ .

Since, regardless of whatever $p_1$ is, the value of ${p_1}^{0}$ will always be $1$ .

$\therefore {p_2}=2$

And, $N=2^{4}=\boxed{16}$ .

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$\bigstar$ An interesting thing to note here is that whenever $D(N)$ is $odd$ , the number $N$ is always a perfect square. This is due to the fact that for any $N$ which is not a perfect square, each of its divisors has a pairing divisor which, when multiplied to the divisor gives us $N$ , so it generates an $even$ number of divisors. But for any perfect square, one of the divisors pairs with itself and thus gives us an $odd$ number of divisors.