# Find the smallest number with four factorizations in the $\mathbb{E}$-zone.

First, let me define the $\mathbb{E}$-zone as the set of all even integers.

I am told to find the smallest number with 4 different prime factorizations in this world.

I should also define what it means to be a prime in the $\mathbb{E}$-zone. A number is prime if it cannot be factored into a product of even numbers. For all intents, odd numbers do not exist.

For example, some of the first few primes are 2,6,10,14,18,22... We can see that a number is prime if it is congruent to 2 mod 4.

Now we also know that prime factorization in this world is not unique. For example, 180 has 3 prime factorizations. Namely, (10,18), (6, 30), and (2,90).

I managed to find a number 2940 that has 4 factorizations: (6,490), (10,294), (2,1470), and (30,98).

However, I do not know how to go about finding the $smallest$ such number.

An even integer $2n$ is prime in the $\Bbb{E}$-zone if and only if $2n \equiv 2 \pmod{4}$, which occurs if and only if $n$ is odd. So, factorization into primes amounts to writing \begin{align} N &= 2n_1 \cdot 2n_2 \cdots 2n_k \\ &= 2^k \cdot n_1 \cdot n_2 \cdots n_k, \end{align} where $n_1, \ldots, n_k$ are odd integers. Since we're looking for small numbers with many prime factorizations, and since the product of odd numbers is odd, we may as well assume that $k = 2$. (You seem to have done this implicitly by writing prime factorizations as pairs.)

If we want $4$ prime factorizations, then we're looking for $4$ pairs of prime factors, or an odd number with $8$ factors. (The factors are automatically odd, too.)

In the integers, the number of factors of the number $p_1^{e_1} \cdots p_r^{e_r}$, where $p_1, \ldots, p_r$ are usual primes, is $$(1 + e_1) \cdots (1 + e_r).$$ So, we need to write $8$ (the number of factors we desire) as $2 \cdot 2 \cdot 2$ or as $4 \cdot 2$. We will consider each possibility in turn.

In the first case, we must consider a product of distinct (odd!) primes $p_1 \cdot p_2 \cdot p_3$. The smallest example of this is $3 \cdot 5 \cdot 7 = 105$, whose factors can be paired off as: $$\begin{array}{ccc} 1 && 3 \cdot 5 \cdot 7 \\ 3 && 5 \cdot 7 \\ 5 && 3 \cdot 7 \\ 7 && 3 \cdot 5 \end{array}$$ Back in the $\Bbb{E}$-zone, these are prime factorizations of $$2^2 \cdot 3 \cdot 5 \cdot 7 = 420$$ (by putting a $2$ on each factor of the pair): $$\begin{array}{ccc} 2 && 210 \\ 6 && 70 \\ 10 && 42 \\ 14 && 30 \end{array}$$

In this case, consider a number of the form $p_1^3 \cdot p_2$. The smallest example of this is $3^3 \cdot 5 = 135$, which is larger than $105$, but I'll follow through the analysis anyway. The factors in pairs are: $$\begin{array}{ccc} 1 && 3^3 \cdot 5\\ 3 && 3^2 \cdot 5 \\ 3^2 && 3 \cdot 5 \\ 3^3 && 5 \end{array}$$ Back in the $\Bbb{E}$-zone, these are prime factorizations of $$2^2 \cdot 3^3 \cdot 5= 540$$ (by putting a $2$ on each factor of the pair): $$\begin{array}{ccc} 2 && 270 \\ 6 && 90 \\ 18 && 30 \\ 54 && 10 \end{array}$$

So, $420$ is the smallest positive integer with $4$ prime factorizations in the $\Bbb{E}$-zone.

• That was extremely helpful. Thank you very much. So if I want the smallest number with 3 distinct factors, for example, I would look for the smallest odd number I can create with 3 distinct odd factors or with 3 total factors, even if some repeat. (trying to make sure I understand this concept). This number would be 180 as $3^2*5 = 45$ and $45*2^2 =180$. Excellent. Thank you. – Tyler Murphy Feb 5 '14 at 8:49