Factorisation Triples

Let $n$ be a positive integer. An integer triple $(a, b, c)$ is called a factorisation triple of $n$ if:$1 \leq a \leq b \leq c$$a \cdot b \cdot c = n$.

Define $f(n)$ to be $a + b + c$ for the factorisation triple $(a, b, c)$ of $n$ which minimises $c / a$. One can show that this triple is unique.

For example, $f(165) = 19$, $f(100100) = 142$ and $f(20!) = 4034872$.

Find $f(43!)$.

The problem requires somewhat advanced combinatorial arguments and the use of factorials.

To solve it, we need to first understand what is 43! (which is 43 factorial).

43 factorial is simply the product of all positive integers from 1 to 43 (i.e., 1*2*3*…*43).

Since we are asked to find a triple $(a,b,c)$ with $a*b*c = 43!$ and we want to minimize the value of $c/a$, it is evident that the numbers $a$, $b$ and $c$ should be as close to each other as possible. This is because the ratio $c/a$ will be smallest when $a$ is as large as possible and $c$ is as small as possible.

To approach this problem, we can start by breaking down 43! into its prime factors. The prime factors of 43 are all the prime numbers up to 43, each multiplied together the number of times they can divide into 43. This gives us a set of prime numbers raised to certain powers.

Let’s denote these prime factors as:

$2^{40} * 3^{20} * 5^{9} * 7^{6} * 11^{3} * 13^{3} * 17^{2} * 19^{2} * 23^{1} * 29^{1} * 31^{1} * 37^{1} * 41^{1} * 43^{1}$

Now, we want to split these into three groups $(a,b,c)$ as evenly as possible. To do that, we can start from the highest powers (here $2^{40}$ and $3^{20}$ etc.) and allocate each prime factor to the groups based on current totals.

We keep filling up the triplets maintaining the order $a \leq b \leq c$ while keeping the values as close together as possible. This can be done by assigning the highest prime powers to the group with the smallest product.

After performing all these steps, we will finally have the values for a, b, and c for which we can simply calculate $f(43!)$ by summing these values.

It should be noted that solving manually this problem requires a lot of effort and careful calculations. The cumbersome part is to distribute the prime factors in order to minimize the ratio, and then multiply those primes to get each part of the triple. These calculations can be demanding and are typically performed using computer algorithms, which can handle such large number operations efficiently.

Even though I haven’t provided the final value for this function, I hope this explanation gives you a good sense of how to approach such problems.

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