Factorial's factors

Firstly, note that $20!=20\times 19\times 18\times 17\times 16\times 15\times 14\times 13\times 12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1$ .

We can then break down each of these numbers into their prime factors, yielding $(2^{2}\times 5)\times (19)\times (3^{2}\times 2)\times (17)\times (2^{4})\times (3\times 5)\times (2\times 7)\times (13)\times (3\times 2^{2})\times (11)\times (2\times 5)\times (3^{2})\times (2^{3})\times (7)\times (2\times 3)\times (5)\times (2^{2})\times (3)\times (2)\times (1)$ .

Further simplification yields $2^{18}\times 3^{8}\times 5^{4}\times 7^{2}\times 11\times 13\times 17\times 19$ .

This is equivalent to $(2^{6}\times 3^{2}\times 5)^{3}\times 3^{2}\times 5\times 7^{2}\times 11\times 13\times 17\times 19$ .

From this, it is clear that the greatest value of $k$ where $k^{3}$ is a factor of $20!$ is $2^{6}\times 3^{2}\times 5$ , that is, $2880$ . So the solution is $\boxed{2880}$ .

I applied a shorter method to solve the problem.

Firstly, it's important to understand that for any $x$ , the expression

$\sum _{ n=1 }^{ k }{ \left\lfloor \frac { 20 }{ { x }^{ n } } \right\rfloor }$

gives the highest power of $n$ such that $x^n|20$ .

Note: $k$ is the last $n$ such that $\left\lfloor \frac { 20 }{ { x }^{ n } } \right\rfloor$ is non-zero.

---

Now, we need to remember that $n$ will always be greater than or equal to 3. Since $\left\lfloor \frac { 20 }{ 6 } \right\rfloor = 6$ , it is clear that 6 is the last $x$ such that $n$ is less than or equal to 3.

But so now we just need to find positive integers under 6 which divide 20! and then find out the largest power.

However, since $4 = 2^2$ and $6=2\cdot3$ , we just need to check for primes, ie 2, 3 and 5.

Using the above formula, we find out that largest power of 2 is $2^{18}$ , 3 is $3^8$ and 5 is $5^3$ . The highest power of 3 is found from the product -

$\left( 2^6\right)^3\left(3^2\right)^3\cdot5^3$

Hence the highest is $2^{ 6 }\cdot 3^{ 2 }\cdot 5=2880$

First we consider how many factors of 2 are there in 20!

$2^{1}$ -> 20/2 = 10

$2^{2}$ -> 20/4 = 5

$2^{3}$ -> 20/8 = 2 R 4

$2^{4}$ -> 20/16 = 1 R 4

There are 10+5+2+1 = 18 factors of 2 within 20!, meaning we will have $2^{6*3}$ .

Then consider how many factors of 3 we have with the same process.

20/3 = 6 R 2

20/9 = 2 R 2

6+2 = 8, meaning at most we can have $3^{2*3}$ .

Next, consider the factors of 5 within 20!.

20/5 = 4

This means at most we can have $5^{3}$

After that, let's consider the amount of 7's.

20/7 = 2 R 6.

Since we do not even have a $7^{3}$ , we know that we will not have any further components of the largest $k^{3}$ .

Therefore, we can now obtain the largest $k^{3}$ by multiplying the perfect cube factors of 20! we previously found.

$k^{3}$ = $2^{6*3}$ * $3^{2*3}$ * $5^{3}$

k = 64 * 9 * 5

$\boxed{k = 2880}$