Greatest Product

Let $x =$ Answer

$x$ has prime factors which when summed with multiplicity will equal $20$

Let's start with the highest prime below $20$

$20 = 19 + 1$

That $1$ isn't very useful is it so we'll just take one from $19$ to get

$20 = 18 + 2$

Now let's turn $18$ into it's prime sum

$20 = (2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2) + 2$

The product of the above values is only $1024$ so how do we make it bigger?

We have nine $2$ s in that bracket which means there's three lots of three $2$ s so

$20 = ((2 + 2 + 2) + (2 + 2 + 2) + (2 + 2 + 2)) + 2$

$20 = ((6) + (6) + (6)) + 2$

The product is only $432$ now but we're not done yet

$20 = ((3 + 3) + (3 + 3) + (3 + 3)) + 2$

So now the product is equal to $1458$

The reason it's so different to before is because

$2^{10} < 3^6 \cdot 2$

This turns into

$2^{10} < (2 + 1)^6 \cdot 2$

Which then turns into

$2^{10} < 2 \cdot (2^6 + 6(2^5) + 15(2^4) + 20(2^3) + 15(2 ^2) + 6(2^1) + 1)$

Which when expanded equals

$2^7 + 3(2^7) + 15(2^5) + 5(2^6) + 15(2^3) + 3(2^3) + 2^1$

Adding together like terms creates

$4(2^7) + 5(2^6) + 15(2^5) + 18(2^3) + 2^1$

We're getting there, we'll put it into a multiple of $2^{10}$ to make it easier

$\frac {2^{10}}{2^1} + (\frac {2^{10}}{2^2} + \frac {2^{10}}{2^4}) + (\frac {2^{10}}{2^2} + \frac {2^{10}}{2^3} + \frac {2^{10}}{2^4} + \frac {2^{10}}{2^5}) + (\frac {2^{10}}{2^3} + \frac {2^{10}}{2^6}) + \frac {2^{10}}{2^9}$

Now we just combine like terms to get

$\frac {2^{10}}{2^1} + \frac {2 \cdot 2^{10}}{2^2} + \frac {2 \cdot 2^{10}}{2^3} + \frac {2 \cdot 2^{10}}{2^4} + \frac {2^{10}}{2^5} + \frac {2^{10}}{2^6} + \frac {2^{10}}{2^9}$

Now we simplify

$2^{10} + \ldots$

Let's stop there we already know that after the dots is a value above zero so we know that it is higher that $2^{10}$

In other questions like this it's best to get as many threes as possible to get the product higher. I'm not sure why it is like this so don't ask me.

trick is that if (2)*(6)=12, which accounts for sum 2+6=8, but we can write 8 as 3+3+2 which makes product as 18,applying same trick we can write 3+3+3+3+3+3+2=20 and product is (3^6) *2=1458