# Maximizing the common value of both sides of an equation (part 2)

Using all the integers from 1 to 12 inclusive (each used exactly once), along with any number of the five arithmetic operations (addition, subtraction, multiplication, division and raising to a power) and parentheses, construct an equation with the largest (and identical) number on both sides.

Clarifications:

$$1+2+3+4+5+6=10 \times (9+8+7) + (12-11)$$
is not a valid equation because the value on the left hand side (21) is not equal to the value on the right hand side (241).

$$(2-1) \times (4-3)=(12-11) \times (10-9) \times (8-7) \times (6-5)$$
is a valid equation because the value on both sides equals $$1$$. However this common value of $$1$$ is not the largest that can be achieved.

You can’t use anything other than what is explicitly allowed. So, for example, you can’t use square roots, concatenation, decimal points, negation, bases other than 10, ...

Related question: Maximizing the common value of both sides of an equation

We can do

$$\huge 6^{12^{3^{4^{11}}}}=(1+5)^{(2+10)^{9^{8^7}}}$$

which is pretty large, indeed.

If I'm getting the (rather tricky) maths right the value is between

$$\huge 10^{10^{10^{2,001,191}}}$$ and $$\huge 10^{10^{10^{2,001,192}}}$$.

• Impressive, looks like it is the best answer so far! I think it would be clearer if you just say it is between $10^{10^{10^{2001191}}}$ and $10^{10^{10^{2001192}}}$. I also did the calculations and I agree with yours. Commented Sep 4 at 9:34
• Wolfram Alpha gives virtually the same answer although it formats it differently. Commented Sep 5 at 12:20

$(2+9)\cdot(1+6)\cdot4\cdot5\cdot12=11\cdot7\cdot3\cdot8\cdot10=18480$

Sorry I missed the fifth arithmetic operation --- power, and thanks for @franck vivien's advice! Here is my update:

$$(1+6)^{(2+9)^{4\cdot 5\cdot 12}}=7^{11^{3\cdot 8\cdot 10}} \approx 10^{10^{249.861...}}$$

And, as a supplement, in order to see the order of magnitude comparisons...

\begin{aligned} 7^{11^{240}} & \approx 10^{10^{249.861...}} \\ 11^{7^{240}} & \approx 10^{10^{202.841...}} \\ 7^{240^{11}} & \approx 10^{10^{26.109...}} \\ 11^{240^{7}} & \approx 10^{10^{16.679...}} \\ 240^{7^{11}} & \approx 10^{10^{9.672...}} \\ 240^{11^{7}} & \approx 10^{10^{7.666...}} \\ \end{aligned}

• Much bigger if we replace the 2 first multiply signs (between 11 and 7, and between 7 and 240 ) in both sides, by exponentiation! Commented Sep 3 at 12:32
• @franckvivien Thank you!!!......!! I've already edited my answer as you suggested.
– tToE
Commented Sep 3 at 13:57

Here is my attempt:

$$\huge 2^{9^{{10}^{11}}} = 8^{3^{\left(\frac{(4+6)^{12}}{5}-1^7\right)}}$$

Both the left hand side and the right hand side are equal to:

$$\huge 2^{3^{\left(2 \times 10^{11}\right)}}$$

which is between

$$\huge 10^{10^{95,424,250,943}}$$ and $$\huge 10^{10^{95,424,250,944}}$$

• Impressive! I love this one!!
– tToE
Commented Sep 4 at 7:24

Pretty sure you can get a lot bigger than this, but..

$$\huge 1\times 12^{11^{(10+8)}}=(3\times 4)^{(5+6)^{(9+7+2)}}$$

which is

between $$10^{10^{18}}$$ and $$10^{10^{19}}$$.

• This uses 8 multiple times. Commented Sep 3 at 11:45
• @Someone I believe this one works, and it's much bigger than the first try (it has about $7\times 10^{13}$ digits). Still think that it could be larger still.
– lulu
Commented Sep 3 at 12:05
• Changed it slightly, latest version has about $10^{18}$ digits.
– lulu
Commented Sep 3 at 12:12

Here are my attempts:

$$\huge 11^{(5\times6)^{8^{(1+7)}}}=(2+9)^{(3\times10)^{4^{12}}}$$

which is between

$$\huge 10^{10^{24,781,982}}$$ and $$\huge 10^{10^{24,781,983}}$$.

• Can you also write it as a power tower like $10^{{10}^\cdots}$ for easier comparison with the other answers? Commented Sep 4 at 6:47
• It is between $10^{10^{24781982}}$ and $10^{10^{24781983}}$ Commented Sep 4 at 9:44