Use 2, 0, 1 and 8 to make 109

Question

Assemble a formula using the numbers $2$, $0$, $1$, and $8$ in any order that equals 109. You may use the operations $x + y$, $x - y$, $x \times y$, $x \div y$, $x!$, $\sqrt{x}$, $\sqrt[\leftroot{-2}\uproot{2}x]{y}$ and $x^y$, as long as all operands are either $2$, $0$, $1$, or $8$. Operands may of course also be derived from calculations e.g. $10+(\sqrt{8*2})!$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $2$ and $8$ to make the number $28$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations which get $109$ will get plus one from me.

Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 \times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get 109, but will not mark them as correct.

Here are some examples to this problem:

• Use 2 0 1 and 8 to make 67
• Make numbers 93 using the digits 2, 0, 1, 8
• Make numbers 1 - 30 using the digits 2, 0, 1, 8

many thanks to the authors of these questions for inspiring this question.

Answer 1

I think...

$\sqrt{\frac{12!}{8!} + 0!} = \sqrt{11881} = 109$

Answer 2

Probably not the intended answer, but, I propose:

$108+\sqrt{\sqrt{\ldots \sqrt{2}}}$, with infinitely many square roots.

Explanation:

Formally: $$\sqrt{\sqrt{\ldots \sqrt{2}}}=\lim_\limits{n\to \infty} s_n=1$$
where $s_0=2$ and $s_{n+1}=\sqrt{s_n}$.

Answer 3

This is technically a solution with n-druple factorials. It is not a good solution.

Solution:

$$\frac{(10!!!)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!}{(2+8)!!!}.$$

Explanation:

First, we note that $2+8=10$. In particular, we can construct the number $10\cdot 7\cdot 4\cdot 1=10!!!=280$ in two different ways.
Once we have two copies of $280$, we can construct the number $$280!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!$$ (that's $171$ factorials, so it equals $280\cdot 109$), and then simply divide by $280$ again. This technique can be used to boringly nuke every problem of this form that allows for n-druple factorials: if you want to get some number $K$, you can use the property that $2+8=10$, so you can get two copies of $N$ for some large $N$ (by repeatedly taking factorials from $10$, for example; in terms of big $N$, any $N\geq 2K$ should do it), and then you can take $N!_{N-K}$ ($N-K$ factorials) to get the number $N\cdot K$ - then you just divide by $N$ and now you have $K$.

Answer 4

If you allow

decimals

then you can do

$$109=(.1)^{-2} + 8 + 0!$$

Answer 5

The trick is to:

... count in hexadecimal: 21 × 8 + 0! = 109

(in decimal: 21h = 33 and 109h = 265)

Answer 6

Hmmm, possibly:

Place a vertical mirror by the $2$ to get a $5$. Concatenate the $5$ with the $0$ to get $50$ and take the $38$-factorial (using $38$ exclamations): $50!^{38}=50\cdot12=600$, and then add $1^8=1$: $50!^{38}+1^8=601$. Turn the $601$ upside-down to get $109$.

and for posterity:

$20!^{15}+1+8$

Answer 7

This probably won't count. It uses $!$, but not in $x!$.

$$108 + !2$$

Explanation

$!n$ is the number of derangements of n objects. In particular $!2 = 1$.

Answer 8

$$8+2^0=9=09\implies 1\{8+2^0\}=109$$

Answer 9

I have a few silly answers :) Though, frustratingly, I haven't been able to solve it yet

$108$++ $= 109$ which you might also write as $108$+^$2$

or if we are allowing transformation of numbers as I've seen above

$(2+8)$ concatenated with $0$, with a vertical line (using the $1$) on the RHS to turn it into a $9$

or

$8-2=6$, which rotated gives $9$, then concatenate $10$ to give $109$

I was thinking about trying to

Change the base of the numbers

but didn't get anywhere with that idea...

Looking forward to seeing the solution!

Answer 10

I break the rules, BUT!

work with 1 and 20 1 - 20 = -19

109 = arccos(sin(-19))

Answer 11

I have another trigonometric answer.

$$8^2 - \arctan (0-1) = 109$$ the base of tangents is $45$, subtract $-45$, add $45$ to $64$

Answer 12

This won't be correct (concatenation of numbers from calculations is not permitted), but this is a way to cheese it, were that allowed

Assuming you can have leading 0's...
$ 2^0 = 1$
$ \sqrt{81} = 09$
Concatenate the two
$109$

Answer 13

This is also just for fun, using $!$ in a different way than factorial:

Uses $!$ as the binary NOT operator (like in C++ and JavaScript) $$108\space+\space!(!2) = 109$$

Explained:

!2 == false, and !false == true. True is numerically represented as $1$. Then you get $108 + 1$, which equals $109$.

Answer 14

We include 3 numbers 1,0,8 to 108, 108 can be written as 107+1
now we have 2 remaining from the list of given numbers we can use as 107+(2*1)=109

Answer 15

Easy, just use a one sided self referencing equation, an ingenious mathematical artefact invented by me just now :

Answer 16

Not correct, because it uses the same digits more than once, but you could get there like this:

$(2 * 8^2) - (8 * 2) - (\frac{1}{2} * 8) + 1^0 = 109$

Answer 17

Even if it was solved, I couldn't help myself and came up with a boring

$$\dfrac{2180}{20}=109$$

Answer 18

My answer is

$\frac {10!} {8!} + (2 * 8) + 1^0$ = 109