# Use 2, 0, 1 and 8 to make 109

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:

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

• Comments are not for extended discussion; this conversation has been moved to chat. Sep 24 '18 at 15:32

I think...

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

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 $s_{n+1}=\sqrt{s_n}$.

• that's an amazing answer --- I never would have thought of that :D Sep 11 '18 at 0:56
• (+1) very nice. Thanks for following my proposal! :) Sep 11 '18 at 6:15
• @Battle The rules also allow for $\sqrt{x}$ any number of times, which is what Surb's done here. Also, $1/x^{1/n}$ does equal 1 when $n$ tends towards infinity. That's how limits work. Without being able to say "the limit of $x$ as $n$ tends towards infinity equals..." calculus (and many other math disciplines) fall apart. We just wouldn't be able to do any calculations involving infinity. Sep 11 '18 at 17:41
• If you want to be "technically", then the question allows taking the square root any number of times, but it doesn't allow taking an infinitely iterated square root, because an taking an infinitely iterated square root isn't taking the square root any number of times. There's no standard definition for literally taking a square root infinitely many times; the best you can do is take the limit of a sequence of increasingly many square roots. Sep 13 '18 at 5:22
• @Surb - well, this answer of yours is all about … surds! Sep 24 '18 at 13:44

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 $K$.

• Great, general solution! I almost want to mark this as the solution for its ingenuity, but there is a simpler way... You get the plus one of course, and I am so glad that I excluded multiple factorials
– tom
Sep 11 '18 at 8:12
• I'm confused, isn't $280!!!!...$ absolutely astronomical compare the the denominator?
– orlp
Sep 14 '18 at 6:58
• @orlp This is not repeatedly applying the factorial operation, this is the multiple factorial. $n!_{(k)}$ is the product of all the positive integers $m \leq n$ so that $m\equiv n\bmod k$. Sep 14 '18 at 13:49

If you allow

decimals

then you can do

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

• Sorry no decimals, but plus one
– tom
Sep 11 '18 at 6:19
• @tom Oh well. Do you know that a solution exists using only the operations you have allowed? Sep 11 '18 at 6:20
• Yes there is a solution.... I will start giving hints soon if people don't get it...
– tom
Sep 11 '18 at 12:19
• When I plug your solution into my calculator, it gives me 7.1. Sep 13 '18 at 14:16
• @AgiHammerthief presumably you are calculating (0.1) - 2 + 8 + 0! rather than raising 0.1 to the -2th power. Sep 15 '18 at 12:02

The trick is to:

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

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

• i like this one. Sep 11 '18 at 10:15
• great! I like this too, plus one, but sorry there is a decimal solution
– tom
Sep 11 '18 at 12:27

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:

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

• +1 that was my first guess also! Sep 11 '18 at 4:51
• Plus one .... Amazing. :-).
– tom
Sep 11 '18 at 6:18
• I'm missing something here... how does 50!^38 = 50 * 12? Isn't 50!^38 a ridiculously large number? Sep 11 '18 at 18:04
• it's a multi-factorial, see my answer to puzzling.stackexchange.com/questions/71748/…
– JMP
Sep 11 '18 at 18:55

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 $$!2 = 1$$.

• Plus one, but sorry not correct
– tom
Sep 11 '18 at 6:21

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

• Nice try, sorry not correct, but plus one
– tom
Sep 11 '18 at 8:14
• @TheSimpliFire Plus one also. It probably can be tagged as lateral thinking (@tom your problem states that you can concatenate numbers, not digits/digit sequences, so technically it's correct) Sep 11 '18 at 9:29
• @trolley813 - will edit the question to make this absolutely clear...
– tom
Sep 11 '18 at 12:20
• $(+1)$ I like this one :) Sep 12 '18 at 12:13

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!

• Great answers, I love the 108++. Good job - plus thanks for posting and using the hidden/reveal in your answer. Hope you enjou Puzzling SE (oh and plus one)
– tom
Sep 11 '18 at 12:30

I break the rules, BUT!

work with 1 and 20 1 - 20 = -19

109 = arccos(sin(-19))

• Ok, you did not use the 8, but that is an interesting and inventive answer - I think it is pretty clever - +1, but not the solution I'm afraid. (I have editted the question to make it clear that you need to use all 4 of the digits)
– tom
Sep 11 '18 at 12:35
• How did you find this answer? Sep 11 '18 at 16:48
• The base of sines and cosines is 90 degrees. You can think of one being plus 90 (sine) and the other being zero (cosine). 109 is 19 different from 90. If you take your difference from 90 (negative number, sine) and then use that as the difference from zero (anticosine), you can solve this. That only applies when your numbers are in the domain of 1 sine and 1 cosine. Sep 12 '18 at 9:34
• This can be improved a little bit, as $1^8 - 20 = -19$ Sep 13 '18 at 5:26

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

• neat solution :-) plus one
– tom
Sep 12 '18 at 23:05

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
$109$

• Nice try, plus one
– tom
Sep 11 '18 at 16:35

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 $109$.

• Nice idea - a bit like the use of the ++ operator in another answer.... :-)
– tom
Sep 12 '18 at 9:03

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

• Welcome to Puzzling! I believe this solution is against the rules; you cannot 'decompose' partial results. Sep 11 '18 at 9:11
• Nice idea, plus one, but not the solution I'm afraid
– tom
Sep 11 '18 at 12:32

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

• ok, plus one for invention
– tom
Sep 12 '18 at 9:02
• Yeah I was bored and maths is boring. Sep 12 '18 at 20:33

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

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

• $109 = 108+2^0$...
– Surb
Sep 28 '18 at 23:28

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

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

• Digits are only allowed to be used once. Sep 12 '18 at 12:31
• Agh, must have glossed over that one. Sep 12 '18 at 12:43

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