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.

17 Answers 17

up vote 58 down vote accepted

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.


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

  • 14
    that's an amazing answer --- I never would have thought of that :D – Hugh Sep 11 at 0:56
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    (+1) very nice. Thanks for following my proposal! :) – TheSimpliFire Sep 11 at 6:15
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    Technically? What is technically? @Battle Mathematics are not a technique, are a science. And x^(1/n) when n tends to infinite is always 1, since x^0 is 1, for any positive x greater than one. – JuanRocamonde Sep 11 at 16:20
  • 3
    @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. – Lord Farquaad Sep 11 at 17:41
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    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. – Tanner Swett Sep 13 at 5:22

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




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 $K$.

  • 4
    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 at 8:12
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    I'm confused, isn't $280!!!!...$ absolutely astronomical compare the the denominator? – orlp Sep 14 at 6:58
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    @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 $m\equiv n\bmod k$. – Carl Schildkraut Sep 14 at 13:49

If you allow


then you can do

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

  • Sorry no decimals, but plus one – tom Sep 11 at 6:19
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    @tom Oh well. Do you know that a solution exists using only the operations you have allowed? – Carl Schildkraut Sep 11 at 6:20
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    Yes there is a solution.... I will start giving hints soon if people don't get it... – tom Sep 11 at 12:19
  • When I plug your solution into my calculator, it gives me 7.1. – Agi Hammerthief Sep 13 at 14:16
  • @AgiHammerthief presumably you are calculating (0.1) - 2 + 8 + 0! rather than raising 0.1 to the -2th power. – IanF1 Sep 15 at 12:02

The trick is to:

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

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

  • 2
    i like this one. – Shahriar Mahmud Sajid Sep 11 at 10:15
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    great! I like this too, plus one, but sorry there is a decimal solution – tom Sep 11 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:


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

$$108 + !2$$


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

  • 1
    Plus one, but sorry not correct – tom Sep 11 at 6:21

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 $9$


$8-2=6$, 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!

  • 3
    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 at 12:30

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

  • Nice try, sorry not correct, but plus one – tom Sep 11 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) – trolley813 Sep 11 at 9:29
  • @trolley813 - will edit the question to make this absolutely clear... – tom Sep 11 at 12:20
  • $(+1)$ I like this one :) – user477343 Sep 12 at 12:13

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 at 12:35
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    How did you find this answer? – hkBst Sep 11 at 16:48
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    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. – ponut64 Sep 12 at 9:34
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    This can be improved a little bit, as $1^8 - 20 = -19$ – Tanner Swett Sep 13 at 5:26

I have another trigonometric answer.

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

  • 1
    neat solution :-) plus one – tom Sep 12 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

  • Nice try, plus one – tom Sep 11 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) $$108\space+\space!(!2) = 109$$


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

  • 2
    Nice idea - a bit like the use of the ++ operator in another answer.... :-) – tom Sep 12 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. – Glorfindel Sep 11 at 9:11
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    Nice idea, plus one, but not the solution I'm afraid – tom Sep 11 at 12:32

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

enter image description here

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

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

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

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

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


  • Digits are only allowed to be used once. – Jaap Scherphuis Sep 12 at 12:31
  • Agh, must have glossed over that one. – Ruben Dijkstra Sep 12 at 12:43

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