Use some mathematical operations to make an expression equal 93.

You must use each one of the digits 2, 0, 2, 2 exactly once, and no other digits.

To score the solutions, your set of operations used must be least powerful.

Least powerful operations

Note that some general constructions like: using a successor function, using square roots and logarithms, using increasing and decreasing operations combined with rounding functions, ... potentially allow to express any number.

Therefore, your goal is to use a set of operations that is least powerful.

Let A and B be two sets of operations. If A can be used to make x integers in an interval [1,N] and B can be used to make y integers in an interval [1,N], then A is less powerful than B if x < y as N goes to infinity. (Unless the operation set can be used to make every number, then it is infinitely powerful and by definition, the worst solution.)

Due to practical constrains, I might estimate the score based on the interval [1,N] for some reasonable N, under some reasonable restrictions such as allowing at most 10 consecutive uses of an unary operation.

The least powerful solution will be accepted.

Established operations

Only established operations are allowed. That is, the operation must appear in a peer reviewed article. For example, double factorial (mathworld) has multiple such references listed on the mathworld website (eg. Meserve, B. E. "Double Factorials." Amer. Math. Monthly 55, 425-426, 1948.).

On the other hand, (a @ b) := 93 if a=b=2 else 0 (defining own operations) would not be allowed.

Obscure sequences

Obscure sequences (I'm sure there are many oeis.org sequences that contain 93 as a constant), are not allowed. This includes almost all oeis.org sequences.

I said almost all, since for example, using parentheses ( ) as the binomial coefficient (n k) is allowed due to lateral-thinking tag, which is then A007318 sequence in the OEIS.

  • 2
    $\begingroup$ You are more than half a year early. $\endgroup$ – Bass Apr 26 at 17:01
  • $\begingroup$ Can you provide more details on the scoring formula? Maybe share the scorer code? $\endgroup$ – justhalf Apr 27 at 11:26
  • $\begingroup$ @justhalf As mentioned in the puzzle, I was computing the percentage of inexpressible integers (given the operation set used in a solution) in some reasonable interval [1,N], as an alternative to estimating the density of expressible numbers as N goes to infinity (see my definition of less powerful operation set). My python code is still messy, but the same result can be achieved by extending any of the many available number formation solvers (for example, search "four fours" on github). So far, I see this maybe was not so clear and attracted some downvotes, so I will retract my score board. $\endgroup$ – Vepir Apr 27 at 13:49
  • 1
    $\begingroup$ What counts as "obscure"? Without defining that, this question is subjective. Additionally, the question seems to be an open-ended question exactly of the type that is not allowed on Puzzling. It's a competition to see who can do the best, rather than a puzzle with a single solution. $\endgroup$ – Deusovi Apr 27 at 15:15
  • $\begingroup$ @Deusovi If we were to prove that a least powerful operation set exists (proving that a best optimal solution exists), would this still be considered open-ended, or would it be a valid optimization problem? (We are minimizing the power of the operation set.); And if we would in addition solve the subjectivity by defining a large source of allowed (non-obscure) operations, would this then be on topic? $\endgroup$ – Vepir Apr 27 at 15:33

I'm unsure how to score this.

$$ \arccos\left(0\right)+\frac{2}{2}+2 $$
arccos(0) (In degrees mode) is 90
So 90 + 1 + 2 = 93

  • $\begingroup$ Nice lateral-thinking! This is definitely better than the previous two solutions! (Your unary function you added is bounded!) In fact, I think this will be hard to beat. The only way of beating this I see, is possibly replacing the factorial with either a binary operation or a bounded unary operation. $\endgroup$ – Vepir Apr 26 at 20:51
  • $\begingroup$ In fact, do you see how you can make this work without factorial? There is a similar arc-ish function you can use, instead of the current one! :) $\endgroup$ – Vepir Apr 26 at 21:01
  • $\begingroup$ @Vepir That works, and is less powerful. $\endgroup$ – A username Apr 26 at 21:05
  • $\begingroup$ Btw, I think you should not spoil what functions you used, outside of the spoiler tag. $\endgroup$ – Vepir Apr 26 at 21:07
  • $\begingroup$ Looks like you can get rid of an operation (and thus several "reachable" numbers) by using the "count elements" function on the twos. $\endgroup$ – Bass May 6 at 4:35

$$93=(\sqrt[.2]{2}-0!)/ \sqrt{.(\dot{!2})}$$

  • $\begingroup$ How do I scored on current score estimation? $\endgroup$ – 伍逸朗 Apr 27 at 9:51
  • $\begingroup$ This looks okay, I will estimate it when I get a chance, probably later today. I would guess your score would be between 3. and 4. on the current list. $\endgroup$ – Vepir Apr 27 at 10:11
  • $\begingroup$ Also, shouldn't there be division instead of multiplication? (.1 repeating is 1/9) $\endgroup$ – Vepir Apr 27 at 10:14
  • $\begingroup$ Oops you are right $\endgroup$ – 伍逸朗 Apr 27 at 10:15
  • 1
    $\begingroup$ I think I will not accept the arccos one as that is actually 90 DEGREES It is a unit more than needed. $\endgroup$ – 伍逸朗 Apr 27 at 10:35

Not sure if this fits the requirements (it probably doesn't). I kind of got lost in them. But here goes.

$\log_{\frac{0!}{2}}\left(\log_{2}\left({\underbrace{\sqrt{\sqrt{\dots\sqrt{2,}\,}\,}}_\text{93 square roots}}\right)\right) = $
$ = 93$

A more general approach

replace 93 square roots with X square roots and you get any number X..

  • $\begingroup$ It works, but it is the worst possible solution to allow logarithms alongside square roots (since every other solution will be less powerful). $\endgroup$ – Vepir Apr 26 at 13:59
  • $\begingroup$ I had a feeling it will come to this. Ok. I will try something else. $\endgroup$ – Marius Apr 26 at 14:55
  • $\begingroup$ You scored 0% on current score estimation :P $\endgroup$ – Vepir Apr 27 at 9:33
  • $\begingroup$ I think your score system is flawed :). If I get the correct answer I should have something more than 0. But I get your point. I didn't put much effort in this. $\endgroup$ – Marius Apr 27 at 9:53
  • 1
    $\begingroup$ If a solution can create every number, then it is by definition 0. I defined a computable score, it is not related to if you put effort in it or not. (Also, if this weren't the case, there would be too much solutions.) $\endgroup$ – Vepir Apr 27 at 10:08

I'm not sure how to score this but I've found a neat little formula using the

Divisor function $\sigma$

As follows

$\sigma(((2+0!)!)^2) + 2 = 93$


Another answer

$$93=p(2)\times p\left(\frac{22}{p(0!)}\right)$$


$p(n)$ (usually written $p_n$) is the $n$th prime number (a commonly used function). Expanding, we get $p(2) \times p\left(\frac{22}{p(1)}\right)=p(2)\times p(\frac{22}{2}) = p(2) \times p(11) = 3 \times 31 = 93$

  • 1
    $\begingroup$ At first glance, it is actually possible to improve this by getting rid of your infinitely powerful unary operation, and replacing it with some other common operation(s). $\endgroup$ – Vepir Apr 27 at 7:27

Well, I have absolutely no idea how this may or may not score, but here are the operations I'll be using:

  • "concatenate some of the original numbers in the order they were given, with an optional decimal point" (a common operation in puzzles)
  • square root
  • subtraction
  • division
  • the "round to nearest integer" operation (denoted by the $\approx$ below)

$$ \frac{20 - \sqrt2}{.2} \approx 93$$

I don't think this can compete with the trigonometrics answer, but since this hits surprisingly close to the target using only a couple of the more commonly seen operations, I thought it might be of interest.

  • $\begingroup$ This is actually close to the trigonometric solution! Even more so if we decide to not count 20 as concatenation operation, but to simply count it as digit concatenation (as it is often the case in four fours puzzle variations). $\endgroup$ – Vepir Apr 27 at 9:09

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