I plan on giving this challenge to my students but I can't seem to solve it myself.

Challenge : write all the integers from $0$ to $100$ using the numbers $2,3,4$ and $5$ exactly once and in this order.

Operations allowed are addition, substraction, multiplication, division, powers and square-roots. Brackets are allowed. Using $23$, $34$ or $45$ is also allowed.

For instance, $18 = 2^3 +\sqrt{4}\times 5$. Every number is used exactly once and in the right order.

So far, I've managed to find every integer from $0$ to $50$ except for $48$ and $49$. From $50$ to $100$, I've only found a good half of them.

I'm not sure this is even doable. In case it is not, I thought about adding one extra operation : concatenation, which I will write $@$ - so that $(2\times 3)@(4+5) = 69$. Even with concatenation, I can't seem to find every integer : I'm missing $9$ of them.

So here's what I propose - even though I'm not sure this fits with this site spirit - :

Help me filling this sheet; by first only trying without concatenation (using $23$ and such does not count as concatenation). You can copy the square root symbol from the cell which contains it. Please be kind and do not modify other answers unless there is an error.

I did not write my answers because I thought some of you could enjoy trying by yourselves.

Since my question does not expect an answer that can be written here (except if somebody finds the strenght to write every solution here...), and since I'm only missing $48$ and $49$ below $50$, here's the secondary question that you can answer here :

Write $48$ and $49$ without using concatenation.

  • $\begingroup$ @DanielRobert-Nicoud : this is already a crucial application. I think some numbers can only be found using this. Plus, another application is using $\sqrt{2}$ (or maybe $\sqrt{3}$) with an even power. Or even using the square root of a perfect square, for instance $\sqrt{2^{3^{\sqrt{4}}-5}}$. $\endgroup$ – krirkrirk Jun 20 '18 at 16:28
  • $\begingroup$ Agreed.${}{}{}{}$ $\endgroup$ – Daniel Robert-Nicoud Jun 20 '18 at 16:38
  • $\begingroup$ "the only thing it can do is to make the 4 into a 2," why on earth to you call that useless? If you need a 2 as the third term and don't need a 4, that'd be useful. $\endgroup$ – fleablood Jun 20 '18 at 16:38
  • $\begingroup$ 49 is a toughie indeed... the only thing I can htink of is $(3 + 4)^2$ but that doesn't even use 5 besides being in the wrong order to boot. $\endgroup$ – Robert Soupe Jun 21 '18 at 2:34
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    $\begingroup$ @RobertSoupe I have a program that attempt to generate all possible groupings for the binary opertors + - * / ^ @ and then replace each left bracket by one of (, -(, sqrt(,-sqrt(. I have found $4960$ combinations that evaluates to a number between $1$ and $100$. Only 6 of them gives $48$ and $49$ and all of them need the @ operator. $\endgroup$ – achille hui Jun 21 '18 at 8:40

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