Part III ... Next year is the 70th anniversary of the publication of the book 1984 by George Orwell. Here is a puzzle to start the anniversary celebrations off a bit early ... I think this is the final one from me on 1984

Warm up

Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$. 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 $1$, $9$, $8$, or $4$. Operands may of course also be derived from calculations e.g. $19*8*(\sqrt{4})$. 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 $8$ and $4$ to make the number $84$) 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 will get plus one from me.

Main Event

The main event is to find two different methods to find 123 without using concatenation. If you did not use concatenation above in the warm up then you only have one more formula to find using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $123$ without using any concatenation. so now, for example, you cannot put $8$ and $4$ together to make the number $84$. The rest of the rules above apply, but concatenation is not allowed. If you didn't use any concatenation above then you have solved the puzzle, but you could try to solve it with concatenation, that is concatenation of the initial numbers only. The two methods must be different; for example, $(9+4)*(1+8)$ and $(8+1)*(4+9)$ would not be considered to be two different methods.

Note (and perhaps hint): For this second part any finite number of functions can be used, though ingenious solutions with infinite numbers of functions will get plus one from me.

Note that in all the puzzles above 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 the required answers, but will not mark them as correct - particularly because a general method was developed by @Carl Schildkraut to solve these puzzles provided that you can make a single number greater than $2$ two times from the numbers you start with - here it could be $9+1$ and $8 + \sqrt{4}$ to get two $10$s, for example.

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

This is part III after the first and second in this series were solved

  • 2 + 2 = 5 ⟹ (1 + 9 + 8 + 4 = 123 ∧ 1 - 9 - 8 - 4 = 123) – Steadybox Sep 23 at 18:10
  • @Steadybox, you have to explain that a bit more for me to understand the implication of the 'assertion/assumption' 2+2=5.... – tom Sep 23 at 18:13
  • It's a joke. 2+2=5 because the party (in the book) says so. And because 2+2=5 is actually false, the statement "2+2=5 ⟹ anything" is true. – Steadybox Sep 23 at 18:17
  • @Steadybox, ok, sorry I didn't understand. – tom Sep 23 at 18:22
  • I could do with an extra $3$. $$\begin{align}(4+1)^\sqrt{9}-\sqrt[3]{8}&=5^3-2 \\ &=125-2 \\ &=\boxed{123}\end{align}$$ But if I could use the floor function $\lfloor\cdot\rfloor$ (round down), then I could write $(4+1)^\sqrt{9}-\lfloor\sqrt{8}\rfloor=123$ instead. $(+1)$ nonetheless :P – user477343 Sep 24 at 6:09
up vote 6 down vote accepted

Here are three possibilities that I have found:

$\sqrt{9\times(8!/4!+1)}=123$

and the other one is (thanks to @PotatoLatte)

$(8-\sqrt{9})!+4-1=123$

the last one

$\sqrt{8+1+9!/4!}=123$

  • Great Job, :-) well done. Personally I think the third and first are very similar as you have effectively multiplied out the brackets - but you definitely have two distinct and I think that you third method is very interesting. --- sorry if I too harsh about `similar methods', but if I was not firm about methods being different then @PotatoLatte would have had two distinct answer before. – tom Sep 23 at 15:50
  • @tom it is for sure similar, but different numbers are used actually. but that's okay if you accept the third as a different answer or not :) it is up to you. – Oray Sep 23 at 15:57
  • Anyway whether 3 or 2 - great answer and the last one is very ingenious! – tom Sep 23 at 16:13

Is it

$(8-\sqrt{9})!+(4-1)$

And

$(8-4+1)! + \sqrt{9}$

  • I like your answer and it will get plus one, but I only consider this to be one method - does that that make sense? – tom Sep 23 at 14:57

I haven't yet seen

$(4+1^8)!+\sqrt9$

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