# 1984 - take the digits 1,9, 8 and 4 and make 369

Part II... 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 ...

Warm up

Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$. 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

If you used concatenation above then make a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $369$ without using any concatenation. so, 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.

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.

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

This is part II after the first in this series was solved

$\frac{(\sqrt9 ! )!}{\sqrt4}+8+1=\frac{(3!)!}{2}+9=\frac{6!}2+9=\frac{720}2+9=360+9=369$

Or, with an infinite number of functions:

$\frac{(\sqrt9 ! )!}{\sqrt4}+8+\sqrt{\sqrt{...\sqrt{1}}}$, which simplifies to the expression above

• brilliant good job, plus one - this looks like the first answer to the main event :-) – tom Sep 20 '18 at 0:56
• If only I didn't use the spare $0$... Eh well, you deserved it :P – Mr Pie Sep 20 '18 at 1:02

Well, for the warm-up

\begin{align}19^{\sqrt{4}}+8&= 19^2+8 \\ &= 361+8 \\ &=\boxed{369}\end{align}

Fun puzzle! And, since $1984=01984$... if I could use a spare $0$...

$$\big((8-1-0!)!\div \sqrt{4}\big)+9=\boxed{369}$$

:D

• Good job - looks like this is the first correct answer to the warm up - great job – tom Sep 20 '18 at 0:58
• @tom Now to not use any concatenation... :P – Mr Pie Sep 20 '18 at 1:00

Perhaps, for the warm-up

$41\times 9\times \sqrt{\sqrt{\sqrt{...8}}}$, with infinite roots
$=369\times \sqrt{\sqrt{\sqrt{...8}}}$.

• Great this is what I was thinking about for the warm up... as the second part could not have infinite functions, but, therefore, the first part could :-) +1 – tom Sep 20 '18 at 0:55
• I added roll over spoiler hider - so that people have to roll over to see your answer – tom Sep 20 '18 at 1:03
• Thanks @tom. As a newbie I am still learning how to format my answers – binu kurian Sep 20 '18 at 1:16
• no problem, but I think @user477343 sorted it out. – tom Sep 20 '18 at 1:22
• I am glad I helped out :) – Mr Pie Sep 20 '18 at 5:49