# 1984 - take the digits 1,9, 8 and 4 and Hard Challenges!

This is kinda follow up question to:

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 ...

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

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

The same rules but there are some different challenges here.

• Challenge No 1: Find 142 with the least amount of operations and parenthesis.
• Challenge No 2: Find 87 with the least amount of operations and parenthesis.
• Challenge No 3: Find 61
• Challenge No 4: Find 71 without using power operation (only ^).
• Challenge No 5: Find 46

Note that infinite square root is not allowed and I will accept the answer which includes all solutions.

• Just double checking square root is counting as a power operation right? Sep 20, 2018 at 13:10
• @gabbo1092 no, I count it as a different operation. adding that info in the question.
– Oray
Sep 20, 2018 at 13:11

Challenge No 1: Find 142 with the least amount of operations and parenthesis.

$148 - \sqrt{9}! =142$

Challenge No 2: Find 87 with the least amount of operations and parenthesis.

$91 - 8 + 4 = 87$

Challenge No 3: Find 61

$1 + \sqrt{9}!\times (8 + \sqrt{4}) =61$

Challenge No 4: Find 71 without using power operation (only ^).

$81 - \sqrt{9}! -4 = 71$

Challenge No 5: Find 46

$\sqrt{9}! \times 8 - \sqrt{4} \times 1 = 46$

• fyi: 142 and 87 are not found with the least amount of operations/parenthesis.
– Oray
Sep 20, 2018 at 13:19
• only Challenge No:2 is not okay, but the rest is done.
– Oray
Sep 20, 2018 at 13:28
• Will focus my efforts there Sep 20, 2018 at 13:29
• $\sqrt 9 !$ master race. Sep 20, 2018 at 22:29

Minimum amount of steps for 87 (Challenge No 2) :

$$\sqrt{841\times9}$$

• yes you got it :)
– Oray
Sep 20, 2018 at 13:34
• I'd argue that that is more than the 91-8+4 answer as they both use two steps (+- and sqrt x respectively) but the second requires an implicit sqrt bracket. Sep 20, 2018 at 15:28

Challenge no. 5 (46):

$$(8-1)^\sqrt{4}-\sqrt{9}=7^2-3=49-3=46$$

For 142:

$$\sqrt4(9\times8-1)=142$$

In order:

$$(-1+9\times8)\sqrt4=142$$

For 87:

$$8(9+\sqrt4)-1$$

• fyi: 142 and 87 are not found with the least amount of operations/parenthesis.
– Oray
Sep 20, 2018 at 13:19