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I'm looking for a solution to make number $75$ with numbers $1$ $9$ $6$ $2$ in that order and the same rules as in Use 2 0 1 and 8 to make 67.
Here a copy of those rules:

  1. You must use all 4 digits. Only the digits $1$, $9$, $6$, and $2$ can be used in that order.
    You can make multi-digit numbers out of the numbers. Examples: $19$, $96.2$

  2. The square function may NOT be used. Nor may the cube, raise to a fourth power, or any other function that raises a number to a specific power. You may use the ^ operation if you use a digit, for example, $(1 + 9)^6 - 2!$ is acceptable (if you're trying to get $999998$), because 1, 9, 6, and 2 are used. However, $19 ^ 2 / 6 + 2$ can't be used to get $62.166...$ because it uses an extra 2.

  3. Sorry, but the integer function may NOT be used. Nor may the round, floor, ceiling, or truncate functions.

  4. $+$, $-$, $\times$, $\div$ or $\frac{\Box}{\Box}$, $()$, $!$, $\sqrt{\Box}$, ${\Box}^{\Box}$, and $!!$ may be used for functions.

Please no brute-force methods. Good luck.

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    $\begingroup$ Questions should be self-contained, so I've edited in the rules from the linked challenge (and added the same tags). $\endgroup$ – Kevin Cruijssen Jun 8 at 22:44
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    $\begingroup$ Let's please not make "near-miss", out-of-intended-order, or otherwise clearly non-pertinent answers here. They do not attempt to answer the posed puzzle, and will be deleted: puzzles with no lateral-thinking tag do not invite such answers. $\endgroup$ – Rubio Jun 9 at 16:27
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    $\begingroup$ agreed, good point @Rubio! $\endgroup$ – Omega Krypton Jun 9 at 16:56
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how about this?

$$1\times 9/.6/.2$$

this is allowed right?

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    $\begingroup$ Wow, this is clever. $\endgroup$ – greenturtle3141 Jun 9 at 1:30
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Loophole that I discovered:

allowing multifactorial also allowed us to multiply $x$ by any integer $<x/2$ using a certain number of $!$s.

(1) Pretty much based on micsthepick's answer...

$(1*9+6)/.2$
$=15/.2$
$=75$

(2) self-innovated:

$(1+9)!!!!!!!/(.6-.2)$
$=(10*3)/.4$
$=75$

(3) self-innovated:

$(-1+9)!!!-\sqrt{\sqrt{\sqrt{...\sqrt{\sqrt{\sqrt{6}}}}}}/.2$ (infinite \sqrt{}s)
$=(8*5*2)-1/.2$, (yes, according to this)
$=75$

(4) self-innovated

$(19+6)!!!!!!!!!!!!!!!!!!!/2$
$=25!^{(19)}/2$
$=25*6/2$
$=75$

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    $\begingroup$ Some notes regarding the multiple-factorials, OP didn't specify whether theya are allowed or not except for single and double factorials. $\endgroup$ – athin Jun 9 at 14:30
  • $\begingroup$ i think it is fine since it is in the linked question. thanks @athin $\endgroup$ – Omega Krypton Jun 9 at 14:37
  • $\begingroup$ Gonna love #3 for insanity :D Good job on it! $\endgroup$ – val Jun 9 at 19:46
  • $\begingroup$ Wow, I actually thought of the infinite square roots one, but didn't post it! +1 :) $\endgroup$ – Duck Jun 10 at 1:58
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I'm not sure if this is allowed, but

$1 9 6 2$

$(1 {\sqrt 9}) + (62)$

$(1 3) + (62)$

$13 + 62$

$75$

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    $\begingroup$ the question does say that you can make multi digit numbers, but I would argue that it does not specifically allow the concatenation operator $\endgroup$ – micsthepick Jun 8 at 23:57
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    $\begingroup$ if 1√9 = 13 then you can simplify it. (9-2)(6-1) = 75 $\endgroup$ – AsifHabib Jun 10 at 7:48
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Here's one that I found:

$.1*((\sqrt{9})!)!+(6/2)$
$= .1 * 720 + 3$
$= 72 + 3 = 75$

Here is one similar to micsthepick's answer:

$1*(9+6)/.2$

And here is a hybrid of the two:

$1*((\sqrt{9})!)!/6!!/.2$
= $720 / (2*4*6) / .2 = 15 / .2 = 75$

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