11
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Questions should be self-contained, so I've edited in the rules from the linked challenge (and added the same tags). $\endgroup$ Commented Jun 8, 2019 at 22:44
  • 6
    $\begingroup$ 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
    Commented Jun 9, 2019 at 16:27
  • 1
    $\begingroup$ agreed, good point @Rubio! $\endgroup$ Commented Jun 9, 2019 at 16:56
  • $\begingroup$ People need to strick this bogus idea of "infinite square roots" of a number being 1. They cannot be written down. If you're going to use the square root symbol, write it a finite number of times, for starters. $\endgroup$ Commented Oct 24, 2021 at 21:33

4 Answers 4

33
$\begingroup$

how about this?

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

this is allowed right?

$\endgroup$
1
  • 11
    $\begingroup$ Wow, this is clever. $\endgroup$ Commented Jun 9, 2019 at 1:30
13
$\begingroup$

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

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

$\endgroup$
5
  • 1
    $\begingroup$ Some notes regarding the multiple-factorials, OP didn't specify whether theya are allowed or not except for single and double factorials. $\endgroup$
    – athin
    Commented Jun 9, 2019 at 14:30
  • $\begingroup$ i think it is fine since it is in the linked question. thanks @athin $\endgroup$ Commented Jun 9, 2019 at 14:37
  • $\begingroup$ Gonna love #3 for insanity :D Good job on it! $\endgroup$ Commented Jun 9, 2019 at 19:46
  • $\begingroup$ Wow, I actually thought of the infinite square roots one, but didn't post it! +1 :) $\endgroup$
    – Duck
    Commented Jun 10, 2019 at 1:58
  • $\begingroup$ No, "infinite square roots" is bogus. 1) They cannot be written down. 2) It is really a form/relative of a floor function. $\endgroup$ Commented Oct 24, 2021 at 21:30
5
$\begingroup$

I'm not sure if this is allowed, but

$1 9 6 2$

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

$(1 3) + (62)$

$13 + 62$

$75$

$\endgroup$
3
  • 13
    $\begingroup$ the question does say that you can make multi digit numbers, but I would argue that it does not specifically allow the concatenation operator $\endgroup$ Commented Jun 8, 2019 at 23:57
  • 3
    $\begingroup$ if 1√9 = 13 then you can simplify it. (9-2)(6-1) = 75 $\endgroup$
    – AsifHabib
    Commented Jun 10, 2019 at 7:48
  • $\begingroup$ No, that is not allowed for standard concatenation of digits. $\endgroup$ Commented Oct 24, 2021 at 21:39
3
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.