6
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All I need is 56-59, 69, 73, 75- 77, 79, 86, 90-94, 99. I've done the rest but I would love to hear other solutions.

Rules:

  1. Use any of the following operations: basic operations (+ - x /), to the power of (^), Square root, factorial, double factorial, concatenation (e.g. 42 is acceptable).
  2. You can write the numbers in any order but you may only use each number once.

Good Luck!


In a comment, the OP said, “Repeating decimals are not allowed as an answer (e.g. 92.22222 ≠ 92). But you may use repeating decimals as part of an equation (don't know any case).”

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20
  • $\begingroup$ Is triple factorial allowed? $\endgroup$
    – WOWOW
    Feb 5 at 8:41
  • $\begingroup$ Triple factorial is not allowed $\endgroup$
    – bob
    Feb 5 at 8:42
  • $\begingroup$ Okay. I found an answer for 56. Should I post a partial answer? $\endgroup$
    – WOWOW
    Feb 5 at 8:42
  • $\begingroup$ What is a partial answer? Any answer is fine $\endgroup$
    – bob
    Feb 5 at 8:44
  • 1
    $\begingroup$ You cant make 41 unless you have something like 42 - 0! $\endgroup$
    – bob
    Feb 5 at 9:38

4 Answers 4

7
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Work in progress

$58 = (4+0!)! / 2 - 2$
$59 = ((4+0!)! - 2) / 2$
$86 = (42 + 0!) \times 2$
$94 = ((4 + 2)!! - 0! ) \times 2$

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5
  • $\begingroup$ Thanks, did you say you had one for 56? $\endgroup$
    – bob
    Feb 5 at 8:48
  • 1
    $\begingroup$ I said it's work in progress :). I hope I can find one for most of them $\endgroup$
    – Marius
    Feb 5 at 8:49
  • 1
    $\begingroup$ @bob that was me, not him $\endgroup$
    – WOWOW
    Feb 5 at 8:51
  • $\begingroup$ Have you found one yet? $\endgroup$
    – bob
    Feb 5 at 9:54
  • 2
    $\begingroup$ Nope. that's all I have. I quit. It's too hard. $\endgroup$
    – Marius
    Feb 5 at 11:54
5
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Partial Answer:

$56 = ((2 + 0!)!)!! + 2 \times 4$
$58 = ((2 + 0!)!)!! + 4!! + 2$ or $(4!!)^2 - (2 + 0!)!$
$69 = \frac{0!}{.\bar{2} - .2} + 4! $
$79 = (4!! + 0!)^2 - 2$ or $79 = (2 + 0!)^4 - 2$
$90 = 42 + ((2 + 0!)!)!!$
$99 = (4!! + 2)^2 - 0!$

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8
  • $\begingroup$ are you sure? I get 58 on this. $\endgroup$
    – Marius
    Feb 5 at 8:52
  • $\begingroup$ I got 56. There might be a mistake though. $\endgroup$
    – WOWOW
    Feb 5 at 8:54
  • $\begingroup$ This is correct, the answer is 56 $\endgroup$
    – bob
    Feb 5 at 8:58
  • 2
    $\begingroup$ @Topwizard29Thegamer 0 factorial is 1, 3 factorial is 6, 6 double factorial is 48. $\endgroup$
    – Sneftel
    Feb 5 at 22:00
  • 2
    $\begingroup$ Wow, congrats for getting 90! $\endgroup$
    – bob
    Feb 7 at 8:35
4
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Partial answer:

$\large 57=((2+0!)!)!!+\frac{\sqrt{4}}{.\bar2}$

$\large 69=((4!)-(0!)) \times \sqrt{ \frac{2}{.\bar2} }$

$\large 71=4! \times \sqrt{\frac{2}{.\bar2}} -0!$

$\large 73=4! \times \sqrt{\frac{2}{.\bar2}} +0!$

$\large 75=((4!)+(0!)) \times \sqrt{ \frac{2}{.\bar2} }$

$\large 76=2 \times (40-2)$

$\large 91=\frac{\sqrt{4}}{.\bar2-.2} + 0!$

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3
  • $\begingroup$ It's just 77 and 93 remaining, right? $\endgroup$
    – WOWOW
    Feb 7 at 22:10
  • $\begingroup$ @WOWOW I believe you are right. Hopefully someone will get them. $\endgroup$ Feb 7 at 22:16
  • $\begingroup$ I'm not good at math lol. I'm quite happy that I even got to write an answer. $\endgroup$
    – WOWOW
    Feb 7 at 22:17
3
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Someone mentioned only 77 and 93 were remaining.

Here is 77:

$77 = ((4!!)!!)*.2 + .2 + 0$
or
$77 = (((4!!)!!) + 2 - 0!) * .2$

And here is 93:

$93 = \frac{4!}{.\bar2} - (\frac{0!}{.2})!!$

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2
  • $\begingroup$ +1 I especially like your first solution for 77. Almost all other answers posted so far use 0 factorial. $\endgroup$ Feb 9 at 5:10
  • $\begingroup$ Congratulations on getting 93. $\endgroup$ Feb 12 at 5:05

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