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You may use math signs you wish, but you must use the numbers

7,3,7,3

in an equation so it equals

24

Is it possible? If so, show how.

No-nos:

  • You are not allowed to combine numbers to make bigger numbers (e.g. you can't have 73)
  • You cannot use numbers more than once, you only have 2 7s and 2 3s
  • You have to use all the numbers given to you.

What is a math sign? Examples: +, -, *, /, (, )

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  • $\begingroup$ Do I have to use both numbers exactly twice, or what? $\endgroup$
    – No. 7892142
    Commented Feb 16, 2015 at 12:44
  • $\begingroup$ @No. 7892142 As expected, yes. $\endgroup$
    – warspyking
    Commented Feb 16, 2015 at 12:46
  • $\begingroup$ Should you just add characters only to the left of the equals sign? $\endgroup$
    – Olive Stemforn
    Commented Jun 26, 2015 at 5:38
  • $\begingroup$ Why are there so many down votes? I mean, I also see this question as a ridiculously simple no-effort quiz, but it's not often too see such a low-scored question recently. (I'm not down voting btw) $\endgroup$
    – William Nathanael
    Commented Jul 16, 2017 at 15:07
  • $\begingroup$ @ Wiliam Nathanael -- He did not define "math signs" and user jscs in comments far below claimed he changed the rules, which is not courteous to the solvers who were already working on the problem. $\endgroup$
    – Olive Stemforn
    Commented Nov 24 at 16:55

7 Answers 7

Reset to default
10
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(3/7+3)*7 =24
=> (3/7 + 21/7) * 7 =24
=> (24/7) * 7 =24
=> 24=24

Took me a while!

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5
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Answer

$$((7/7)+3)*(3!) = 4*(3!) = 24 $$

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3
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There's an implied $1$, but I didn't actually write it:

$7 \times 3 + 3 + \int_{\{7\}}dx$

And here's another solution (I think you can argue that $7$ is used twice):

$7 \times 3 - 3 = 24$ in base $7$

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  • 1
    $\begingroup$ I love the second one! $\endgroup$
    – William Nathanael
    Commented Jul 16, 2017 at 15:03
  • $\begingroup$ I love the second one! $\endgroup$
    – William Nathanael
    Commented Jul 16, 2017 at 15:03
2
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Using not all of them:

3*7 + 3 = 24

Using all of them:

$\lceil 7/3 \rceil + 3*7 = 24$

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11
  • $\begingroup$ "math signs" who said you could use ceil? $\endgroup$
    – warspyking
    Commented Feb 16, 2015 at 12:49
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    $\begingroup$ Who said I couldn't? "Math sign" isn't exactly a rigorous definition. $\endgroup$
    – No. 7892142
    Commented Feb 16, 2015 at 12:50
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    $\begingroup$ @warspyking "Examples" still does not exclude ceil(). $\endgroup$
    – No. 7892142
    Commented Feb 16, 2015 at 12:53
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    $\begingroup$ Changing the rules to disallow an answer after it's posted sucks, @warspyking; it's incredibly frustrating for someone who's trying to participate in your challenge. Futhermore, the ceiling operation does have what you call a "math sign": $\lceil x \rceil$. Please please review your challenges before you post them and deal with things like this then, rather than snubbing your answerers. $\endgroup$
    – jscs
    Commented Feb 16, 2015 at 20:11
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    $\begingroup$ "Math signs" has no standard definition, @warspyking -- there's nothing to be "not understood". It's up to you to make the rules clear before posting. And, as I showed you, $\lceil\rceil$ is a "math sign". $\endgroup$
    – jscs
    Commented Feb 16, 2015 at 23:58
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((7*7)^{1/2})*3+3
$\sqrt{7*7}*3+3$
$=(7)*3+3$
$=21+3$
$=24$

(equivalent to original answer but w/o using "1/2")
Of course it's unclear if a radical is covered by "math signs" ...

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0
0
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square root of 7*7 with is 7 * by 3 + 3

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1
  • $\begingroup$ Hi archie, Welcome to Puzzling SE. Please ensure that your answers make sense to immediate readers, and hide your answers using the spoiler tag >! Alsoif a question already has an accepted answer, be sure your answer adds something good to the puzzle. $\endgroup$
    – Spencerkatty
    Commented Jun 22, 2016 at 1:00
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I guess square root and factorial are allowed:

$$(\sqrt{7-3} +\sqrt{7-3})! = (2+2)! = 4$$

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  • $\begingroup$ You need to edit the right-hand side to be 24, not 4. $\endgroup$
    – Olive Stemforn
    Commented May 22, 2019 at 2:46

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