0
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$6 6 6 6 = 58$

$+ - * /$ and $()$ only

58 must remain as 58, (not 5 + 8, etc.)

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  • 1
    $\begingroup$ Why it's getting downvoted ? $\endgroup$ – user27395 Oct 1 '16 at 13:53
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    $\begingroup$ Are we allowed to turn the sixes upside-down? If so, we have this curious identity that uses 5 sixes: 696/(6+6) = 58 $\endgroup$ – The Turtle Oct 1 '16 at 15:52
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    $\begingroup$ @TheTurtle: If we are using 5 sixes, without turning any upside down we can do [(6+6)/6]^6-6 = 58. $\endgroup$ – Xenocacia Oct 1 '16 at 17:27
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    $\begingroup$ It's definitely an impossible task without some lateral thinking. $\endgroup$ – greenturtle3141 Oct 1 '16 at 17:30
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    $\begingroup$ It's not possible. Not even with decimal points and exponentiation. So this is "guess how I'm cheating". $\endgroup$ – Rosie F Oct 1 '16 at 17:37

10 Answers 10

15
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$66 - 6 - 6 = 58$
(at least in the base-14 system).

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    $\begingroup$ Also 6*(6+6)+6 in same base and 6*6*6+6 in another nice base. And 6*(6+6+6) in 20. $\endgroup$ – ypercubeᵀᴹ Oct 1 '16 at 22:33
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Lateral thinking solution #507:

$66-6-6+4=58$
The $4$ is a $/$ over a $+$.

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9
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Here's a fun possibility:

Adding just '/': $6666 \ne 58$

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    $\begingroup$ Upvoting, but technically the op explicitly stated he wants the sixes to equal 58. $\endgroup$ – greenturtle3141 Oct 1 '16 at 17:28
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    $\begingroup$ @greenturtle3141 Why upvote if it doesn't answer the question? $\endgroup$ – Rand al'Thor Oct 1 '16 at 17:54
8
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My guess is that the desired answer is:

6()-(6+6)/6 = 58 meaning 60 - 12/6 = 58

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    $\begingroup$ I interpret those empty brackets as them containing a zero, since, by definition, zero is nothing. $\endgroup$ – EKons Oct 1 '16 at 19:00
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    $\begingroup$ He's using the two brackets as either side of a 0 not as actual brackets. $\endgroup$ – gtwebb Oct 1 '16 at 19:34
5
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Maybe

$-(-6-(6*6)-16)=58$ ; where the $1$ is a sideways minus sign

Explanation

$-(-6-36-16) = -(-42-16) = -(-58) = 58$

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  • $\begingroup$ You seem to be using the $1$ as a digit, not as you specified. $\endgroup$ – EKons Oct 1 '16 at 19:02
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    $\begingroup$ @ΈρικΚωνσταντόπουλος The 1 is not a 1, it is a minus sign rotated 90 degrees. $\endgroup$ – wizzwizz4 Oct 1 '16 at 20:32
  • $\begingroup$ @wizzwizz4 Oh, I was confused. So, this is a lateral-thinking answer, huh? $\endgroup$ – EKons Oct 2 '16 at 6:24
4
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You can run my code here which takes as input the number of test cases on one line, and then for each test case there are four lines, each the value of one of $a,b,c,d$ which can have any of the operations occur on them as long as the result is an integer. It prints out all of the resulting possibilities. It was marked correct when submitted to this online judge. This is not a full answer, but I believe this means that there is no valid way to do this without lateral-thinking, for which there is no tag.

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    $\begingroup$ I think you could improve this by removing the emphasis on your code and instead describing what it does and what the result means. The list of integers returned are all the possible integer solutions for the problem, and the fact that the list does not include the number 58 when given 6 6 6 6 as input shows that this is impossible without lateral-thinking. I would like to upvote this, but at the moment it is not an answer so I cannot. $\endgroup$ – wizzwizz4 Oct 1 '16 at 20:30
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    $\begingroup$ Agreed: as it stands, this would be better off as a comment than an answer. $\endgroup$ – Rand al'Thor Oct 1 '16 at 21:47
3
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The closest I can get is

(6**6)/(6!)-6 = 58.8
it needs to be cast to an integer

But yes, factorials and rounding are not allowed.

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2
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Probably invalid, but a creative attempt:

$(66-66)_1=58$

Note that the number is: $(0000000000000000000000000000000000000000000000000000000000)_1$
Using leading zeroes...

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  • $\begingroup$ I would edit out the "please don't downvote" if I were you; that will prime users to downvote your answer. I actually think it's pretty good, but that's just my opinion. $\endgroup$ – wizzwizz4 Oct 1 '16 at 20:27
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    $\begingroup$ I downvoted because this doesn't work in any level; 0 is not equal to 58 in any base and where did 1 come from? $\endgroup$ – ffao Oct 1 '16 at 21:20
  • $\begingroup$ @ffao came from lateral thinking? $\endgroup$ – Evan Carslake Oct 2 '16 at 0:03
  • $\begingroup$ @ffao 1 came from unary. $\endgroup$ – EKons Oct 2 '16 at 6:25
  • $\begingroup$ @EvanCarslake In base 1, there is only 1 digit, $0$. Since $0_1$ is $0_{10}$, and leading zeroes are ignored though, we can assume that, in a sense, this is invalid (50% of people think it's valid, 50% think it's invalid.) $\endgroup$ – EKons Oct 3 '16 at 12:41
1
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I'm going for:

$\dfrac{-6-6-6}{6}=5-8$ and $\dfrac{6+6+6}{6}=-5+8$

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    $\begingroup$ According to the updated question, this answer is invalid. $\endgroup$ – Rand al'Thor Oct 1 '16 at 17:54
1
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How about:

$6+6+(6/6)=5+8$

This works because

$6 + 6 = 12$ and $6 / 6 = 1$, so $12 + 1 = 13 = 5 + 8$.

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    $\begingroup$ According to the updated question, this answer is invalid. $\endgroup$ – Rand al'Thor Oct 1 '16 at 17:54
  • $\begingroup$ @randal'thor, I'll leave it here as an example $\endgroup$ – Beastly Gerbil Oct 1 '16 at 18:02

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