2
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This question already has an answer here:

Part one: You are presented with ten incomplete equations:

  • 0 0 0 = 6

  • 1 1 1 = 6

  • 2 2 2 = 6

  • 3 3 3 = 6

  • 4 4 4 = 6

  • 5 5 5 = 6

  • 6 6 6 = 6

  • 7 7 7 = 6

  • 8 8 8 = 6

  • 9 9 9 = 6

Your job is to fill in mathematical operators in order to make all equations true.

  1. You can not connect two digits into one number. So for example you cannot connect two fives into fifty five. Each digit is it's own separate number.
  2. You cannot add in more numbers, including in operators. So squaring is invalid, because in order to square you need to add in the number two. A square root is allowed, but not any other root that requires additional numbers.
  3. A square root of a square root is a fourth root, which is also not allowed
  4. You cannot change anything but the left side of the equation
  5. Functions such as sine, tangent etc. Are not allowed.
  6. Using ^ is not accepted

Part two: using the same rules, can you fill in

  • n n n n = 6
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marked as duplicate by boboquack, Wen1now, Glorfindel, ABcDexter, Engineer Toast Oct 25 '17 at 12:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @boboquack I wasn't aware of this, but my version is a bit tweaked $\endgroup$ – Lavigo Oct 25 '17 at 8:16
  • $\begingroup$ @boboquack If it does count as a duplicate, do as you please $\endgroup$ – Lavigo Oct 25 '17 at 8:19
  • $\begingroup$ That comment was automatically generated by the system. My observation is that the only difference is part two - if you changed part two to be the whole question, it would be distinct enough, but at the moment most of the question is the same is the linked question. $\endgroup$ – boboquack Oct 25 '17 at 8:23
  • $\begingroup$ The rules are also different $\endgroup$ – Lavigo Oct 25 '17 at 8:40
  • $\begingroup$ All the solutions in the accepted answer work here. $\endgroup$ – boboquack Oct 25 '17 at 9:01
5
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Part One:

(0! + 0! + 0!)! = 6
(1 + 1 + 1)! = 6
2 + 2 + 2 = 6
3! + 3! - 3! = 6
4 + 4 - √4 = 6
5 + (5 / 5) = 6
6 + 6 - 6 = 6
7 - (7 / 7) = 6
(√(8 + (8 / 8)))! = 6
(√9)! + (√9)! - (√9)! = 6

Part Two:

((n!/n!)+(n!/n!))# = 6

# is being used as Primorial, which is the product of the first x prime numbers. In this case, the first two primes, which is 2*3=6

Alternatively:
((n! + n! + n!) / n!)! = 6

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  • $\begingroup$ A. I said a square root of a square root is not allowed B. What about part two? $\endgroup$ – Lavigo Oct 25 '17 at 8:13
  • $\begingroup$ Try typing √a (√a) instead of va or even $\sqrt a$ ($\sqrt a$). $\endgroup$ – boboquack Oct 25 '17 at 8:13
  • $\begingroup$ second equation requires a correction as either of the below shown ways: (1) (1 + 1 + 1) ! OR (2) (1! + 1! + 1!) ! but not the given one! $\endgroup$ – Mea Culpa Nay Oct 25 '17 at 8:15
  • $\begingroup$ Thanks for spotting my slight error with the 1s, Mea. What about now, @Lavigo ? $\endgroup$ – Timoris Oct 25 '17 at 8:39
  • 1
    $\begingroup$ It took me a while to think of it. The tricky part was trying to make an equation that would work for 0 as well as all other integers. I knew I'd have to use factorials, but then I also thought that might screw with larger numbers. Though now I think about it, I think ((n!+n!+n!)/n!)! would also work. $\endgroup$ – Timoris Oct 25 '17 at 8:57
3
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In continuation to earlier answer, here is one for part two:

((n+n+n)/n) ! = 6 for all n >0.

And if n = 0,

((n! + n! + n!))! * n! or ((n! + n! + n!))! / n!

And here is one more, perhaps

( log n (n * n *n ) )! that is log to the base n (subscript) of n cubed.

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  • $\begingroup$ The previous answer was wrong though. $\endgroup$ – Lavigo Oct 25 '17 at 8:20
  • $\begingroup$ Division by 0 alert! Please deal with the special case. $\endgroup$ – boboquack Oct 25 '17 at 8:20
  • $\begingroup$ Ok it was fixed now $\endgroup$ – Lavigo Oct 25 '17 at 8:21
  • $\begingroup$ The last one is the one I originally came up with. Well done. $\endgroup$ – Lavigo Oct 25 '17 at 9:01

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