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Given these four equations/inequalities:

1x=2
2x=4
3x=3
4x>8

What is x?

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46
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The answer is:

The "<" symbol

Why?

In some programming languages, "<=" means "is less than or equal to", and "<>" means "is not equal to".

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    $\begingroup$ From here: "Just for trivia, <> was chosen in some languages because it can be interpreted as "less than or greater than", or in other words, not equal." (The longer you live ...) $\endgroup$ – tomd Jul 25 at 21:30
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    $\begingroup$ And <> is not necessarily the same as !=. Some objects it doesn't make sense to compare < or >. $\endgroup$ – qwr Jul 25 at 22:11
  • $\begingroup$ Precisely correct! $\endgroup$ – Ed Marty Jul 25 at 23:16
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    $\begingroup$ Hah. And here I though it was about x being another representation of >< in a smaller print. $\endgroup$ – ChatterOne Jul 27 at 10:31
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Answer: x is

a variable (as in programming)

That's because

the given "equations/inequalities" are in fact (some dialect of) BASIC code, so 1,2,3,4 at the beginning of the each line are just labels/line numbers. So, it's a perfectly valid program consisting of 3 assignments (lines 1-3) to the same variable x and 1 comparison in line 4 (the result of which is not assigned to any variable). Better formatting and comments:


 1 x = 2 'assign 2 to x (creating a new variable, declaration is not required in BASIC)
 2 x = 4 'assign 4 to x
 3 x = 3 'assign 3 to x
 4 x > 8 'compare x with 8, returning false (0)
 

P.S.

If x is required to be a number, my answer is x = 3 (this will be the value of x after running this code).

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    $\begingroup$ Very clever! Headed in the right direction, but not what I was going for. Also, your P.S. is not necessary, because rot13(k vf abg n ahzore) $\endgroup$ – Ed Marty Jul 25 at 7:55
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    $\begingroup$ rot13(vs vg jnf onfvp gurer jbhyq unir orra fcnprf orgjrra gur ynoryf/yvar ahzoref naq gur pbzznaqf) $\endgroup$ – melfnt Jul 25 at 10:56
4
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X could be operator (IS NOT)

!

Because

1x=2
2x=4
3x=3
4x>8

are always either true or false conditions in any programming language.

1!=2    //Always True
2!=4    //Always True
3!=3    //Always False
4!>8    //Always True
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    $\begingroup$ Huh? What language supports !> ? $\endgroup$ – Loren Pechtel Jul 28 at 0:14
  • $\begingroup$ @LorenPechtel You are right .. thank you for ponting out mistake. $\endgroup$ – Nikhil Sharma Jul 28 at 2:19
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    $\begingroup$ @LorenPechtel T-SQL $\endgroup$ – gszavae Jul 28 at 8:13
2
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x could be:

(π/2)!

where the

! is a not operator

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  • $\begingroup$ ! seems more like a factorial in the last context at least. $\endgroup$ – Ed Marty Jul 26 at 16:03
  • $\begingroup$ Seconded--! as negation precedes, not follows. Trailing it only means factorial. $\endgroup$ – Loren Pechtel Jul 28 at 0:15
  • $\begingroup$ @LorenPechtel In the context, the first line would be read 1(p/2)!=2 where "!=" means "not equals". $\endgroup$ – gszavae Jul 28 at 8:10

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