You have the following false “equation”, made with six toothpicks arranged on a table in the following way:

XI = I

You have to re-establish the equality by changing the position of a single toothpick. Common-sense rules apply: you are not allowed to bend, break, or remove any toothpicks.

Re-establish the equality? As in I need an equals sign?

One could move the "I" in "XI" and make a != sign (in math, the = sign with a slash over it).

Or, just make 1/1 = 1.

Well, technically, I suppose the != (I can’t remember the ASCII for a not equals sign, if there even is one…) is an equality operator.

You could make a V = V (lopsided V’s: | = |/)

1/1 = 1 is a good solution, but not the only one

The final solution must be stil an equality: != is not a good answer…

Well, the one X would look funny. Kind of like my V’s.

I’m looking at it at 90 degrees and I can’t see anything yet. I think I need to have a few more beers, that seemed to make programming better today anyhow.

You can move the "I" in "XI" over to the other side and make it X = X.

I know!

Make // 1 = 1

Then it would be a comment, so the compiler wouldn’t look at it anyways.

Move a toothpick above the X so that

Sqrt of 1 = 1

That would be a funky sqrt sign, but it works.

Well we know XI is 11 in roman numerals. How about moving one of the ‘=’ to the right to make

XI – 11 ?

These are all good solutions. I like it.

it is already true

XI=I

X=1

OK, move the I pick from XI to the right hand side and get

X = II

this is obviously true. why, X = 10 in binary (msdn here or what?) is 2, which is the same as II in roman encoding.

I would go with either X = X, II = II or I/I = I

And there’s always:

/I != I

Where the ! and = overlap – you take one of the picks from the X and move it over the = sign

I have an IQ of like…uh 30? I’m like a total moron but even I know that…

II = II

Roman numerals.

oops text didnt come out right…lemme try it another way! haha X != 1 that should be more readable…

heck actually my typo is legal…. II = II by "changing the position" of one and tilting the other! That not a bend, break, or a remove…and I thik common sense states that changing the axis is not changing the position.

X means 10 in Roman numeral while + also means 10 in Chinese numeral

X is non equal to I

So leave the number X at the right side and cross the equality sign

the non equal sighn is when you cross the equal sign.

Hey Adi, some of your maths related stuff has been pretty funky, I remember the stuff about Godel numbers

Why not blog about Lax winning the Abel prize instead of silly puzzles (or both, if you feel that way inclined)?

Move the left I over the =, so that it reads:

X <> I (where <> is = with a dash).

Equation is correct then.

OMG. The solution was already here. Sorry ’bout that.

What about Wesner’s solution of sqrt(1) = 1?

(Move the one part of the X to make the sqrt symbol)

"it is already true

XI=I

X=1 "

How so? You lose a tootpick in the second line.

Roman numeral XI is 11 in base 10, and that does not equal 1.

I’ve got it!

X’ = 1

(x prime, the derivative of x, taken with respect to x, is one)

Almost as good as | = |

Precisely!

The prime symbol is too short for it to be a toothpick…

yes, i still insist on

X = II

(10) = (2)

Judging from the position of the toothpicks, it’s rather obvious that you can’t move the toothpicks that make up the equality sign or the right side one.

That means you could only move the three toothpicks that make up the lvalue. Unfortunatly, that X sign is rather difficult to alter (/| or | don’t really mean anything) so the only one left that could be moved is the vertical toothpick.

Since X = II or IX = I or -X = I don’t really make sense, I believe another answer (besides 1/1 = 1) is X = T. Thinking a bit outside the box, X and T could hold the same value and therefore be equal.

The actual answer, derivative, is less satisfying than the other creative alternatives suggested.

Anyway, how about square root of -1 is the imaginary number:

/-1 = I

Since X1 = 1, then X = 1, so X = 1 ^^ 1 (1 raised to the first power). Or you could raise X to the first power to get the same result.

Here’s another one, sort of:

1^x = 1

where the 1 is moved so that x becomes its exponent in superscript. The positioning is a little off, and you could argue about 1^.5 having multiple values, for example, but otherwise it works.

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