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Complete the equality to make it true:

1 1 1 1 = 5
  • You can add any math operation or symbol to the left side.
  • You cannot change right side nor the = (adding a previous < or > is considered changing it).
  • You can't add any number. Not even implicitly: i.e. √x is x0.5 so that would be adding numbers.
  • You can't combine the 1s (not for integers like 11 and not for decimals like 1.1).
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4 Answers 4

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One possible answer:

$(1+1+1)!-1 = 5$

because

$ (1+1+1)!-1= 3!-1 = 6-1 = 5$

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    $\begingroup$ I read that ! as not... but it's not D: $\endgroup$
    – n_plum
    Jun 16, 2017 at 21:28
  • $\begingroup$ Is this a common notation? $\endgroup$
    – greenturtle3141
    Jun 16, 2017 at 21:31
  • $\begingroup$ @greenturtle3141 Fixed. $\endgroup$
    – MikeQ
    Jun 16, 2017 at 21:55
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How about:

$( 1 / .1 ) / ( 1 + 1 ) = 5$

And with only three $1$s on the left-hand side:

$1 / (.1 + .1) = 5$

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  • $\begingroup$ You can't add any number. Not even implicit. I think this implies a 0 before the dot. $\endgroup$
    – Roman Gräf
    Jun 17, 2017 at 13:36
  • $\begingroup$ @RomanGräf Don't see how that implies a 0, but if you're going to stretch it that far you could say factorials implicitly imply $1..n - 1$ $\endgroup$
    – Paul Evans
    Jun 17, 2017 at 13:45
  • $\begingroup$ i would say that you normally write $0.1$ instead of $.1$. Most calculators mal let you write it as $.1$ but any math teacher would give you a bad mark on that. (Sorry for my english) $\endgroup$
    – Roman Gräf
    Jun 17, 2017 at 13:47
  • $\begingroup$ @RomanGräf saying any math teacher is quite a stretch as well! $\endgroup$
    – Paul Evans
    Jun 17, 2017 at 13:50
  • $\begingroup$ You can't tell me your math teacher, (if you have any, if not take the last math teacher that you had) would let you write $.1$ for $0.1$? $\endgroup$
    – Roman Gräf
    Jun 17, 2017 at 13:51
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Using the set of operators found in typical programming languages (so no factorial, square root, etc.), it's possible to do this by inserting just one operator in each gap, plus parentheses to control precedence:

(1 << (1 + 1)) + 1 = 5

And here's a link verifying the result of the calculation using an actual programming language.

This makes use of the following operation:

The "left shift" operation << multiplies its left argument by 2 to the power of its right argument. In this case, we're calculating $1 \times 2^{1+1}$, i.e. 4, then adding 1.

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  • $\begingroup$ Clever. But that is not a math operation, it's a bitwise operator. $\endgroup$
    – Rubio ♦
    Jun 17, 2017 at 5:04
  • $\begingroup$ I think it's a matter of semantics whether bitwise operations are maths, right? They're certainly mathematical in the broader sense (and they've proven to be more useful than, say, factorials in practice). $\endgroup$
    – ais523
    Jun 17, 2017 at 8:01
  • $\begingroup$ Although I'm not sure if it's a math operation I was hoping this solution to come up. I really like it! $\endgroup$
    – Diego
    Jun 17, 2017 at 16:46
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Possibility:

$1\times\left((1+1)^2 +1\right)= 5$

Because:

$\begin{align}1 \times (2^2 + 1) &= 5\\ 1 \times (4 + 1) &= 5\\ 1 \times 5 &= 5\\ 5 &= 5 \end{align}$

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  • $\begingroup$ Can you square things since you can square root them? Or does that count as adding a number $\endgroup$
    – n_plum
    Jun 16, 2017 at 21:36
  • $\begingroup$ Actually it does count as adding a number $\endgroup$
    – Diego
    Jun 16, 2017 at 22:08
  • $\begingroup$ Sadface (btw you may want to show where you got this from, it's not original) $\endgroup$
    – n_plum
    Jun 16, 2017 at 22:09
  • $\begingroup$ I haven't thought of that, but you're right. It was shown to me by some guy, I'll try to track de original source of it. Thanks! $\endgroup$
    – Diego
    Jun 19, 2017 at 12:43

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