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This puzzle is all about making 4 with 4 ones, but with certain constraints.

Allowed operations:

  • $-$ Subtraction
  • $-a$ Negation
  • $\times$ Multiplication
  • $\div$ Division
  • $\sqrt{a}$ Square Root
  • $\sqrt[b]{a}$ Arbitrary Roots
  • $!$ Factorial
  • $.\!a$ Decimal
  • $.\!\overline{a}$ Recurring decimal

Note: Arbitrary roots must use 1s.

Easy: Give me four examples of making 4 with 4 ones without addition.

Medium: Give me two examples of making 4 with 4 ones without addition, negation, or factorial.

Hard: Give me two examples of making 4 with 4 ones without addition and factorial, using only one subtraction and one negation.

(In each of the above cases, do not just circumvent the ban on addition by doing $a -- b$.)

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  • $\begingroup$ Request for clarification (see my answer to the "hard" question and comments thereunder): 1. Does the "Do not just circumvent this ..." apply to the Medium and Hard questions or only to the Easy one? (I suppose it applies vacuously to the Medium one.) 2.Is it intended to forbid lightly disguised circumventions where $-b$ is, say, multiplied by 1 before subtracting? $\endgroup$
    – Gareth McCaughan ♦
    Commented Mar 22, 2017 at 0:12
  • $\begingroup$ @GarethMcCaughan Yes to both. $\endgroup$
    – noneuclideanisms
    Commented Mar 22, 2017 at 0:14
  • $\begingroup$ Damn. I'll add a note to my answer saying it's invalid according to the intended rules, then. $\endgroup$
    – Gareth McCaughan ♦
    Commented Mar 22, 2017 at 0:15
  • $\begingroup$ Further request for clarification: in the "hard" question, do you mean (1) exactly one sub and one neg between the two examples, or (2) exactly one of each in each example, or (1') at most one of each between the two examples, or (2') at most one of each in each example? $\endgroup$
    – Gareth McCaughan ♦
    Commented Mar 22, 2017 at 0:18
  • 1
    $\begingroup$ @randal'thor Partly because I feel reluctant to delete something for being "invalid on a technicality", partly because I referred to it in comments above (though only as motivation), partly because I prefer to leave my mistakes on public view :-). I guess deletion probably is best, but it feels kinda weird. $\endgroup$
    – Gareth McCaughan ♦
    Commented Mar 22, 2017 at 0:27

7 Answers 7

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Give me four examples of making 4 with 4 ones without addition.

  1. $(1\div0.1)-(\sqrt{1\div0.\bar{1}})!=10-(\sqrt{9})!=10-6=4$

  2. $-(-\sqrt{1\div0.\bar{11}}-1)=-(-\sqrt{9}-1)=-(-4)=4$

  3. $(\sqrt{1\div0.\bar{1}})!\times(1-\sqrt{0.\bar{1}})=(\sqrt{9})!\times(1-\tfrac{1}{3})=6\times\tfrac{2}{3}=4$ (thanks @elias)

  4. $(\sqrt{1\div0.\bar{1}})!-1-1=(\sqrt{9})!-2=6-2=4$ (thanks @elias)

Give me two examples of making 4 with 4 ones without addition, negation, or factorial.

  1. $(1-0.\bar{1})\div(\sqrt{0.\bar{1}}-0.\bar{1})=\tfrac{8}{9}\div\tfrac{2}{9}=4$

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  • $\begingroup$ I used one of yours. I hope thats ok! $\endgroup$
    – user35295
    Commented Mar 22, 2017 at 1:27
  • $\begingroup$ @AllanCao Of course! $\endgroup$
    – Rand al'Thor
    Commented Mar 22, 2017 at 1:35
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+100
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A concept for the last question,

$(1\div1\%)+\sqrt[{\sqrt{0.\bar{1}}}](1\div\sqrt{0.\bar{1}})! = 100 + 6^3= 316$

I know I used 5 ones. Can't think of a way to reduce it yet.

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  • 1
    $\begingroup$ You are actually correct, because you can use arbitrary index radicals. You don't need to use a 1 for it. $\endgroup$
    – Marius
    Commented Mar 24, 2017 at 8:09
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    $\begingroup$ Awarded the bounty to you since you've come the closest to Medium. $\endgroup$
    – noneuclideanisms
    Commented Mar 30, 2017 at 23:40
  • $\begingroup$ @Marius whoa, what? Are you saying (and do we have the OP's sanction for saying) that we can put anything in the index of a radical (1) even if it isn't made out of 1s and (2) no matter how many 1s it uses? Surely that can't be the intent. $\endgroup$
    – Gareth McCaughan ♦
    Commented Apr 3, 2017 at 13:48
  • $\begingroup$ I was saying that but I got corrected by the op. I was wrong $\endgroup$
    – Marius
    Commented Apr 3, 2017 at 16:08
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Easy no.1.:

$\Big(\sqrt{\frac{1}{.\overline{1}}}\Big)!\times\Big(1-\sqrt{.\overline{1}}\Big)=\Big(\sqrt{\frac{1}{\frac19}}\Big)!\times\Big(1-\sqrt{\frac19}\Big)=\big(\sqrt{9}\big)!\times\Big(1-\frac13\Big)=3!\times\frac23=6\times\frac23=4$

Easy no.2.:

$\Big(\sqrt{\frac{1}{.\overline{1}}}\Big)!-1-1=\Big(\sqrt{\frac{1}{\frac19}}\Big)!-2=\big(\sqrt{9}\big)!-2=3!-2=6-2=4$

An idea towards medium no.1.:

Using the idea
$\sqrt[\frac32]{8}=4$
, but I managed to get only
$1/.\overline{1}-1=8$
and
$1/(1-\sqrt{.\overline{1}})=3/2$
, which contain way too many 1s.

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    $\begingroup$ I incorporated these (crediting you) along with another two in my answer - hope you don't mind! $\endgroup$
    – Rand al'Thor
    Commented Mar 22, 2017 at 0:06
  • $\begingroup$ Please feel free to do so anytime. $\endgroup$
    – elias
    Commented Mar 22, 2017 at 0:07
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Give me four examples of making 4 with 4 ones without addition.

  1. $(1\div0.1)-(\sqrt{1\div0.\bar{1}})!=10-(\sqrt{9})!=10-6=4$ (thanks @rand al'thor)

  2. $(-1-1)\times(-1-1)=4$

  3. $(\sqrt{1\div0.\bar{1}})!\times(1-\sqrt{0.\bar{1}})=(\sqrt{9})!\times(1-\tfrac{1}{3})=6\times\tfrac{2}{3}=4$ (thanks @elias)

  4. $(\sqrt{1\div0.\bar{1}})!-1-1=(\sqrt{9})!-2=6-2=4$ (thanks @elias)

Give me two examples of making 4 with 4 ones without addition, negation, or factorial.

  1. noneuclideanisms caught that this one doesn't work

  2. $(1-0.\bar{1})\div(\sqrt{0.\bar{1}}-0.\bar{1})=\tfrac{8}{9}\div\tfrac{2}{9}=4$ (Thanks @rand al'thor)

Give me two examples of making 4 with 4 ones without addition and factorial, using only one subtraction and one negation.

  1. Work in progress

Make 316 with 4 ones.

  1. Work in progress

Work in progress!

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2
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Extra question to make 316 from four ones:

$$(\sqrt{.1})/(.1*.1*.1)$$

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    $\begingroup$ Nice solution, but I need an exact result. $\endgroup$
    – noneuclideanisms
    Commented Mar 22, 2017 at 21:43
  • $\begingroup$ Exact result? It gets a decimal of around 316 $\endgroup$
    – user35295
    Commented Mar 23, 2017 at 2:54
  • $\begingroup$ Maybe I'm stupid, but I get 243 from this? $\endgroup$
    – Gintas K
    Commented Mar 23, 2017 at 10:00
  • $\begingroup$ The answer is exactly 316 with no decimal. $\endgroup$
    – noneuclideanisms
    Commented Mar 23, 2017 at 23:14
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Easy 4 with 4 ones with out addition

$(-1-1) \times (-1-1)$
$\sqrt{\frac{1}{.\bar1}}! - 1 - 1$
$\sqrt{\frac{1}{.\bar1}}! \times (1- \sqrt{.\bar1}) $
$\frac{-1}{-.1-.1} - 1$

Medium: 4 with 4 ones without addition, negation, or factorial.

$\sqrt[\frac{3}{2}]{\frac{1}{.1}-1 - 1}$
$\sqrt[\frac{3}{2}]{\frac{1}{.\bar{1}}-1} \times 1$

Hard: 4 with 4 ones without addition and factorial, using only one subtraction and one negation.

$\sqrt[\frac{1}{2}]{-1-1} \times 1 \times 1$ (same as the easy one)
$\sqrt[\frac{1}{2}]{-1-1} \times 1 \div 1$ (I guess easy is the new hard)

Legend 316 with 4 ones.

$\sqrt[\frac{1}{3}]{(\sqrt{1\div.\bar{1}})!} + 1 \div 1\% = \sqrt[\frac{1}{3}]{(\sqrt{9})!} + 100 = 6^3 + 100 = 316$

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    $\begingroup$ To my understanding you're not allowed to do $\sqrt[b]{a}$ with arbitrary $b$, you have to produce them from ones as well. $\endgroup$
    – elias
    Commented Mar 24, 2017 at 9:22
  • $\begingroup$ @elias. I'm not challenging this, but can you point me where this says so? I think I missed it. $\endgroup$
    – Marius
    Commented Mar 24, 2017 at 9:30
  • $\begingroup$ I can't see it written, but if it was allowed, any $x\gt0,x\ne1$ could be produced from a single 1 as $\sqrt[y]{.1}$ with a well-chosen $y=-\frac{1}{log_{10}{x}}$, which would make the question uninteresting. $\endgroup$
    – elias
    Commented Mar 24, 2017 at 13:42
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    $\begingroup$ Elias is correct, arbitrary roots use up 1s. $\endgroup$
    – noneuclideanisms
    Commented Mar 24, 2017 at 14:48
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Eas  y   ier:   Make $4$ with  4  3 ones without addition.

$\require{begingroup}\begingroup \displaystyle\kern4em{}\def \@ #1{\sqrt{ #1 \small\raise2.9ex\, }}{} { -1 - \@ { .\overline1 } \, \over -\@ { .\overline1 } }{} ~~ = ~~ { -1 - \sqrt{ \tfrac 19 } \, \over -\sqrt{ \tfrac 19 } }{} ~~ = ~~ { -1 - \tfrac 13 \, \over -\tfrac 13 }{} ~~ = ~~ { -\tfrac43 \, \over -\tfrac13 \, }{}~~\equiv~~ ~ 4{}\endgroup$

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