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As the title suggests, the puzzle here is making the number 103 by using exactly four 0s and the following operations:

  • $+$ Addition
  • $-$ Subtraction
  • $\times$ Multiplication
  • $\div$ Division
  • $\sqrt{a}$ Square Root
  • $\sqrt[b]{a}$ Arbitrary Roots
  • $!$ Factorial
  • $\%$ Percent (e.g. $4\% = 0.04$)

The solution is slightly difficult, but it shouldn't be too hard for you crazy puzzlers. Have fun!

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  • $\begingroup$ For arbitrary roots, can we pick any b we want, or must we substitute a result for b? $\endgroup$
    – greenturtle3141
    Mar 19, 2017 at 18:25
  • $\begingroup$ @greenturtle3141 If you use an arbitrary root, you must substitute a result. $\endgroup$
    – noneuclideanisms
    Mar 19, 2017 at 18:37
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    $\begingroup$ 0 / 0! = 103, right? $\endgroup$
    – Jodrell
    Mar 21, 2017 at 12:15

5 Answers 5

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As rand al’thor points out in the solution built upon here, there must be a way to formulate a $\small 3$ with only two $\small 0 \kern1mu$s.   How promising that...

             ...$~~ 3 = \sqrt{10-1 \, \small\raise.3ex\strut} ~$...

                        ...uses only two $\small 1$s.

And how convenient that $\small \surd$ and $ \small \% $ can combine to scale $\small 1 $ by any positive power of $ \small 0.1 \,$.   So (with ever fewer operations, at 18  15  14) ...

$\require{begingroup}\begingroup{} \def \@ #1{{ \sqrt { #1 \, \scriptsize\raise.4ex\strut } }}{} \def \sp {{ \kern 3mu \% }}{} \def \fpp {{ \small \kern 2mu \% \kern1mu \% }}{} \def \fp {{ \small \kern 2mu \% \! \raise.4ex\strut }}{} \def \p {{ \scriptsize \kern 1mu \% }}{} \def \E { ~~ \equiv ~~ }{} \def \= { ~~ = ~~ }{} \def \- { \, - \, }{} \kern 5em \small\begin{align}{}\normalsize { 0! + \@{\@ { 0!\fp }\sp \- 0! \fpp } \over 0!\fp }{} &\= { 1 + \@{ \@ { 1\p \! } \sp \- 1 \p\p } \over 1 \p }{}\\[3ex] &\= { 1 + \@{ \@{ .01 \! } \sp \- 1 \p\p } \over 1 \p }{}\\[2ex] &\= { 1 + \@{ .1 \p \- 1 \p\p } \over 1 \p }{}\\[2ex] &\= { 1 + \@{ .001 - .0001 } \over .01 }{}\\[2ex] &\= { 1 + \@{ .0009 } \over .01 }{}\\[2ex] &\= { 1 + .03 \over .01 }{}\\[2ex] &\E \normalsize 103{}\end{align}\endgroup$


Initial solution.   More operations but also more symmetry.

$\begingroup \displaystyle \kern6em{} \def \@ #1{{ \sqrt { #1 \, } }}{} \def \fp {{ \small \kern 2mu \% \raise.4ex\strut }}{} \def \p {{ \small \kern 1mu \% }}{} { 0! + \@{ \@{ 0!\fp\p\p } \, - \, \@{ 0!\fp\p\p\p } } \over 0!\fp }{}\endgroup$


Backpuzzle.   How about $\boldsymbol{.4}$ with only 3 (three) $\small 0 \kern1mu$s and just 11 operations?

$\begingroup{} \def \@ #1{{ \sqrt { #1 \scriptsize\raise.4ex\strut } }}{} \def \fp {{ \small \kern 2mu \% \scriptsize\raise.9ex\strut }}{} \def \p {{ \scriptsize \kern 1mu \% }}{} \kern 5em \small\begin{align}{}\normalsize \sqrt { 0!\fp } + \sqrt {\sqrt{ 0!\fp } - 0!\fp \, }{} & ~~ = ~~\@ { 1\p } + \@ { \@ { 1\p } - 1\p \, }{}\\[2ex] & ~~ = ~~ \@ { .01 } + \@ { \@ { .01 } - .01 \, }{}\\[2ex] & ~~ = ~~ \@ { .01 } + \@ { .1 - .01 \, }{}\\[2ex] & ~~ = ~~ \@{ .01 } + \@ { .09 }{}\\[2ex] & ~~ = ~~ .1 + .3{}\\[2ex] & ~~\equiv ~~ \normalsize .4{}\end{align}\endgroup$

$\large($And a ${0! \over 0!\%\%\%}$ more thanks to rand al’thor for fixing up $\scriptsize\sqrt{\%\%\%\%\,\tiny\raise1ex\strut}$ from the initial solution! $\large)$

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    $\begingroup$ ^vote with a note: you can replace $\sqrt{0!%%%%}$ by simply $0!%%$. Nice answer! $\endgroup$
    – Rand al'Thor
    Mar 19, 2017 at 19:16
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    $\begingroup$ Thus the cleanup begins, thank you @rand al'thor! $\endgroup$
    – humn
    Mar 19, 2017 at 19:17
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    $\begingroup$ The humnification is spreading: first manshu, now me too. $\endgroup$
    – Rand al'Thor
    Mar 19, 2017 at 21:22
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    $\begingroup$ (Pps., @rand al'thor, the direction is humn $\to$ manshumn $\to$ manshu. See you$\llap{\raise1ex{\underline{~~~~~}}}$ m$\,$e$~~\llap{\raise1ex{\underline{\,}}}$anshu there once all is complete!) $\endgroup$
    – humn
    Mar 19, 2017 at 21:27
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My intended answer:

$103 = \frac{\sqrt{\sqrt{0!\%}\,- \,0!\%}\,\% + \sqrt{0!\%}}{\sqrt{0!\%}\,\%}$

(The strategy used isn't too far off from humn's answer)

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Incomplete answer

I can do it using five zeroes:

$\Big(0!+(0!)\%+(0!)\%+(0!)\%\Big)\div(0!)\% = 103.$

In order to reduce this to four, all we need to do is

find a way to construct the number $3$ using only two zeroes,

because then the final solution will be

$(0!+3\%)\div(0!)\% = 103.$

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    $\begingroup$ Alternatively, $\frac{0!}{0! \%} + 3$ $\endgroup$
    – greenturtle3141
    Mar 19, 2017 at 18:38
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Will this work(I know it may be a cheating, but thought of giving this). Most important or unimportant it is with five zeros (if in case op changes his/her mind to have with 5 zeros, then I am ready with my solution after @Rand)

$$(0!)0(0!)+0!+0!=103$$

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  • $\begingroup$ Is that first term supposed to be 101? $\endgroup$
    – Jacob Jones
    Mar 20, 2017 at 7:55
  • $\begingroup$ Also, isn't that too many zeroes? I would downvote, but then your score wouldn't be 0 any more. $\endgroup$
    – Mr Lister
    Mar 20, 2017 at 8:42
  • $\begingroup$ @MrLister I wrote but I don't know where I missed the message. By the way, I wrote there are actually 5 zeroes instead of 4. I wrote this when I was going to sleep.(you need to have 125 rep to downvote :) $\endgroup$
    – L.K.
    Mar 20, 2017 at 9:07
  • $\begingroup$ @Jacob indeed (I wrote it before but I don't know where is my message) $\endgroup$
    – L.K.
    Mar 20, 2017 at 14:51
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I think this is easier:

103 = 0! + 0! + 0! + 0!/%

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    $\begingroup$ % is an operation, you can't divide by it. $\endgroup$
    – Mike Earnest
    Mar 20, 2017 at 20:36

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