# Making 103 from 4 zeroes

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!

• For arbitrary roots, can we pick any b we want, or must we substitute a result for b? – greenturtle3141 Mar 19 '17 at 18:25
• @greenturtle3141 If you use an arbitrary root, you must substitute a result. – noneuclideanisms Mar 19 '17 at 18:37
• 0 / 0! = 103, right? – Jodrell Mar 21 '17 at 12:15

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! $\large)$

• ^vote with a note: you can replace $\sqrt{0!%%%%}$ by simply $0!%%$. Nice answer! – Rand al'Thor Mar 19 '17 at 19:16
• Thus the cleanup begins, thank you @rand al'thor! – humn Mar 19 '17 at 19:17
• The humnification is spreading: first manshu, now me too. – Rand al'Thor Mar 19 '17 at 21:22
• (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!) – humn Mar 19 '17 at 21:27

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

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

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

• Alternatively, $\frac{0!}{0! \%} + 3$ – greenturtle3141 Mar 19 '17 at 18:38

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)

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

• Is that first term supposed to be 101? – Jacob Jones Mar 20 '17 at 7:55
• Also, isn't that too many zeroes? I would downvote, but then your score wouldn't be 0 any more. – Mr Lister Mar 20 '17 at 8:42
• @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 :) – L.K. Mar 20 '17 at 9:07
• @Jacob indeed (I wrote it before but I don't know where is my message) – L.K. Mar 20 '17 at 14:51

I think this is easier:

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

• % is an operation, you can't divide by it. – Mike Earnest Mar 20 '17 at 20:36