22
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Belated 20th Happy Birthday to Google (600613)!

Challenge: Use 6, 0, 0, 6, 1, 3 to make the number 20.

Rules:

  • Use all numbers, in that exact order.

  • Allowed operations: +, -, x, /, sqrt, (); no factorials, nth roots or powers

  • No concatenation of final result (e.g. 2 || 0) but concatenation allowed for original numbers

  • All numbers in base 10.

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  • 1
    $\begingroup$ Are we allowed to use powers? $\endgroup$ – Wais Kamal Sep 30 '18 at 19:13
  • $\begingroup$ @WaisKamal Nope, only basic arithmetic operations and a square root. $\endgroup$ – TheSimpliFire Sep 30 '18 at 19:14
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    $\begingroup$ Which bases can i use? Is it allowed to use the ternary system? $\endgroup$ – 12431234123412341234123 Oct 1 '18 at 14:31
  • $\begingroup$ I'm a fool! Haha. Okay I think the easiest is noticing that 20 is the square root of 400, this is the third solution in the accepted answer. $\endgroup$ – Apollys supports Monica Oct 1 '18 at 18:40
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    $\begingroup$ Yes, G=6, but g=9 :) So the number should be 600913. $\endgroup$ – Zizy Archer Oct 2 '18 at 12:31
23
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How about

$\frac{60}{0+6-1\times3}$

another one

$60\times(0\times6+\frac{1}{3})$

last one

$\sqrt{\frac{600}{6}\times(1+3)}$

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  • 1
    $\begingroup$ (+1) that was quick! I'll wait for more answers before ticking :) $\endgroup$ – TheSimpliFire Sep 30 '18 at 18:54
  • $\begingroup$ The first one evaluates to 30, not 20. $\endgroup$ – Wais Kamal Sep 30 '18 at 19:08
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    $\begingroup$ @WaisKamal I think it is still 20. You may want to check it again. $\endgroup$ – Oray Sep 30 '18 at 19:10
  • $\begingroup$ Oh yeah, sorry for that :) $\endgroup$ – Wais Kamal Sep 30 '18 at 19:11
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    $\begingroup$ This is one of the few types of problems where $\div$ actually looks better than the fraction; it makes the sequence of inputs more obvious. $\endgroup$ – jpmc26 Oct 2 '18 at 9:40
9
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Here is a solution:

$\frac{60 + (0 × 6 × 1)}3 = 20$

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8
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How about:

-(60-0)/6*(1-3)

I rolled back the MathJax edit as, although it looks prettier, I feel that the in-line solution is required to precisely meet the "in that exact order." requirement.

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  • 8
    $\begingroup$ answers need to be of a certain length to be posted $\endgroup$ – Nobody Oct 1 '18 at 7:34
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    $\begingroup$ This is basically the same as JonMark Perry's (second) solution. $\endgroup$ – Jaap Scherphuis Oct 1 '18 at 12:02
6
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Not quite an answer, but a kind of lateral (or literal) thinking one

$6\times0+\frac{x}{x}+0+6+13$ (assuming $x$ non-zero)

It's because

Allowed operations: +, -, x, /, sqrt, (); so, x is allowed (despite it's used as a variable, not operation; there is definitely a lowercase x, not a multiplication sign in the question text)

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  • $\begingroup$ Nice thinking, but you use both x and . $\endgroup$ – Ken Y-N Oct 2 '18 at 4:20
5
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How about:

$6-0^0+(6-1)\times3$.

Without a power:

$\frac{60+0}{6}\times(-1+3)$

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  • 2
    $\begingroup$ "no factorial" it seems. $\endgroup$ – Oray Sep 30 '18 at 19:03
  • $\begingroup$ @oray; thx, I'll go with 0*0 (I assume this is power?) $\endgroup$ – JMP Sep 30 '18 at 19:04
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    $\begingroup$ I have no idea what * is :) $\endgroup$ – Oray Sep 30 '18 at 19:05
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    $\begingroup$ Sorry, no powers! $\endgroup$ – TheSimpliFire Sep 30 '18 at 19:06
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    $\begingroup$ Isn't $0^0$ not really well-defined? $\endgroup$ – LegionMammal978 Sep 30 '18 at 19:21
5
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I wonder why this very obvious solution has not yet been mentioned:

(60+0×61)/3

It's only 11 characters long. I doubt you can make it any shorter.

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  • $\begingroup$ Even though this exact solution hasn't been mentioned, it is almost the same as xhienne's. $\endgroup$ – Jaap Scherphuis Oct 1 '18 at 11:00
4
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How about $\sqrt{600/6}\times\sqrt{1+3}$?

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  • 2
    $\begingroup$ Nice. Note that @Oray has already got that one :P $\endgroup$ – TheSimpliFire Oct 1 '18 at 6:21
  • $\begingroup$ Heh. I thought I checked them all before posting. Hmmm... wait, that's a subtly different solution (assuming we're looking at the same). He's taking the square root of 400, I am multiplying 10 and 2. $\endgroup$ – Viktor Toth Oct 1 '18 at 12:15
4
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How about this:

$-(6^0+0-6)(1+3)$

edit new one without power:

$\frac{\sqrt{600\times6}}{(1\times 3)}$

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  • 1
    $\begingroup$ Great answer, but sorry, no powers! :) $\endgroup$ – TheSimpliFire Oct 1 '18 at 6:21
  • $\begingroup$ ah I thought it meant you can't square the number as in you can't add another number. new one should do the job $\endgroup$ – Naia Suzuki Oct 1 '18 at 8:12
0
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a very simple way to go would be-

$6+0!+0+6+1+3!=20$

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  • 2
    $\begingroup$ No factorials :P $\endgroup$ – TheSimpliFire Oct 1 '18 at 6:47
  • $\begingroup$ oh, missed that, sorry. $\endgroup$ – Shahriar Mahmud Sajid Oct 1 '18 at 6:48

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