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.

  • 1
    Are we allowed to use powers? – Wais Kamal Sep 30 at 19:13
  • @WaisKamal Nope, only basic arithmetic operations and a square root. – TheSimpliFire Sep 30 at 19:14
  • 1
    Which bases can i use? Is it allowed to use the ternary system? – 12431234123412341234123 Oct 1 at 14:31
  • 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. – Apollys Oct 1 at 18:40
  • 1
    Yes, G=6, but g=9 :) So the number should be 600913. – Zizy Archer Oct 2 at 12:31
up vote 22 down vote accepted

How about

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

another one

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

last one

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

  • 1
    (+1) that was quick! I'll wait for more answers before ticking :) – TheSimpliFire Sep 30 at 18:54
  • The first one evaluates to 30, not 20. – Wais Kamal Sep 30 at 19:08
  • 2
    @WaisKamal I think it is still 20. You may want to check it again. – Oray Sep 30 at 19:10
  • Oh yeah, sorry for that :) – Wais Kamal Sep 30 at 19:11
  • 1
    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. – jpmc26 Oct 2 at 9:40

Here is a solution:

$\frac{60 + (0 × 6 × 1)}3 = 20$

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.

  • 8
    answers need to be of a certain length to be posted – casualcoder Oct 1 at 7:34
  • 1
    This is basically the same as JonMark Perry's (second) solution. – Jaap Scherphuis Oct 1 at 12:02

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)

  • Nice thinking, but you use both x and . – Ken Y-N Oct 2 at 4:20

How about:

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

Without a power:

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

  • 2
    "no factorial" it seems. – Oray Sep 30 at 19:03
  • @oray; thx, I'll go with 0*0 (I assume this is power?) – JonMark Perry Sep 30 at 19:04
  • 1
    I have no idea what * is :) – Oray Sep 30 at 19:05
  • 1
    Sorry, no powers! – TheSimpliFire Sep 30 at 19:06
  • 3
    Isn't $0^0$ not really well-defined? – LegionMammal978 Sep 30 at 19:21

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.

  • Even though this exact solution hasn't been mentioned, it is almost the same as xhienne's. – Jaap Scherphuis Oct 1 at 11:00

How about $\sqrt{600/6}\times\sqrt{1+3}$?

  • 2
    Nice. Note that @Oray has already got that one :P – TheSimpliFire Oct 1 at 6:21
  • 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. – Viktor Toth Oct 1 at 12:15

How about this:

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

edit new one without power:

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

  • 1
    Great answer, but sorry, no powers! :) – TheSimpliFire Oct 1 at 6:21
  • 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 – Naia Suzuki Oct 1 at 8:12

a very simple way to go would be-

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

protected by Community Oct 1 at 5:18

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