# Google's 20th Anniversary: Make 20 using 6 0 0 6 1 3

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.

• Are we allowed to use powers? Commented Sep 30, 2018 at 19:13
• @WaisKamal Nope, only basic arithmetic operations and a square root. Commented Sep 30, 2018 at 19:14
• Which bases can i use? Is it allowed to use the ternary system? Commented Oct 1, 2018 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. Commented Oct 1, 2018 at 18:40
• Yes, G=6, but g=9 :) So the number should be 600913. Commented Oct 2, 2018 at 12:31

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

another one

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

last one

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

• (+1) that was quick! I'll wait for more answers before ticking :) Commented Sep 30, 2018 at 18:54
• The first one evaluates to 30, not 20. Commented Sep 30, 2018 at 19:08
• @WaisKamal I think it is still 20. You may want to check it again.
– Oray
Commented Sep 30, 2018 at 19:10
• Oh yeah, sorry for that :) Commented Sep 30, 2018 at 19:11
• 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. Commented Oct 2, 2018 at 9:40

Here is a solution:

$$\frac{60 + (0 × 6 × 1)}3 = 20$$

-(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.

• answers need to be of a certain length to be posted Commented Oct 1, 2018 at 7:34
• This is basically the same as JonMark Perry's (second) solution. Commented Oct 1, 2018 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 ✕. Commented Oct 2, 2018 at 4:20

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

Without a power:

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

• "no factorial" it seems.
– Oray
Commented Sep 30, 2018 at 19:03
• @oray; thx, I'll go with 0*0 (I assume this is power?)
– JMP
Commented Sep 30, 2018 at 19:04
• I have no idea what * is :)
– Oray
Commented Sep 30, 2018 at 19:05
• Sorry, no powers! Commented Sep 30, 2018 at 19:06
• Isn't $0^0$ not really well-defined? Commented Sep 30, 2018 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. Commented Oct 1, 2018 at 11:00

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

• Nice. Note that @Oray has already got that one :P Commented Oct 1, 2018 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. Commented Oct 1, 2018 at 12:15

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

edit new one without power:

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

• Great answer, but sorry, no powers! :) Commented Oct 1, 2018 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 Commented Oct 1, 2018 at 8:12

a very simple way to go would be-

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

• No factorials :P Commented Oct 1, 2018 at 6:47
• oh, missed that, sorry. Commented Oct 1, 2018 at 6:48