A very interesting problem.

Question

Complete the following table of mathematical operations, $$\begin{align*}2+2+2&=6\\ 3 \ \quad 3\ \quad 3&=6 \\ 4 \ \quad 4\ \quad 4&=6 \\ 5 \ \quad 5\ \quad 5&=6 \\ 6 \ \quad 6 \ \quad 6&=6 \\ 7 \ \quad 7\ \quad 7&=6 \\ 8 \ \quad 8\ \quad 8&=6 \\ 9 \ \quad 9\ \quad 9&=6\end{align*} $$

I know that in the second there is a $ \cdot $ and then a $ - $, and it is true that $ 3 \cdot 3-3 = 6 $, but in the others I do not know another complex mathematical operation that helps me complete that table, someone help me.

I suggest frequent use of square roots and possibly factorials. — RobertTheTutor, Apr 16 '21 at 20:24
3*3-3 = 6, sqrt(4)+sqrt(4)+sqrt(4) = 6, 7-7/7 = 6, 9 - 9 /sqrt(9) = 6 — A.A., Apr 16 '21 at 20:25
I define the mathematical operation $w$ which maps any ordered pair onto $6$. $ 4 w 4 w 4 = 6 w 4 = 6$. Done. You're welcome. — B. Goddard, Apr 16 '21 at 20:26
Hmm, this reminds me of the "24 game" we played back in school, where you draw 4 random cards and try to make them = 24. Pretty fun! I wonder if there's a way to do 4, 8, and 9 without any roots though... — Y. S., Apr 16 '21 at 20:37
The title of this question is utterly useless for anyone wanting to search on MSE - and (IMHO) false. — Rob Arthan, Apr 16 '21 at 22:14
https://www.cut-the-knot.org/arithmetic/funny/6by3digits.shtml uses the solution proposed by @BStar below in his comment for 8 (and adds (0! + 0! + 0!)! = (1! + 1! + 1!)! = 6) — mau, Apr 17 '21 at 07:50

Star Bright · Answer 1 · 2021-04-18T18:32:51.930

3

$$\begin{align*}2+2+2&=6\\ \sqrt 3 \cdot \sqrt 3\ +3&=6 \\ \sqrt4 +\sqrt 4\ +\sqrt 4&=6 \\ 5\div 5+ 5&=6 \\ 6 \cdot 6\div 6&=6 \\ -7\div{7}+7&=6 \\ \sqrt[3]8 +\sqrt[3]8 +\sqrt[3]8 &=6 \\ \sqrt 9 \cdot\sqrt 9\ -\sqrt 9&=6\end{align*} $$

edited Apr 18 '21 at 18:32

answered Apr 16 '21 at 20:32

Star Bright

2,338

cube root kinda seems like cheating imo – Rushabh Mehta Apr 16 '21 at 20:42
How about 8 – √(√(8 + 8)) = 6? We can argue either way for this - it is square root of square root and fourth root at the same time:-). – Star Bright Apr 16 '21 at 21:17
The last line in your answer is not right. $\sqrt9 + \sqrt9 + \sqrt9$ is 9 not 6 – A.A. Apr 18 '21 at 18:09
@Alaleh Really - no one noticed till now:-). Thanks! – Star Bright Apr 18 '21 at 18:33

score 2 · Answer 2 · answered Apr 16 '21 at 20:32

2

$$ 2 + 2 + 2 = 6 $$

$$ 3 \times 3 - 3 = 6 $$

$$ \sqrt4 + \sqrt4 + \sqrt4 = 6 $$

$$ 5\div5 + 5 = 6 $$

$$ 6 + 6 - 6 = 6 $$

$$ 7 - 7\div7 = 6$$

$$ \sqrt[3]8 + \sqrt[3]8 + \sqrt[3]8 = 6$$

$$ 9 - 9 \div \sqrt9 = 6 $$

answered Apr 16 '21 at 20:32

A.A.

534

A very interesting problem.

2 Answers2