Puzzle needs to be solved

**tilake1** · August 10th 2010, 05:25 AM

Put Any Mathematic sign:

2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6
1 1 1 = 6

Let me know the answer.. 2,3 and 6 are solved..

**Archie Meade** · August 10th 2010, 05:57 AM

Originally Posted by tilake1

Put Any Mathematic sign:

2 2 2 = 6
3 3 3 = 6
4 4 4 = 6

$\sqrt{4}+\sqrt{4}+\sqrt{4}$

5 5 5 = 6

$5+\frac{5}{5}$

6 6 6 = 6
7 7 7 = 6

$7-\frac{7}{7}$

8 8 8 = 6

$\sqrt[3]{8}+\sqrt[3]{8}+\sqrt[3]{8}$

9 9 9 = 6

$9-\frac{9}{\sqrt{9}}$

1 1 1 = 6

$(1+1+1)!=6$

Let me know the answer.. 2,3 and 6 are solved..

.

**tilake1** · August 10th 2010, 06:06 AM

Thanks...

**Soroban** · August 10th 2010, 07:50 AM

Hello, tilake1!

I couldn't find an elementary expression for the 8's . . .

. . $\begin{array}{ccc} (1 + 1 + 1)! &=& 6 \\ \\[-3mm] 2 + 2 + 2 &=& 6 \\ \\[-3mm] 3! \times 3 \div3 &=& 6\\ \\[-3mm] \sqrt{4}+\sqrt4+\sqrt4 &=& 6 \\ \\[-3mm] 5 + (5 \div 5) &=& 6 \\ \\[-3mm] 6 \times 6 \div 6 &=& 6 \\ \\[-3mm] 7 - (7 \div 7) &=& 6 \\ \\[-3mm] \log_{\sqrt{\!\!\sqrt{8}}}(8\sqrt{8}) &=& 6\\ \\[-2mm] (9 + 9) \div \sqrt9 &=& 6 \end{array}$

**Wilmer** · August 10th 2010, 01:08 PM

2 other ways for the 4's: 4 + 4 - sqrt(4) = 6 and 4/.4 - 4 = 6

On the 8's; since sqrt is allowed, then cubert should be allowed: cubert(8) + cubert(8) + cubert(8) = 6

**Archie Meade** · August 10th 2010, 03:18 PM

$8^0\left(log_28+log_28\right)=6$

**Wilmer** · August 10th 2010, 04:53 PM

Originally Posted by Archie Meade

$8^0\left(log_28+log_28\right)=6$

That's cheating Archie; you're using extra numbers, 0 and 2 !!

**Archie Meade** · August 10th 2010, 05:14 PM

Originally Posted by Wilmer

That's cheating Archie; you're using extra numbers, 0 and 2 !!

That sinks the factorial then!
cos that's a multiplication by 2, written with a footprint.

**Archie Meade** · August 11th 2010, 09:47 AM

$\displaystyle\huge\left[\left(\frac{d}{dx}[8]\right)!+\left(\frac{d}{dx}[8]\right)!+\left(\frac{d}{dx}[8]\right)!\right]!=6$

which holds for all constants.

**Wilmer** · August 11th 2010, 09:56 AM

Take this Archie:
(8 + 8 + 8) / CEILING(pi) = 6

**Archie Meade** · August 11th 2010, 11:05 AM

Define S to be the "Sum of the digits of a number".

$S(8+8+8)=6$

**Unbeatable0** · August 11th 2010, 12:02 PM

Originally Posted by Archie Meade

Define S to be the "Sum of the digits of a number".

$S(8+8+8)=6$

Define $G\equiv 6$ . Then $G(a+a+a) = 6$ and this
holds for all constant $a$ 's

**simplependulum** · August 11th 2010, 06:18 PM

HaHa , and

$a^{ \log_a(2) } + a^{ \log_a(2) } + a^{ \log_a(2) } = 6$

**Wilmer** · August 12th 2010, 08:23 AM

Originally Posted by Soroban

. . $\begin{array}{ccc} 3! \times 3 \div3 &=& 6\\ \\[-3mm] \end{array}$

3 * 3 - 3 = 6 ; keep it simple!

**Wilmer** · August 12th 2010, 08:28 AM

Another for the three 1's; arrange 1st two as a V:
V1 = 6

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