One of our regular visitor Aisharya Rana contributed this puzzle, which I found interesting. Hence I’m posting here to be pondered upon by all. Post your answers as comments below –

**Arithmetic Puzzle**

In this puzzle you need to insert any arithmetic sign in between the same digit (from 1 to 9) repeated thrice. The final result should be 6 in each case. I’m doing one for illustration –

2 2 2 = 6

can be expressed as –

2 + 2 + 2 = 6

Solve the following yourself –

1 1 1 = 6

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

Remember, each equation can be solved in more than one way. Post your answers in the comment section below.

Quite interesting puzzles. Had difficulty with some, but anyway solved all. was not able to solve (1+1+1)!=6. Answer is provided by Ritesh

8^(1/3)+8^(1/3)+8^(1/3) = 6

(1+1+1)!=6

Every puzzle has been solved except first one so I am posting that solution:

(1+1+1)!=6

2+2+2=6

3*3-3=6

4+4-√4=6

5÷5+5=6

6÷6*6=6

7-7÷7=6

8-8^0-8^0=6

√9*√9-√9=6

2+2+2=6

3*3-3=6

4+4-√4=6

5÷5+5=6

6÷6*6=6

7-7÷7=6

8-8^0-8^0=6

√9*√9-√9=6

2+2+2=6

3*3-3=6

4+4-√4=6

5÷5+5=6

6÷6*6=6

7-7÷7=6

8-8^0-8^0=6

√9*√9-√9=6

3×3-3 = 6

6+6-6 = 6

Other you please solve.

3×3-3 = 6

6+6-6 = 6

Other you please solve.