I give you 3 digits and a result and you must put all the signs necessary to restore the equality.

I give you an example. The remainder you solve by yourself.

2 + 2 + 2 = 6

Easy! Isn't this? It is the same for the remainder.

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
i poste answer tommorow

• (1+1+1)! = 3! = 6
  [(2/2)+2]! = 3!
  (3+3-3)! = 3!
  (4-(4/4))! = 3!
  5+(5/5) = 6
  6+6-6 = 6
  7-(7/7) = 6
  (8^0+8^0+8^0)! = 3!
  (9^0+9^0+9^0)! = 3!
• 12 years agoHelpfull: Yes(4) No(0)
• (1+1+1)!=6
  2+2+2=6
  3*3-3=6
  root(4)+root(4)+root(4)=6
  (5/5)+5=6
  6-6+6=6
  (-7/7)+7=6
  Cube root(8)+Cube root(8)+Cube root(8)=6
  root(9)*root(9)-root(9)=6
• 12 years agoHelpfull: Yes(1) No(0)
• (1+1+1)!
  2+2+2
  3*3-3
  square root of 4+square root of 4+square root of 4
  5/5+1
  6+6-6
  7-7/7
  cube root of 8+cube root of 8+cube root of 8
  square root of 9*square root of 9-square root of 9
• 12 years agoHelpfull: Yes(0) No(0)
• (1+1+1)!
  2+2+2
  3+3-3
  4+4-√4
  5+(5/5)
  6-6+6
  7-(7/7)
  ³√8+³√8+³√8
  (9+9)-√9
  
  
• 12 years agoHelpfull: Yes(0) No(0)