Here's a fun (albeit difficult) one:
Make these equations true using arithmetic operations:
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
For example: 6 + 6 - 6 = 6
(I hope I did not spoil some of you :D)
Allowed operators are:
+, -, *, /, ! , ^, %
Setting parenthesis is also allowed.
The ^
operator is an exception as you are permitted to supply a second argument to it which may be any positive integer or the multiplicative inverse of it.
$x^{1/y}$ is always positive and real.
If you find an alternative solution using other operators you may post it but please also provide a solution using only these 7 operators.
For those of you who think this was easy, here is a bonus:
0 0 0 = 6
Answer
1.
$(1+1+1)! = 6 $
2.
$2+2+2 = 6$
3.
$3*3-3 = 6$
4.
$\left(4-\frac 4 4\right)! = \sqrt 4+\sqrt 4+\sqrt 4=6$
5.
$5+\frac 5 5 = 6$
6.
$6*\frac 6 6 = 6 + 6 -6=6$
7.
$7-\frac 7 7 = 6$
8.
$\left(\sqrt{8+\frac 8 8}\right)! = 6$
9.
$\left(\frac{\sqrt{9}\sqrt{9}}{\sqrt 9}\right)! = 6$
Bonus:
$(0!+0!+0!)! = 6$
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