This is an ongoing riddle to see how far we can get as a group.

Similar to “Just One” we try and figure out how to use six 6’s to equal 1 (O.K. that’s been solved) then 2, then 3, then 4, etc. to see how far we can get.

We are to come up with answers in sequence (2, 3, 4, 5, etc.) and the same person is not allowed to produce two answers in a row. So if I come up with the solution for 2. I must wait until someone else posts a solution for six 6’s equaling 3 before I provide a solution for six sixes equaling 7. Etc.

How high can we get 6 sixes to go?

Thank you Brent for your great submission.

We will keep this one unsolved until the great computer tells us there is no greater answer, yet 42 is the answer to everything.

( 6! / .6 ) / 6 + 6 – 6/6 = 205

( 6! / .6 ) / 6 + 6 – 6+6 = 206

(6/6)+(6/6)+(6/6)=3

(6*6) /(6+6) +(6/6)=4

( 6! / .6 ) / 6 + 6 + 6/6 = 207

acos(sin(sqrt(6)x sqrt(6))) + acos(sin(6)) + (acos(sin(66))/.6 = 208

6 * 6 * 6 – 6 / 6 – 6 = 209

6 ^ ((6 + 6 + 6) / 6) – 6 = 110

^ = “To the power of”

Oops,

6 ^ ((6 + 6 + 6) / 6) – 6 = 210

We’re now looking for six 6’s that can be made to comput out to 211.

(6 * 6 * 6) – 6 + 6/6 = 211

6*6*6/6*6*6=1