Make an expression that equals 11 using exactly six zeroes and ...

[lookagain]

certain other symbols.

How would you solve this using exactly six digits of zeroes and
no other digits to make 11?

No concatenation of zeroes or concatenation of
factorial signs, except for this form: (X!)!, where
X is a number, or expression in general.
No more than five addition signs
No more than one subtraction/negative sign
No multiplication signs
No division signs
No exponentiation
No more than two square root symbols
No more than two pairs of parentheses
No more than eight factorial symbols
(They are just to be regular factorials, no subfactorials,
derangements, multifactorials, etc.)
No other operators/functions/symbols are permitted.


Two examples are:

(0! + 0!)(0! + 0! + 0!)! - 0! = 11

-0! + (0! + 0!)(0! + 0! + 0!)! = 11



(There are a handful of "parent"/more unique solutions, and then there are several relatively trivial ones
using one extra pair of parentheses in various places, different orderings of subexpressions, or both.)

 

[lookagain]

Sqrt[(0! + 0! + 0! + 0! + 0!)! + 0!] = 11

The above expression on the left, or its equivalent, \(\displaystyle \ \ \sqrt{(0! + 0! + 0! + 0! + 0!)! + 0!} \)

is a solution. Your job is done.


As a bonus, try to come up with more (read: different) "parent" solutions still using my rules.

 

[jonah2.0]

WARNING: Beer soaked rambling/opinion/observation/reckoning ahead. Read at your own risk. Would be readers can take it seriously or take it with a grain of salt. In no event shall the wandering quixotic math knight-errant Sir jonah in his inebriated state (usually in his dead tired but mentally revived inebriated state) be liable to anyone for special, collateral, incidental, or consequential damages in connection with or arising out of the use of his "enhanced" beer (and tequila/absinthe) powered views.

certain other symbols.

How would you solve this using exactly six digits of zeroes and
no other digits to make 11?

No concatenation of zeroes or concatenation of
factorial signs, except for this form: (X!)!, where
X is a number, or expression in general.
No more than five addition signs
No more than one subtraction/negative sign
No multiplication signs
No division signs
No exponentiation
No more than two square root symbols
No more than two pairs of parentheses
No more than eight factorial symbols
(They are just to be regular factorials, no subfactorials,
derangements, multifactorials, etc.)
No other operators/functions/symbols are permitted.


Two examples are:

(0! + 0!)(0! + 0! + 0!)! - 0! = 11

-0! + (0! + 0!)(0! + 0! + 0!)! = 11



(There are a handful of "parent"/more unique solutions, and then there are several relatively trivial ones
using one extra pair of parentheses in various places, different orderings of subexpressions, or both.)

Have you made no attempt to solve the problem yourself?

 

[lookagain]

Have you made no attempt to solve the problem yourself?

Two examples are:
.
.

Those are examples I worked out.

 

[lookagain]

That's cheating: "no multiplication signs" implies
no multiplication allowed; if not, then it should be
specified that multiplication is allowed.

Similarly, this would be a division:
(0! + 0!)
(0! + 0! + 0!)!

I will report this Major Trespass to Sir Jonah
who shall decide on the appropriate penalty.

No, it's not cheating. "No multiplication signs" means I am cutting down on the
number of solutions by limiting how I want any multiplications to be shown.

I don't know where you get off that being a division, because it isn't.
It's an expression "floating" above another expression.

You've got another candidate for a solution!? Then make that your next post.

 

[greg1313]

\(\displaystyle \sqrt{((0! + 0! + 0!)! \ - \ 0!)! \ + \ 0! \ } \ + \ 0 \ = \ 11 \)

 

[lookagain]

\(\displaystyle \sqrt{((0! + 0! + 0!)! \ - \ 0!)! \ + \ 0! \ } \ + \ 0 \ = \ 11 \)

Good. That is one of the correct ones.

There are still others.

 

[v8archie]

\(\displaystyle \sqrt{(\sqrt{(0! + 0! + 0! + 0!)! \ + \ 0! \ } \ ) ! \ + \ 0! \ } \ \ = \ 11\)