Revisiting making 120 using exactly 5 zeroes and other symbols/operators

lookagain

Elite Member
Joined
Aug 22, 2010
Messages
3,186
Original forum page for reference:

http://www.freemathhelp.com/forum/t...-certain-restrictions-)?highlight=zeroes+make


Partial amendments:
-----------------------


You must use exactly 5 zeroes.

No exponentiation is allowed.

The factorial sign is allowed (and is expected to be used.)

The use of the square root is allowed.

\(\displaystyle \times, \ or \ \ *, \ \div, \ \ /, \ \ or \ a \ fraction \ bar, \ \ +, \ and \ - \ \)are allowed. I edited again.

Grouping symbols are allowed and can also be used for multiplication.

No concatenation (if it were to be possible) is allowed.

No decimal points are allowed.

No floor functions, ceiling functions, log functions, antilog functions,
no gamma functions, no trig functions, no double factorials, or any
other characters/words are allowed.


Clarifications/corrections on my instructions and/or solutions of mine
or solutions of other users are welcome and encouraged.

--------------------------------------------------------------------------------

In the link above, a person can read that Subhotosh Khan already gave
a couple of correct solutions (that follow those former rules as well as
these newer rules).


Two of his are:


(0! + 0! + 0! + 0! + 0!)! = 120

[(0! + 0!)(0! + 0!) + 0!]! = 120


_____________________________________________


For me, certain/select solutions (expressions) will be deemed more elegant if they can
be streamlined with fewer operators, symbols, and/or operators.
 
Last edited:
I presume the capitalized "X" means multiplication.
May I humbly suggest using the now standard "*".

Certainly. I will edit that post for that.


Plus I am curious as to why simply five 1's are not used,
since 0! = 1.

Well, there are possibilities, for example, where I want one zero to do the job of multiplying
something else to make 0, but two ones might be required to do it instead as (1 - 1)(whatever).
That would involve one more 1 than the single 0 to do the same job. I want to free up a 0
in certain circumstances.



Thank you.

...
 
Since "re-shaping" is not on the prohibited list, push each 0
from the sides going inward, forming an 8 with each.
8 * (8 + 8 - 8/8) = 120

Loud applause please (while I'm leaving the building) :rolleyes:


Just in case any puzzle solvers wonder, the "solution" by Denis does not count
(as eights can't be used) and he knows it because he's joking.
(Anyway, Denis is an actual expert on solving these anyway.)

However, I do support it for its creativity. Consider it being given
The Honorable Mention Award.


--------------------------------------------------------------------------


Here is another one (by me) that also does not count:

Take two zeroes and put them respectively atop two other zeroes to
make two eights in a sort of "partial snowman" style. Leave the fifth
zero as a zero.


\(\displaystyle \bigg(\sqrt{(\sqrt{ \ 8 + 8 \ } \ )! + 0! \ }\bigg)! \ = \ 120\)


-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.


Okay, please let us now present serious solutions after the above "creativity."
 
For typing purposes only, I'll use 1 for 0!

[(1 + 1 + 1)! - 1 / 1]! = 120 Y

[(1 + 1 + 1)! - 1 * 1]! = 120 Y

{[(1 + 1 + 1) / 1]! - 1}! = 120 Y

{[(1 + 1 + 1) * 1]! - 1}! = 120 Y

[(1 + 1 + 1)! - 1 + 0]! = 120 *** Y

[(1 + 1 + 1)! - 1 - 0]! = 120 Y

[(1 + 1 + 1 + 0)! - 1]! = 120 Y

[(1 + 1 + 1 - 0)! - 1]! = 120 Y

[(1 + 1 + 1 / 1)! - 1]! = 120 Y

[(1 + 1 + 1 * 1)! - 1]! = 120 Y

Got sleepy and stopped !


Hard to know where to draw the line; like:

Y - - - -> [(1 + 1 + 1)! - 1 + 0]! = 120 *** former

could also be:

Y - - - - > 0 + [(1 + 1 + 1)! - 1]! = 120 latter


. . . I see these last two as being different, because with the former, there is an (outermost)
factorial function that includes 0 (acting as a 0) as part of its argument.
But in the latter, there is a 0 added to the (outermost) factorial without a 0 (acting as a 0)
as part of that argument.


- - - - -- - - - - - - - - -


This response post is under general construcion for further editing.


Also:

\(\displaystyle \bigg(\sqrt{( 0! + 0! + 0! + 0! )! + 0! \ }\bigg)! \ = \ 120\)


and


\(\displaystyle \bigg(\sqrt{( (0! + 0!)(0! + 0!) )! + 0! \ }\bigg)! \ = \ 120\)


and



\(\displaystyle \bigg[(0! + 0! + 0!)! - \bigg(\dfrac{0}{0!}\bigg)!\bigg]! \ = \ 120\)



- - - - - -- - - - - - -



Can other examples be found?
 
Last edited:
That is incorrect. Another week in the corner.


No, Denis, you are incorrect. Here are some of the simplifying steps:


\(\displaystyle \bigg[(0! + 0! + 0!)! - \bigg(\dfrac{0}{0!}\bigg)!\bigg]! \ = \)


\(\displaystyle \bigg[(1 + 1 + 1)! - \bigg(\dfrac{0}{1}\bigg)!\bigg]! \ = \)



\(\displaystyle \bigg[(3)! - (0)!\bigg]! \ = \)



\(\displaystyle [6 - 1]! \ = \)



\(\displaystyle 5! \ = \)


\(\displaystyle 120\)
 
Last edited:
Here is a new one:


\(\displaystyle \bigg[(0! + 0! + 0!)! + (\sqrt{-0!})(\sqrt{-0!})\bigg]! \ = \ 120\)
 
Top