EinsteinJr

Et pourquoi cette question ?

You're right.

Anyway, I don't like cookies.

Then nobody will get a cookie, because it is impossible to do with only one operator.

And I don't give a d**n if I don't get a cookie, since I have got a full pack.

Perfect.

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Here's your cookie:

Another solution:

[cos(0)+cos(0)+cos(0)+cos(0)+cos(0)]!

You're right Melody, when I posted it I realized how stupid my remark was

I found a fourth way :

[(0! + 0!)^{(0! + 0!)} + 0!]!=(2²+1)!=(4+1)!=5!=120

But it seems to be impossible to do with only one symbol.

He didn't say only one mathematical operator, so it means I can use two ones.

And, if you're so smart Anonymous, find a way you can do it using one mathematical operator.

Good luck !

I found two other ways (the second is a little bit tricky, since a 8 is formed by a 0 put atop another 0)

[(0!+0!)(0!+0!)+0!]!=[(1+1)(1+1)+1]!=(4+1)!=5!=120

$$({\sqrt{({\sqrt{{\mathtt{8}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}}}){!}{\mathtt{\,\small\textbf+\,}}{\mathtt{0}}{!}}}){!}$$=$$({\sqrt{({\sqrt{{\mathtt{16}}}}){!}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}}}){!}$$=$$({\sqrt{{\mathtt{4}}{!}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}}}){!}$$=$$({\sqrt{{\mathtt{25}}}}){!}$$=5!=120

(0!+0!+0!+0!+0!)!=(1+1+1+1+1)!=5!=1*2*3*4*5=120

The operator is the factorial (!).

Sorry, I was too busy to answer back to you.

OK, you both have

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And one cookie for each: