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I know a way by which i can make i can get total of 120 by using five zeros 0,0,0,0,0 and any one mathematical operator. Do you?

Trincent  May 19, 2015

Best Answer 

 #1
avatar+886 
+16

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

The operator is the factorial (!).

EinsteinJr  May 19, 2015
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16+0 Answers

 #1
avatar+886 
+16
Best Answer

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

The operator is the factorial (!).

EinsteinJr  May 19, 2015
 #2
avatar+4663 
+5

Nice one EinsteinJr, you really know your way round factorials!

MathsGod1  May 19, 2015
 #3
avatar
+10

The plus sign (+) is also a mathematical operator,  so you've used two operators. No cookie for you!

Guest May 20, 2015
 #4
avatar+886 
+10

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

EinsteinJr  May 20, 2015
 #5
avatar+91038 
+5

"any one mathematical operator"     -      any one means ONLY one 

 

That is a quote from the question.  

 

It doesn't matter though - I liked your answer  :)

I like you second answer too.

I would like to see anon do it with ony one operator as well    LOL

Melody  May 20, 2015
 #6
avatar+886 
+15

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.

EinsteinJr  May 20, 2015
 #7
avatar+91038 
+5

It wasn't that stupid - I can understand your annoyance.

You have presented some really clever answers there :)

Melody  May 20, 2015
 #8
avatar+886 
+15

Another solution:

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

EinsteinJr  May 20, 2015
 #9
avatar
+5

Morgan Tud’s solutions:

 

web2.0calc . com  /questions/a-question-you-can-answer#r6

 

Technically his second answer is only two operators because “NOT” is a logical operator. 

 

You still don’t get a cookie!

Guest May 20, 2015
 #10
avatar+886 
+5

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.

EinsteinJr  May 20, 2015
 #11
avatar
+5

Unearned cookies are not as good!

Guest May 20, 2015
 #12
avatar+886 
+5

Anyway, I don't like cookies.

EinsteinJr  May 20, 2015
 #13
avatar
+5

Is that why you give them away?

 

The one you gave CPhill looked like it had roach droppings on it. 

Guest May 20, 2015
 #14
avatar+886 
+5

You're right.

EinsteinJr  May 20, 2015
 #15
avatar+91038 
+5

LOL   

The Morgan Tud link doesn't work :(

Melody  May 21, 2015
 #16
avatar+886 
+10

The first answer [cos(0)+cos(0)+cos(0)+cos(0)+cos(0)]! is right since cos(0)=1, but the second one is wrong (not(0)=-1 so [not(0)+not(0)+not(0)+not(0)+not(0)]!=(-1-1-1-1-1)!=(-5)!=MATH ERROR)

EinsteinJr  May 21, 2015

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