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@vipandeep-n6kj1q • Feb 22, 2011
2^0+2^1+2^2+2^3 = 15 -
@abrakadabra • Feb 22, 2011
11 raised to 11? 😲 -
@thebigk • Feb 22, 2011
I'm not math expert. But I think 11[sup]11[/sup] would be quite a big number. -
@ishan-nohePN • Feb 22, 2011
Phew! both of you.You were really quick.Hats off...
correct answer AKD,Biggie.
@Vipandeep:This is a wrong answer as you have used '2' also.Answer is 11^11.
😁 -
@civilprincess-308hDv • Feb 22, 2011
Okie can you use three zeros and get 6:?: -
@ishan-nohePN • Feb 22, 2011
But why do you want to use 0? when you can get biggest number greater than 6?CIVILPRINCESSOkie can you use three zeros and get 6:?:
😁
@CP,Biggie,AKd and all ceans: What do you think about 1111! i.e; 1111factorial? -
@thebigk • Feb 22, 2011
(0! + 0! + 0!)! => 3! = 3x2x1 = 6.CIVILPRINCESSOkie can you use three zeros and get 6:?: -
@thebigk • Feb 22, 2011
I guess the number of operators need to be defined in this case. I can go on adding factorial signs and keep expanding the number.ishutopreBut why do you want to use 0? when you can get biggest number greater than 6?
😁
@CP,Biggie,AKd and all ceans: What do you think about 1111! i.e; 1111factorial? -
@praveenkumar-66Ze92 • Mar 2, 2011
Yes then what about 11! x 11! ??? Goes somewhere more than 11[SUP]11[/SUP]! -
@thebigk • Mar 2, 2011
That's my point, the number of operators need to be restricted.😉praveenscienceYes then what about 11! x 11! ??? Goes somewhere more than 11[SUP]11[/SUP]! -
@praveenkumar-66Ze92 • Mar 2, 2011
Oki! Even though it can be only upto max of 3... My point is proved! Lolz 😁 -
@harshajd-F0ZGFM • Mar 14, 2011
if the factorial can be used then (11!)^(11!) will be the biggest number