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ganeshie8
 4 years ago
\(11_{11}^2 = ?_{11}\)
ganeshie8
 4 years ago
\(11_{11}^2 = ?_{11}\)

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mathslover
 4 years ago
Best ResponseYou've already chosen the best response.0What is this? @ganeshie8 can u explain this to me?

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1that looks correct... actually i thought of asking as we move left in base 11, 11^2 becomes 100 ? right ?

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1@mathslover subscript means its base 11

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.3@Zekarias check that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You made algebra mistake I think @UnkleRhaukus

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well then the answer must be less than 121

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1im getting 121 too.. but the q i thought of asking was different

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.3\[11^2_{11}=(11+1)^2=12^2=144\] \[144=121+22+1=121_{11}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1349083712778:dw

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.3http://openstudy.com/study#/updates/50694b4ce4b0e3061a1d953f

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1349083784134:dw

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1\(11_{10}^2\) = \(100_{11}\) \(11_{10}^3\) = \(1000_{11}\) \(11_{10}^4\) = \(10000_{11}\)

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1does that mean the positional value changes as smooth as it changes in base 10 ? therez no advantage of base 10... i use to think we use base 10 coz the progression of positional value is smooth in base 10

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.3im not sure what you mean by "positional value changes as smooth"

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1in base 10, the position value increases in powers of \(10_{10}\), like 10, 100, 1000, 100000 .... . same is happening wid base 11 also, its changing as powers of \(11_{10}\) : 10, 100, 1000, 100000... only if i think in base 11, i see its changing smoothly... else, in base 10 i see it changing as, 11, 121, 1331, 14641...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since there are 9 digits only....... So I think base is better

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1as long as i think in one base system only, i dont see any problem. all systems look equal to me nw. @sauravshakya 9 digits or 10 digits hw does it matter... u mean 9/10 is easy to remember for us... and since, around 5 digits would be too small, and again, around 20 would be too many to remember... so 10 digits look fairly good to remember so we stuck wid base 10... . looks good rationale to me :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I mean it is easy to do calculation using base 10..... like division and multiplication

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This surely takes time: A^2_11 = ?_11

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1349086304417:dw

ganeshie8
 4 years ago
Best ResponseYou've already chosen the best response.1surely it doesnt look easy..but thats becoz we didnt learn multiplication tables in base 11, we learned them in base 10. but if we lived in base 11, we would have done it in a snap. i dont think we would have missed anything.. . but im really not sure.. im just inclining towards thinking like this, not fully sure yet .. .

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.3dw:1349086853953:dw
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