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3 solutions
Nice solution without using complex equations :). How far have you done calculus? @Satvik Golechha
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Truthfully, I have just learnt the u + v ) rule of differentiation...... I see that you have learnt many new things. From where are you studying at present?
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@Satvik Golechha -Extremely sorry for the unintentional late rpely due to poor internet connections :( . New things? Many? Seriously? Man, I guess you need to look up a dictionary and see the definition of "new" and " things" :P. I actually haven't learnt anything at all except a bit of calculus. I by-passed limits since I found them too tough and you call me a perosn who's recenlty learnt a lot :P....BTW, What's u+v rule and where did you learn that from? I also see that you've learnt to solve for maximium and minimum values of expression...how?
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@Krishna Ar – Your sudden increase @Krishna Ar in levels clearly indicated that you've learnt too much lately.... I mostly solve max and min by luck....I try it mostly for all equal values... u+v rule states that d(u+v)/dx =du/dx + dv/dx
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@Satvik Golechha – @Satvik Golechha - I Think you;re referring to my levelling up in geo and combi. Well that's again bcos I solved quite a few easy level 3/4 questions and that aint got anything to do with learning a lot....though I wish I cud and also do u know limits? Luck? srsly?
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@Krishna Ar – @Krishna Ar Yup, luck..srsly. I also want to learn limits, but I dunno from where.~ What were those lvl#4's you solved?
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@Satvik Golechha – @Satvik Golechha - One of them was abou advertising in combinatorics and in geometry a few level 3's like Something to do with an equilateral trigon, two of your problems!!! and a problem by Fahim S- to do with circle areas i guess
@Satvik Golechha
–
satvik golecha you can learn limits from '' mathematics class 11 NCERT '' in this there is very nice explanation of solving and using limits other books are
i)experimental mathematics in action you will find many interesting probs in this book
ii) also try 12th NCERT for learning calculus
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@Parth Lohomi – Thank you very much.
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@Satvik Golechha – So you are using them @Satvik Golechha
@Krishna Ar – this is actually the chain rule of differention this kind of rule is also aplicable in integration
awesome and brilliant and elegant and whatever other good thing there is !!!!!!!
I did it the same way ........... by the way nice solution Satvik
i will say simple using logarithms consider , l o g 2 2 0 1 4 = 2 0 1 4 l o g 2 = 2 0 1 4 ( 0 . 3 0 1 0 ) = 6 0 6 . 2 1 4 since the charecterestic is 606 therefore 2 2 0 1 4 contains 607 digits hence proved
It needs a better rating, this is a nice question.
By the way, there is a formula ceiling function( 0.30103*n) [0.30103 = log base 10 n ]
0.30103 * 2014 = ceiling function(606.27) = 607
I don't know what's the derivation, by the way, this is a good discussion:
https://math.stackexchange.com/questions/177973/is-there-a-way-to-determine-how-many-digits-a-power-of-2-will-contain
Okay, so we want the number of digits in 2 2 0 1 4 . The number of digits in x ∗ 1 0 n is n + 1 , and 2 2 0 1 4 is 1 0 6 0 6 . 2 7 4 4 1 1 2 6 7 2 5 8 1 3 6 8 , which can be easily seen by the rules of logarithm. So the number of digits is the smallest integer greater than 6 0 6 . 2 7 7 4 , which is 6 0 7 .