What a beautiful year!
S = ( 0 2 0 1 5 ) + ( 1 2 0 1 5 ) + ( 2 2 0 1 5 ) + ⋯ + ( 2 0 1 5 2 0 1 5 )
For S as defined above, how may zeros does the representation of S have in binary system?
The answer is 2015.
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1 solution
Sir, I think you forgot ( 1 + x ) n = k = 0 ∑ n ( k n ) 1 k ⋅ x n − k and in the last equality (in binary) 2 2 0 1 5 = 1 0 0 0 0 0 0 0 . . . (2015 zeros) . Anyway, everything is all right for me(Upvoted) . I have seen some solutions from you and they are really elegant,great. Thank you for these solutions
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Thanks. I have amended the solution.
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Sir i guess you had a typo at the last line it will be 1 2 0 1 5 z e r o e s 0 0 0 . . . 0 0 0
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@Md Zuhair – Agreed. The last number should start with a 1 and it should be clear that it is in base 2.
@Md Zuhair – Thanks. It should be 1.
Thanks. It should be 1.
Thanks it should be 1
S = k = 0 ∑ 2 0 1 5 ( k 2 0 1 5 ) = ( 1 + 1 ) 2 0 1 5 in base 10 = 2 2 0 1 5 in base 10 = 1 2 0 1 5 zeros 0 0 0 ⋯ 0 0 0 in base 2 Note that ( 1 + x ) n = k = 0 ∑ n ( k n ) x k Putting x = 1