2118 ?
If " N " be the number of natural numbers <2009 which can be expressed in the form of [x[x]] for some positive Real " x " then the sum of digits of " N " is -
{ where [.] is GIF }
-
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6 Answers
21let
x= p+f,p is integer ≥1 (because 0 is not natural no.) and f<1
[x]=p
[x[x]] = p2+ [pf]
there exists some f<1 for which pf = t (p<t<0)
so [pf] gives all integers from the set {0,1,...., p-1}
therefore [x[x]] gives exactly p integers (+ve) for any selected p≥1)
therefore number of natural nos <2009, of the form [x[x]] is summation of p , p=1 to 44
there 990 numbers
71By the way, the answer was given like this 18 (990) so I thought that answer would the product of two.
But I have got the same answer too but not their product as the answer. :)
21"By the way, the answer was given like this 18 (990) so I thought that answer would the product of two."
There is no point of such a confusion.
how can the digital sum be 18 * 990 ?
There can be at most 2008 numbers. Can any number between 1 to 2008 have such a BIG digital sum ?/:D