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for the 1st 9 natural nos it will be sum of them itself i.e 1+2+3+....9 = 45 (i)
for all the two digit numbers containg non zero digits the value will be ,
(1+2+3+.......9)(1+2+3+....9) =452 (ii)
similarly for all 3 digit number containing non zero digits the value will be,
(1+2+3+.......9)(1+2+3+....9)(1+2+3+.......9) = 453 (iii)
now time for the digits containg zero,
for 10 to 90 => (1+2+3+.......9)
for 110 to 190 => (1+2+3+.......9)
for 210 to 190 => 2(1+2+3+.......9)
:::::::::::::::::::::::::::::::::::::::::::::::::::::
910 to 990=> 9(1+2+3+.......9)
summation of these values = (1+2+3+.......9) +(1+2+3+.......9)(1+2+3+.......9) = 45+452 (iv)
101 to 109 => (1+2+3+.......9)
201 to 109 => 2(1+2+3+.......9)
::::::::::::::::::::::::::::::::::::::::::::
901 to 909 =>9(1+2+3+.......9)
summation= (1+2+3+.......9)(1+2+3+.......9)=452 (v)
lastly for 100,200,300...900 the value will be (1+2+3+.......9)=45 ( vi)
adding (i)+(ii)+(iii)+(iv)+(v)+(vi),
=> 3.45+3.452+453 = 97335 = f(1) + f(2) + ...... + f(999)
let each + ve integer less then 1000 be a 3 digit number by putting the requisite number of 0s in from of it.
the sum is simply -
(0.0.0 + 0.0.1 + .... 9.9.9)^3 - 0.0.0
we replce 0s here by 1s. ca3use multiplying by ones here is equivalent to ignoring the 0s.
so we get , (1 + 2 + .. + 9)^3 - 1 = 46^3 - 1 = 3^3* 5 * 7 * 103.
so the answe is 103.
i will explain why i cubed this.
suppose we have 3 digits - 4 , 5 , 6.
the numbers formed will be
456 , 564, 465, 546 645, 654.
the product is just the same in each case.
which can be written as 6*4*5*6.
or, 6abc.
also terms like 111, 222, 333, contribute to terms of the form x^3, y^3 ...
terms like 223, 556 cotribute to 3(n^2)m ,....
so i used the cubic expansion. :)
let each + ve integer less then 1000 be a 3 digit number by putting the requisite number of 0s in from of it.
the sum is simply -
(0.0.0 + 0.0.1 + .... 9.9.9)^3 - 0.0.0
we replce 0s here by 1s. ca3use multiplying by ones here is equivalent to ignoring the 0s.
so we get , (1 + 2 + .. + 9)^3 - 1 = 46^3 - 1 = 3^3* 5 * 7 * 103.
so the answe is 103.
i will explain why i cubed this.
suppose we have 3 digits - 4 , 5 , 6.
the numbers formed will be
456 , 564, 465, 546 645, 654.
the product is just the same in each case.
which can be written as 6*4*5*6.
or, 6abc.
also terms like 111, 222, 333, contribute to terms of the form x^3, y^3 ...
terms like 223, 556 cotribute to 3(n^2)m ,....
so i used the cubic expansion. :)
let each + ve integer less then 1000 be a 3 digit number by putting the requisite number of 0s in from of it.
the sum is simply -
(0.0.0 + 0.0.1 + .... 9.9.9)^3 - 0.0.0
we replce 0s here by 1s. ca3use multiplying by ones here is equivalent to ignoring the 0s.
so we get , (1 + 2 + .. + 9)^3 - 1 = 46^3 - 1 = 3^3* 5 * 7 * 103.
so the answe is 103.
i will explain why i cubed this.
suppose we have 3 digits - 4 , 5 , 6.
the numbers formed will be
456 , 564, 465, 546 645, 654.
the product is just the same in each case.
which can be written as 6*4*5*6.
or, 6abc.
also terms like 111, 222, 333, contribute to terms of the form x^3, y^3 ...
terms like 223, 556 cotribute to 3(n^2)m ,....
so i used the cubic expansion. :)
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the value of f(1) + f(2) + ...... + f(999) comes out to be 97335 and thus 103 is the greatest prime divisor.