let for all n , natural numbers

f(n) = product of non zero digits of n.

find he largest prime divisor of

f(1) + f(2) + ...... + f(999).

the answer is 103. think about it. :)

 the value of   f(1) + f(2) + ...... + f(999)  comes out to be 97335 and thus 103 is the greatest prime divisor.

show the procedure. that is what would matter..

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. :)