Let for all n , natural numbers f(n) = product of non zero digits of n.find the largest prime divisor of

f(1) + f(2) + ...... + f(999). (the answer is 103)

The question is the largest prime no that divides f(n)

Lets simplify f(1)+f(2)+f(3)+f(4)+...+f(999)

= (1+2+3+4...+9)+1+(1+2+3+4+...+9)+2+2(1+2+3+4+...+9) +3+3(1+2+3+4+...+9) +...

On simpifying and using formula for summation of n natural no n(n+1)/2we get that the summation is 

Full solution is not given as u needed hints fr the question

((9*10)/2+1)((9*10)/2+1)((9*10)/2+1) -1 = (46*46*46)-1  97335

On prime factorization os 97335 we getthe last prime as 103 and hence we said tht the largest prime tht devides

Approved

Thnks for ur reply

But,what does 

product of non zero digits of n  

mean?

The product of non zero digits of n is like  f(999) = 9*9*9 , f(998)= 9*9*8, f(101)= 1*1 (0 not included)

How did u do this?

 f(1)+f(2)+f(3)+f(4)+...+f(999)= (1+2+3+4...+9)+1+(1+2+3+4+...+9)+2+2(1+2+3+4+...+9) +3+3(1+2+3+4+...+9) +...

This can be done in following way

This can be done in f(1)= 1, f(2)=2, ... ,f(9)=9 , f(10) = 1 , f(11) =(1*1), f(12)= (1*2), f(13)=(1*3),...f(19),f(20)= 2 , f(21)= (2*1) and so on