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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)

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)

Grade:10

5 Answers

Swapnil Saxena
102 Points
10 years ago

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

Sathya
35 Points
10 years ago

Thnks for ur reply

 

But,what does 

product of non zero digits of n  

mean?

Swapnil Saxena
102 Points
10 years ago

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)

Sathya
35 Points
10 years ago

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) +...

Swapnil Saxena
102 Points
10 years ago

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

Think You Can Provide A Better Answer ?