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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
9 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
9 years ago

Thnks for ur reply

 

But,what does 

product of non zero digits of n  

mean?

Swapnil Saxena
102 Points
9 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
9 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
9 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

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