Sum of two digit numbers | PRMO-2016 | Problem 7

Let s(n) and p(n) denote the sum of all digits of n and the products of all the digits of n(when written in decimal form),respectively.Find the sum of all two digits natural numbers n such that \(n=s(n)+p(n)\)

Let \(n\) is a number of two digits ,ten's place \(x\) and unit place is \(y\).so \(n=10x +y\).given that \(s(n)\)= sum of all digits \(\Rightarrow s(n)=x+y\) and \(p(n)\)=product of all digits=\(xy\)

now the given condition is \(n=s(n)+p(n)\)

Can you now finish the problem ..........

From \(n=s(n)+p(n)\) condition we have,

\(n=s(n)+p(n)\) \(\Rightarrow 10x+y=x+y+xy \Rightarrow 9x=xy \Rightarrow y=9\) and the value of\(x\) be any digit....

Can you finish the problem........

Therefore all two digits numbers are \(19,29,39,49,59,69,79,89,99\) and sum=\(19+29+39+49+59+69+79+89+99=531\)

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