For any positive integer \(n\) , let \(S(n)\) denote the sum of digits of \(n\). Find the number of \(3\) digit natural numbers \(n\) for which \(S(S(n))=2\). What is the answer ?

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TopNewest100 – Anshul Sanghi · 1 year, 10 months ago

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The crux move is to notice all such numbers are congruent \(2\) modulo \(9\) – Souryajit Roy · 1 year, 10 months ago

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Answer is \(\boxed{100}\) – Karthik Sharma · 1 year, 10 months ago

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– Nishant Singh · 11 months ago

Can i know how,it came 100? i have tried the question but couldn't get the answerLog in to reply

and let \(a+b+c = xy\) Where \(xy\) is a number and x and y are digits.

So, we have \(S(S(n)) = S(xy) = x+y = 2\)

Then make three cases - x=0, y=2 , x=2, y=0 and x=y=1.

Do a little bit of P & C to get the answer. Good luck.

(I did this question in RMO 2014 the same way I mentioned. There are other quicker methods to do this too.) – Karthik Sharma · 9 months, 3 weeks ago

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