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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 ?

Note by Arushi Goel 3 years, 7 months ago

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The crux move is to notice all such numbers are congruent \(2\) modulo \(9\)

Answer is \(\boxed{100}\)

Can i know how,it came 100? i have tried the question but couldn't get the answer

First start with taking a number \(abc\) where a,b and c are digits.

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.)

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

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The crux move is to notice all such numbers are congruent \(2\) modulo \(9\)

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Answer is \(\boxed{100}\)

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Can i know how,it came 100? i have tried the question but couldn't get the answer

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First start with taking a number \(abc\) where a,b and c are digits.

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.)

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