clearly it is easy to see that the numbers which satisfy $S(S(n)) = 2 $ also satisfy $S(n)$must be one of $= 2,11,20$ but than $S(n) = 2 mod 3 $ and hence $n = 2 mod 3 $ thus we have that all 3 digit numbers of form $3k+2 $ for natural $k$ satisfy our need. clearly it is an AP with $101$ as first term and common difference $3$ hence total number of solutions are $ 101+3(n-1) < 1000 $ and thus $ n = 100 $

Please explain the last step.....

in the last step we apply the formula of $n$ term of AP and since it must be less than 100 we involved the inequality to it

Just replace $3$ with $9$ in the above, since $n\equiv S(n) \pmod{3}$.

What about 104, 107, 113, 116, 122, etc.....all these are of 3k + 2 type but does not satisfy the given conditions.

The answer is 85 by the official solution. What they did was manually list out all the numbers and count them!!! I was wasting my time trying to find out a nice solution.

1. Since $S(S(n))\equiv S(n)\equiv n\bmod{9}$, we deduce from $S(S(n))=2$ that $n\equiv 2\bmod{9}$. 2. Now consider an arbitrary 3-digit integer $n$ with $n\equiv 2\bmod{9}$. Since $n$ is 3-digit, we have $S(n)\le3\cdot9=27$ and of course $S(n)\ge1$. From $S(n)\equiv n\equiv 2\bmod{9}$ and $1\le S(n)\le 27$ we deduce $S(n)\in\{2,11,20\}$. This implies $S(S(n))=2$. 3. Hence the question boils down to counting the 3-digit integers $n$ with $n\equiv 2\bmod{9}$. We need to find all $k$ for which $100\le 9k+2\le 999$, which implies $98\le 9k\le 997$ and $11\le k\le 110$. 4. Hence the answer to the problem is $110-10=100$.

So is the answer 100 or 85??

I think it should be $100$ as in the official solution there are many numbers like $425$ missing

Plus it is easy to see that all numbers which are $2 (mod 9)$ satisfy $S(S(n)) =2$

THE OFFICIAL SOLUTION IS WRONG

Wavefunction wrote: Plus it is easy to see that all numbers which are $2 (mod 9)$ satisfy $S(S(n)) =2$ Yeah true

Ya The official solution is wrong !!! First time I saw something like this...I am also getting answer as $100$ ($n$ will simply satisfy $\equiv 2 \mod 9$ ).What must be the plight of the guys who gave the exam lol

Sum of digits of a + b + c = 3, such that a, b, c are (a nonzero) digits. Maximum (a + b + c) is 27. Min is 1. We list out all cases: 3, 12, 21 Now we think of 'sticks' that make up the specific columns a, b, c. a must have 1, so the remaining stacks are made out of: 2, 11, 20. The first stack is: 3 ways for double-add, 3 ways to pair-add, thus 6 ways/numbers. For the second, we have : a {0, 1, 2, ..., 9}. For a -> +0, we have b {9, ..., 0} and c is uniquely determined. This continues up until a -> +2, than as a increases linearly, b decreases linearly. Thus we have 3*10 + 9 + 8 + 7 + ... + 1 = 30 + 45 = 75. For 21, when a -> +3, b -> +9. For a -> +4, b -> +{8, 9}. Thus we have 1 + 2 + 3 + 4 + 5 + 6 + 7 = 21. Thus there are 75 + 21 + 6 = 107 numbers.

Sorry misread the question...I wonder if there's a pattern...

JohnHankock wrote: Sorry misread the question...I wonder if there's a pattern... Yes there is. $n \equiv \, 2 \, (\text{mod} \, 9)$ Therefore we are counting the numbers that are 2 mod 9 between 102 and 992. They form an arithmetic progression with $a_1 = 103$ and $a_n = 992$ and difference 9. To find n, $992 = 101 + 9(n-1)$ $n = 100.$.

I am getting 279 numbers I took cases for s(n) = 11 20 and 02 and then I took n to be of the form abc and solved the equation by no of non integral solutions

I am getting 279 numbers

I am getting 103. You got 279 because in some of the solutions of your equation, digits are $\geq10$

aditya21 wrote: clearly it is easy to see that the numbers which satisfy $S(S(n)) = 2 $ also satisfy $S(n)$must be one of $= 2,11,20$ but than $S(n) = 2 mod 3 $ and hence $n = 2 mod 3 $ thus we have that all 3 digit numbers of form $3k+2 $ for natural $k$ satisfy our need. clearly it is an AP with $101$ as first term and common difference $3$ hence total number of solutions are $ 101+3(n-1) < 1000 $ and thus $ n = 100 $ This is partially incorrect as in this we will also include numbers like 224 and 332 which doesn't satisfy the condition given in the question