$n$ has digit sum 100; $2n$ has digit sum 110 My question is:
A $n$-digit number is given whose digit sum is $100$, the number when doubled gives digit sum as $110$ then what is this $n$-digit number?
My approach:
I assumed $n$-digits to be $x_{1},x_{2},\cdots x_{n}$ and $n$-digits after doubling the original number to be $y_{1},y_{2},\cdots y_{n}$, so the equation comes out to be,
$$\sum_{i=1}^{n}x_{i}=100$$
And another equation,
$$\sum_{i=1}^{n}y_{i}=110$$
I'm not able to proceed futher after this.
 A: The number $999999999922222$ satisfies the required condition. 
I got to this number by noting that:


*

*If you multiply any multiple of $9$ by $2$, the sum of digits remains the same.

*If you multiply any number containing all digits less than $5$ by $2$, then the sum of digits doubles.
Therefore, in the above number, on doubling, the sum of the digits of the first part containing $10$ $9$'s remains the same, and the second part which contains only $2$s, sees its sum of digits doubled. So the new sum of digits will be $110$.
A: For any $n$-digit integer $a$, let 
$[a_{n-1} a_{n-2} \cdots a_1 a_0 ]$ be its decimal representation. i.e. an ordered list of numbers $a_{n-1}, \ldots, a_0$ from $\{ 0, \ldots, 9 \}$ such that
$$a = \sum_{k=0}^{n-1} a_k \times 10^k$$
Let $X(a) \stackrel{def}{=} \sum\limits_{k=0}^{n-1} a_k$ be its sum of digits.
When we add two numbers $a = [a_{n-1} \cdots a_0 ]$, $b = [b_{n-1} \cdots b_0 ]$,
the digits of its sum $d = [d_n d_{n-1} \cdots d_0 ]$ can be determined by following algorithm.


*

*init carry $c$ to $0$ and index $k$ to $0$.

*compute $v = a_k + b_k + c$. if $v \ge 0$, set $d_k$ to $v - 10$ and $c$ to $1$ otherwise, set $d_k$ to $v$ and $c$ to $0$.

*increase $k$ by $1$. If $k < n$ repeat step 2. otherwise
set $d_{n}$ to carry $c$.


As one can see, everytime a carry is triggered at step 2. the sum of digits for $d$ will be decreased by $9$. From this, we can deduce
$$X(a) + X(b) - X(a+b) = 9 \times \text{ number of carries triggered in step 2 }$$
When $a = b$, it is easy to see a carry will be triggered when and only when $a_k = b_k$ is a digit $\ge 5$. This implies
$$2X(a) - X(2a) = 9 \times \text{ number of } a_k \ge 5$$
For the given $n$-digit integer, let call it $m$, we known $X(m) = 100$ and $X(2m) = 110$. This implies it contains $\frac{2\cdot 100 - 110}{9} = 10$ digits $\ge 5$.
If we want $m$ to be as small as possible, we will make all these $10$ digits to be $9$ and push them to the slot of $a_0,\ldots,a_9$. The account for $90$ out of $100$ in the sum $X(m)$. We are left with digits $\le 4$ to cover them. The smallest possible choice is pushing $2,4,4$ to $a_{12},a_{11},a_{10}$. 
In short, the $13$-digit number $$m = 2,449,999,999,999$$
is a solution (in fact the smallest solution) of the problem.
A: A number with 10 fives followed by fifty ones will do. Obviously it is not a small number as some have already given, but thought of posting it as a simple solution. Actually any arrangement of these should also work and you may throw in any amount of zeroes too. 
A: It's easy to see that the $30$-digit number
$$145{,}145{,}145{,}145{,}145{,}145{,}145{,}145{,}145{,}145$$ 
has digit sum $100$, while its double, 
$$290{,}290{,}290{,}290{,}290{,}290{,}290{,}290{,}290{,}290$$
has digit sum $110$.
Added later: Following up on Christian Blatter's answer, let $N$ be any number, let its digit sum be $S(N)$, and let $m$ be the number of digits in $N$ that are greater than or equal to $5$.  Let $N'$ be the number in which each of those $m$ digits is decreased by $5$ (so that each digit of $N'$ is between $0$ and $4$.  Then clearly $S(N)=S(N')+5m$.  But we also have $S(2N)=2S(N')+m$, since we can obtain $2N$ by adding the $m$ carried $1$'s to the appropriate digits of $2N'$, none of which are greater than $8$ (so there are no additional carries). It follows that
$$2S(N)-S(2N)=9m$$
So if $S(N)=100$ and $m=10$, we have $S(2N)=2S(N)-9m=2\cdot100-9\cdot10=110$, and if $S(N)=100$ and $S(2N)=110$, we have
$$m={2S(N)-S(2N)\over9}={2\cdot100-110\over9}=10$$
A: Any number comprised of repeating $4$s and $5$s will be the same sum when doubled. We then just need a series of digits that when doubled adds $10$.
$4545454545454545454522222$ will do it.
