Fine with Five?

Let $S(n)$ denote the digit sum of $n$ . We can see fairly easily that, given $S(n)$ , $S(2n)$ can be calculated in the following way:

• $8 \cdot 2 = 16$ , so for $k$ $8$ 's in $n$ , $(1+6)k = 7k$ will be added to the digit sum of $2n$ .

• $7 \cdot 2 = 14$ , so for $k$ $7$ 's in $n$ , $(1+4)k = 5k$ will be added to the digit sum of $2n$ .

• $6 \cdot 2 = 12$ , so for $k$ $6$ 's in $n$ , $3k$ will be added.

• $5 \cdot 2 = 10$ , so for $k$ $5$ 's in $n$ , $k$ will be added.

• For all other digits less than $5$ , the digit sum will double per digit.

Since we are trying to look for the number of $5$ 's in $n$ , let the number of $5$ 's be equal to $x$ , and let the sum of the digits below $5$ be equal to $y$ . The problem gives that

$S(n) = 104 = 4(8) + 3(7) + 2(6) + 5x + y,$

so $5x + y = 39$ . We also know, by our double-digit-sum calculator, that

$S(2n) = 100 = 4(7) + 3(5) + 2(3) + x + 2y,$

so $x + 2y = 51$ . Solving for $x$ , we find that our answer is $\boxed{3}$ .

First, note that it does not specify where the $8,7,6$ 's are, so we can assume that they go in the front in the order $888877766$ and hope that it works out later.

We have that the current digit sum is $65$ , so we must have $104-65=39$ more. In addition, the digit sum of $2n$ so far would be $1+7+7+7+7+5+5+5+3+(2\text{ or }3)=50\text{ or }49$ where the sum is unclear because we don't know if the next digit will carry or not.

Now let us assume that the next digit is a $5$ . That means that the sum so far of $2n$ is $50$ because $2\times 5$ carries over $1$ .

Now, note that for all digits $0\rightarrow 4$ , we have that the digit sum of $2n$ is always twice the digit sum of $n$ . Because $2\times 39\ne 50$ , we must have at least one $5$ in $2n$ to balance it out.

Now let's do some simple casework: if there is one $5$ , then the digit sum of $n$ not including the digits given is $34$ and the digit sum of $2n$ (again not including the digits given) is $50$ . Obviously $34\times 2\ne 50$ so we consider the next case.

If there are two $5$ 's, then the digit sum of $n$ is $29$ and the digit sum of $2n$ is $49$ . Since $29\times 2\ne 49$ , we move on to the next case.

If there are three $5$ 's, then the digit sum of $n$ is $24$ and the digit sum of $2n$ is $48$ . Since $24\times 2=48$ , this is our working case and so the answer is $\boxed{3}$ fives.

Well I did a combination of 1,2,3,4,5 to get 104... I found 3 5s, 2 4s and 1 1,2,3 each sums up to 104

Since location of digits are unknown, hit and trial worked and I had no idea how to use other info - 2n and 100