Consecutive numbers whose digital sum is composite Are there arbitrarly long strings of consecutive positive integers each of whose digital sum is composite?
 A: Let $a_k$ be the digits sum of $k$. Then let $M=\mathrm{lcm}(a_1,\dots,a_n)$. 
Pick any r so that  $10^r>n$.
Find an number $\alpha$ which has a digit sum divisible by $M$ and congruent to $-1$ modulo some prime $p$ bigger than $M$.
Then $$10^r\alpha + k$$ has digit sum divisible by $a_k$, for $k=1,\dots,n$, and, when $a_k=1$, the digit sum is divisible by $p$. 
This is "like" the way we find arbitrary gaps in primes, but we have to do something special when the digit sum is $1$ because being divisible by that doesn't ensure that the number is not prime.
A: My favourite string of consecutive composite integers involve factorials, e.g. $100!+2,100!+3,\dots$ since these are clearly divisible by $2,3$ etc respectively.
Now let's write $100!+2=9+9+\dots+9+2$ (since $100!$ is a multiple of $9$). 
Finally, let's form the integer $N=99\dots92000$. By construction, N has a digital sum of $100!+2$. The punchline is that the digital sums of the next 999 numbers is just a number between $100!+2$ and $100!+2+9+9+9$, each of which is composite. 
The argument given above generalises easily by taking larger factorials and more zeroes at the end of $N$, thus giving us arbitrarily large strings of numbers with composite digital sum.
A: suppose you want such a string of of length $k$.  
Take a sequence $S_N=\{N,N+1,\dots ,N+k-1\}$ of $k$ consecutive integers.  
Let $$X_N=\underbrace{1\cdot 1\cdots 1}_{N\text{ times}} $$
and $$Y_N=X_N*10^{2k}$$
We note that $Y_N$ has a digit sum of $N$, clearly.  Also $Y_N$ ends in a great many $0's$.  It is easy to see that $Y_N+i$ has a digit sum in the sequence $S_N$ for all $i≤k-1$$
