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WLOG let $ M$ be a positive integer (doesn't matter since we can just replace $ M$ with $ int(|M|) + 1$ and secure a result that is not weaker). Let the common difference be $ n = 10^k - 1$. Then I claim that for a sufficiently large value of $ k$, there is a starting value $ a$ such that no number with $ M$ or fewer non-zero digits can be congruent to $ a$ mod $ n$, showing that each value in the arithmetic progression $ a,a + n,a + 2n,...$ has at least $ M + 1$ non-zero digits, so has digital sum exceeding $ M$. Each number with $ M$ or fewer non-zero digits can be written as $ \sum_{i = 1}^Ma_i\cdot 10^{b_i}$, with each $ a_i \in \{0,1,2,3,4,5,6,7,8,9\}$, and each $ b_i \in \mathbb{N}$. This expression can the take on at most $ 10^Mk^M$ different values mod n, since each $ a_i$ can take on at most 10 values, and each $ 10^{b_i}$ can only take on the values $ 10^0,10^1,...,10^{k - 1}$ ($ k$ different values) mod n. If $ 10^Mk^M < 10^k - 1$ then this guarantees an $ a$ so that each value in the arithmetic sequence has to be written using $ M + 1$ or more non-zero digits, securing the result. But this inequality will hold true for some large enough $ k$ since exponential functions grow faster than polynomial functions. Seemed far too easy for an N6... maybe it's because it is 1999. More likely a mistake on my part, but I haven't found an error... please point out if there is one, I am very prone to making them. Cheers, Rofler

It obviously suffices to find arithmetic sequences for all positive integers $M$. Let $s(n)$ denote the sum of the digits of $n$ in base 10. We claim that for any positive integer $k$, $s((10^M - 1)k) \geq 9M$ (so the arithmetic sequence with kth term $(10^M - 1)k$ would satisfy the conditions of the problem.) We will prove this by using strong induction on $k$. For the base case $k = 1$, $s(10^M - 1) = 9M$. We now wish to show that for any $k > 1$, if our claim is true for all $j < k$, then $s((10^M - 1)k) \geq 9M$. If the base $10^M$ representation of $(10^M - 1)k$ is $a_0 10^0 + a_1 10^M + \cdots + a_p 10^{Mp}$, then $0 \equiv (10^M - 1)k = a_0 10^0 + a_1 10^M + \cdots + a_p 10^{Mp} \equiv a_0 + a_1 + \cdots + a_p \pmod{10^M - 1}$. Since $k > 1$, $p > 0$, so $a_0 + a_1 + \cdots + a_p < a_0 10^0 + a_1 10^M+ \cdots + a_p 10^{Mp}$, so we can find some $j < k$ such that $(10^M - 1)j = a_0 + a_1 + \cdots + a_p$. But \begin{align*} s((10^M - 1)k) &= s(a_0 10^1 + a_1 10^M + \cdots + a_p 10^{Mp}) \\ &= s(a_0) + s(a_1) + \cdots + s(a_p) \\ &\geq s(a_0 + a_1 + \cdots + a_p) \\ &= s((10^M- 1)j) \\ &\geq 9M, \end{align*} which completes our proof.

Solution: We consider the number $c=\frac{10^k-1}{10-1}=111111111.....11$(with $k$ ones) where $k$ is a positive integer with $k>M$. Consider the AP $a_n=n*c$, where we put the said number we considered. We claim that $a_n$ has the desired property. Lemma: For any $n$ if we try to write $a_n$ as sums of powers of $10$, at least $k$ powers of $10$ are used. This is equivalent to saying that sum of digits of $a_n$ are always at least $k$. Proof: Suppose there exist nonnegative integers $p_1,p_2,p_3,...p_l$ such that $l \leq k-1$ and $c|\sum_{i=1}^{l} 10^{p_i}$. If so, we can suppose that all $p_i$ are less than or equal to $k-1$ since we know that $c|10^a-10^{a-k}$ trivially we can always replace $p_i$ with a smaller number if some $p_i$ is more than $k-1$. This trivially implies the existence of one number among $a_1, a_2, a_3,...a_9$ which has digit sum strictly less than $k$. This is a contradiction as trivially all $a_1, a_2,...a_9$(basically $111....1,222...2,....,999...9$ are the $a_1$ to $a_9$, each having $k$ digits) have digit sum at least $k$. So, we can see that all terms of the sequence have digit sum at least $k$. Proved. Since $k>M$, this proves the problem.

Take any positive integer $M.$ Define a common difference $d = \overbrace{11 \dots 11}^{n \ 1s}$ for a very large $n.$ Consider any number that has sum of digits less than or equal to $M.$ Then it can be represented as the sum $$\sum_{i=1}^M a_i$$ where each of $\{ a_1,a_2, \dots a_M \}$ is an integer power of $10,$ or $0.$ So each term may take on only $n+1$ different values $\pmod{d}$ since $10^n \equiv 10^{0} \pmod{d}.$ So there are at most $(n+1)^M$ different values this sum can take, but this is less than $d$ for large choices of $n.$ So there is some residue $a$ where any positive integer $\equiv a \pmod{d}$ has sum of digits exceeding $M$ as desired. $\square$

Simply take $a(10^k-1)$ for large $k$. It is easy to observe by writing $a_1a_2a_3…a_n000…0 - a_1a_2a_3…a_n$ that this has some of digits atleast $9k$. Use the fact $S(a) + S(b) \geq S(a+b)$ for this. And we are done!

Consider the arithmetic progression $(10^n-1)\cdot a+k$ for $a\ge 1$. The common difference is $10^n-1$ which is not divisible by $10$. Suppose for all $n$ there exists $a$ such that for all $k$, $(10^n-1)\cdot a+k$ has sum of digits less than $M$ then \[(10^n-1)\cdot a+k = \sum_{i=1}^{\lfloor M\rfloor}b_i\]Where $b_i$ is $0$ or a power of $10$. Note that the $b_i$ takes $n+1$ different residues $\pmod {10^n-1}$ and so the right hand side takes at most $(n+1)^{\lfloor M\rfloor}$ values. However, the left hand side, varying $k$, takes $10^n$ different values. For large enough values of $n$, this is false. Therefore, there exists values $k$ and $n$ such that the sequence $(10^n-1)\cdot a+k$ satisfies the conditions, for all $M$.

I claim that the sequence of multiples of $10^n-1$ always has digit sum at least $9n$. Clearly this finishes. Denote the digit sum function as $s$. Note that $s(a)+s(b)\ge s(a+b)$ for all $a,b$. We will induct on $m$, where the number is $m(10^n-1)$. Base cases $m=1,2$ are trivial. Note that a number is equivalent to the sum of the chunks of $n$ digits, starting counting from the least significant digit, mod $10^n-1$. Therefore, $10^n-1$ divides the sum of these chunks. Now note that the sum of these chunks is divisible by $10^n-1$, so it has digit sum at least $9n$, so the digit sums of these chunks, added together, is at least $9n$. $\blacksquare$