Problem 6

Suppose $\gcd(m,10)=1$, and fix any residue $b\in\{0,1,\dots,m-1\}$ modulo $m$. For any $M$ divisible by $\varphi(99m)$, Euler's theorem gives $100^M\equiv 1\pmod{99m}$, so the alternating number $A_M=10(100^M-1)/99$ consisting of $M$ copies of $10$ satisfies $A_M\equiv 0\pmod m$. Moreover, for any integer $r\ge 1$ with $2k_r-1\equiv 0\pmod{\varphi(m)}$, adding $2\cdot 10^{2k_r-1}$ to $A_M=10(100^M-1)/99$ flips one of the odd-positioned digits $1$ to $3$ (preserving the alternating odd-even pattern) while shifting the residue modulo $m$ by $+2$ since $10^{2k_r-1}\equiv 1\pmod m$. Choosing $M$ large enough and flipping $c\equiv 2^{-1}b\pmod m$ such digits at distinct exponents $2k_r-1\equiv 0\pmod{\varphi(m)}$ gives an even alternating number $f(b)$ with $f(b)\equiv b\pmod m$.