Problem 6

For even $w\ge 2$, we construct by induction an even alternating $w$-digit multiple $g(w)$ of $2^{w+1}$ and an even alternating $w$-digit multiple $h(w)$ of $5^w$ (both having odd leading digit since they have even length and end in an even digit). For the base $w=2$, take $g(2)=32$ (divisible by $8$) and $h(2)=50$ (divisible by $25$). To pass from $w$ to $w+2$, prepend a two-digit block from the $25$-element set $\{10,12,\dots,18,30,32,\dots,98\}$, i.e. add $10^w(10a+b)$ for $a\in\{1,3,5,7,9\}$ odd and $b\in\{0,2,4,6,8\}$ even: since $10^w(10a+b)=2^{w+1}\cdot 5^w(5a+b/2)$ is divisible by $2^{w+1}$ and covers all four residues of $2^{w+1}$ modulo $2^{w+3}$ as $a\in\{1,3,5,7,9\}$,$b\in\{0,2,4,6,8\}$ vary, some choice extends $g(w)$ to $g(w+2)$ divisible by $2^{w+3}$; similarly, the $25$ choices of $10^w(10a+b)$ cover all $25$ multiples of $5^w$ modulo $5^{w+2}$ (as $\gcd(2^w,25)=1$), so some choice extends $h(w)$ to $h(w+2)$ divisible by $5^{w+2}$.