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  text: 'In general, given enough $n$-digits, you can emulate an $m$-digit, for any $m, n \\in$ [45h $\\mathbb N$]. If $m < n,$ you can emulate an $m$-digit using just one $n$-digit &mdash; in other words, you can use a [-42d] like a $7$-digit if you want to, by just ignoring three of the possible ways to set the digit wheel. If $m > n,$ things are a bit more difficult, but only slightly.\n\nBasically, with 2 $n$-digits, you can emulate a $n^2$-digit, as follows. Using your two $n$-digits, encode a number $(x, y)$ where $0 \\le x < n$ and $0 \\le y < n$. Interpret $(x, y)$ as $xn + y.$ You have now encoded a number between 0 (if  $x = y = 0$) and $n^2 - 1$ (if $x = y = n-1$). Congratulations, you just used two $n$-digits to make an $n^2$ digit!\n\nYou can use the same strategy to emulate $n^3$-digits (interpret $(x, y, z)$ as $xn^2 + yn + z$), $n^4$-digits (you get the picture), and so on. Now, to emulate an $m$-digit, just pick an exponent $a$ such that $n^a > m,$ collect $a$ copies of an $n$-digit, and you're done.\n\nThis isn't necessarily the most efficient way to use $n$-digits to encode $m$-digits. For example, if $m$ is 1,000,001 and $n$ is 10, then you need seven 10-digits. Seven 10-digits are enough to emulate a 10-million-digit, whereas $m$ is a mere million-and-one-digit &mdash; paying for a 10-million-digit when all you needed was an $m$-digit seems a bit excessive. For some different methods you can use to recover your losses when encoding one type of digit using another type of digit, see [44l] and [3ty]. (These techniques are fairly useful in practice, given that modern computers encode everything using [3p0 bits], i.e. 2-digits, and so it's useful to know how to efficiently encode $m$-messages using bits when $m$ is pretty far from the nearest power of 2.)',
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