Base k vs Modulo k

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Below is a message I sent to our first year students back in 2017 about a connection between modular arithmetic and working in other bases. I don't know if this helps, but somehow arithmetic modulo 10 always seems so easy to explain!


In our introduction to modular arithmetic, I suggested a connection between "modulo \(k\)" and "base \(k\)". For positive integers, when  working modulo \(10\), using decimal notation, we only care about the last decimal digit. (That is the same as the remainder when you divide by \(10\).)

If you work in other bases, the connection is similar: base \(8\) (octal) is not the same thing as modulo \(8\), but for positive integers, the remainder when you divide by \(8\) is the same as the last octal digit. For example \(25\) in base \(10\) is congruent to \(1\) modulo \(8\), and \(25\) base \(10\) is written as \(31\) in octal. Any power of \(31\) (octal) still ends in \(1\) (octal). Returning to decimal notation, \(25^n\) is congruent to \(1\) (mod \(8\)) for all positive integers \(n\), and so \(25^n-1\) is always divisible by \(8\).

Exercise: for which positive integer values of \(n\) is \(6^n+1\) divisible by \(7\)?

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