## Divisibility of "112" by 3

*Originally presented in tweet form, inspired by this tweet.*

Hm. After having tried a few numbers, I *think* "112", as a number expressed in any base, is not divisible by 3. It's surprising to me, so now I need to find a proof of this statement—the proof is found and presented below.

Suppose "112" (hereafter represented as \([112]\)) is expressed in base \(a\), where \(a \ge 3\). So, \[ [112] = a^2 + a + 2 = a(a+1)+2.\] Since if \( a\%3=0\) or \(a\%3=2\) we are done (the first term is divisible by 3 and the remainder is not divisible by 3), suppose that \(a\%3=1\). Since \(a\%3=1\), there exists an integer \(b\) such that \(3b+1=a\).

Plugging in this expression for \(a\) in the expression for \([112]\), we get, \[\begin{align*} [112] & = (3b+1)(3b+1+1)+2 \\ & = 9b^2+9b+4. \end{align*}\] The first two terms are divisible by 3, and the last term is 4, so \([a(a+1)+2]\%3=1\). That is, \([112]\) in base \(a\) is not divisible by 3, as we wanted to prove.