# BK's PROOFS

A miscellaneous collection of proofs—some light; some significant; others completely frivolous

## Divisibility of "112" by 3

Thu Jan 7 20:12 PST 2021

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.