Problem #PRU-5082

Problem

Let’s "prove" that the number $1$ is a multiple of $3$. We will use the symbol $\equiv$ to denote "congruent modulo $3$". Thus, what we need to prove is that $1\equiv 0$ modulo $3$. Let’s see: $1\equiv 4$ modulo $3$ means that $2^1\equiv 2^4$ modulo $3$, thus $2\equiv 16$ modulo $3$, however $16$ gives the remainder $1$ after division by $3$, thus we get $2\equiv 1$ modulo $3$, next $2-1\equiv 1-1$ modulo $3$, and thus $1\equiv 0$ modulo $3$. Which means that $1$ is divisible by $3$.