# 2021 AMC 10A Problems/Problem 11

## Problem

For which of the following integers $b$ is the base-$b$ number $2021_b - 221_b$ not divisible by $3$?

$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$

## Solution 1

We have \begin{align*} 2021_b - 221_b &= (2021_b - 21_b) - (221_b - 21_b) \\ &= 2000_b - 200_b \\ &= 2b^3 - 2b^2 \\ &= 2b^2(b-1), \end{align*} which is divisible by $3$ unless $b\equiv2\pmod{3}.$ The only choice congruent to $2$ modulo $3$ is $\boxed{\textbf{(E)} ~8}.$

~MRENTHUSIASM

## Solution 2 (Easy)

Vertically subtracting $$2021_b - 221_b$$ we see that the ones place becomes 0, and so does the $b^1$ place. Then, we perform a carry (make sure the carry is in base $b$!). Let $b-2 = A$. Then, we have our final number as $$1A00_b$$

Now, when expanding, we see that this number is simply $b^3 - (b - 2)^2$.

Now, notice that the final number will only be congruent to $$b^3-(b-2)^2\equiv0\pmod{3}$$ if either $b\equiv0\pmod{3}$, or if $b\equiv1\pmod{3}$ (because note that $(b - 2)^2$ would become $\equiv1\pmod{3}$, and $b^3$ would become $\equiv1\pmod{3}$ as well, and therefore the final expression would become $1-1\equiv0\pmod{3}$. Therefore, $b$ must be $\equiv2\pmod{3}$. Among the answers, only 8 is $\equiv2\pmod{3}$, and therefore our answer is $\boxed{\textbf{(E)} ~8}.$

- icecreamrolls8

## Solution 3 (Educated Guess)

Note that choices $\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},\textbf{(E)}$ are congruent to $0,1,0,1,2$ modulo $3,$ respectively. Since only one of the choices is correct, we pick $\boxed{\textbf{(E)} ~8}$ due to its uniqueness.

~MRENTHUSIASM

## Video Solution (Simple and Quick)

~ Education, the Study of Everything

## Video Solution

~North America Math Contest Go Go Go

~savannahsolver

~IceMatrix