# 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 $$2021_b - 221_b = 2000_b - 200_b = 2b^3 - 2b^2 = 2b^2(b-1).$$ This expression 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

Vertically subtracting $$2021_b - 221_b$$ we see that the ones place becomes 0, the $b^1$ place becomes 0 as well. Now, at the $b^2$ place, we must perform a carry, but instead of incrementing the place's value by 10 like we normally would in base 10, we do so by $b$, and make the $b^3$ place in $2021_b$ equal to 1. Thus, we have our final number as $$1100_b$$

Now, when expanding, we see that this is simply $b^3 + b^2$, which factors into $$b^2(b+1)$$

Now, notice that the final number will only be congruent to $$b^2(b+1)\equiv0\pmod{3}$$ if either \$b\equiv2\pmod{3}, or

## Video Solution (Simple and Quick)

~ Education, the Study of Everything

## Video Solution

~North America Math Contest Go Go Go

~savannahsolver

## Video Solution by TheBeautyofMath

~IceMatrix

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 