# Difference between revisions of "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.

## 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. 