# 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 (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 $$b^2(b+1)\equiv0\pmod{3}$$ 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 $\equiv\pmod{3}$, and therefore our answer is $\boxed{\textbf{(E)} ~8}.$

- icecreamrolls8

## Video Solution (Simple and Quick)

~ Education, the Study of Everything

## Video Solution

~North America Math Contest Go Go Go

~savannahsolver

~IceMatrix

## See Also

 2021 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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

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