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 (Factor)

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 (Vertical Subtraction)

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 (Residues)

By the definition of bases, we have $2021_b - 221_b = \left(2b^3+2b+1\right) - \left(2b^2+2b+1\right).$

For two values of $b$ with the same residue modulo $3,$ each of the expressions $2b^3+2b+1$ and $2b^2+2b+1$ have the same residue modulo $3.$ If $2021_b - 221_b$ is divisible by $3,$ then $2b^3+2b+1$ and $2b^2+2b+1$ are both or neither divisible by $3.$

Note that choices $\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},\textbf{(E)}$ are congruent to $0,1,0,1,2$ modulo $3,$ respectively. This means $\textbf{(A)}$ and $\textbf{(C)}$ are either both correct or both incorrect. Since there is only one correct answer, $\textbf{(A)}$ and $\textbf{(C)}$ are both incorrect. Similarly, we conclude that $\textbf{(B)}$ and $\textbf{(D)}$ are both incorrect. This leaves $\boxed{\textbf{(E)} ~8},$ the answer choice with a unique residue.

~MRENTHUSIASM (revised by emerald_block)

Video Solution (Simple and Quick)

https://youtu.be/1TZ1uI9z8fU

~ Education, the Study of Everything

Video Solution

https://www.youtube.com/watch?v=XBfRVYx64dA&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=10

~North America Math Contest Go Go Go

Video Solution 3

https://youtu.be/zYIuBXDhJJA

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/t-EEP2V4nAE

~IceMatrix

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 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

Page

Toolbox

Search