Difference between revisions of "2021 AMC 10A Problems/Problem 11"

Revision as of 13:30, 5 May 2021

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 $$ (Error compiling LaTeX. Unknown error_msg)1100_b.

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
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All AMC 10 Problems and Solutions

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

@@ Line 10: / Line 10: @@
 ==Solution 2==
-Vertically subtracting <cmath>2021_b - </cmath>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 <cmath>2021_b equal to 1. Thus, we have our final number as </cmath>1100_b.
+Vertically subtracting <cmath>2021_b - 221_b</cmath> we see that the ones place becomes 0, the b^1 place becomes 0 as well. Now, at the <cmath>b^2</cmath> 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 <math>b</math>, and make the <cmath>b^3</cmath> place in <cmath>2021_b</cmath> equal to 1. Thus, we have our final number as <math></math>1100_b.
 ==Video Solution (Simple and Quick)==

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