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

Revision as of 13:48, 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 (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 $\equiv2\pmod{3}$ .

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

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

@@ Line 9: / Line 9: @@
 ~MRENTHUSIASM
-==Solution 2==
+==Solution 2 (Easy)==
-Vertically subtracting <cmath>2021_b - 221_b</cmath> we see that the ones place becomes 0, the <math>b^1</math> place becomes 0 as well. Now, at the <math>b^2</math> 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 <math>b^3</math> place in <math>2021_b</math> equal to 1. Thus, we have our final number as <cmath>1100_b</cmath>
+Vertically subtracting <cmath>2021_b - 221_b</cmath> we see that the ones place becomes 0, and so does the <math>b^1</math> place. Then, we perform a carry (make sure the carry is in <math>base b</math>!). Let <math>b-2 = A</math>. Then, we have our final number as <cmath>1A00_b</cmath>
-Now, when expanding, we see that this is simply <math>b^3 + b^2</math>, which factors into <cmath>b^2(b+1)</cmath>
+Now, when expanding, we see that this number is simply <math>b^3 - (b - 2)^2</math>.
-Now, notice that the final number will only be congruent to <cmath>b^2(b+1)\equiv0\pmod{3}</cmath> if either $b\equiv2\pmod{3}, or
+Now, notice that the final number will only be congruent to <cmath>b^2(b+1)\equiv0\pmod{3}</cmath> if either <math>b\equiv0\pmod{3}</math>, or if <math>b\equiv1\pmod{3}</math> (because note that <math>(b - 2)^2</math> would become <math>\equiv1\pmod{3}</math>, and b^3 would become <math>\equiv1\pmod{3}</math> as well, and therefore the final expression would become <math>1-1\equiv0\pmod{3}</math>. Therefore, <math>b</math> must be <math>\equiv2\pmod{3}</math>.
 ==Video Solution (Simple and Quick)==

Page

Toolbox

Search