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

(Solution 2)
(Solution 2)
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==Solution 2==
 
==Solution 2==
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 <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, 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>
  
 
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 is simply <math>b^3 + b^2</math>, which factors into <cmath>b^2(b+1)</cmath>
  
Now, notice that the final number will only be
+
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
  
 
==Video Solution (Simple and Quick)==
 
==Video Solution (Simple and Quick)==

Revision as of 14:37, 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 \[1100_b\]

Now, when expanding, we see that this is simply $b^3 + b^2$, which factors into \[b^2(b+1)\]

Now, notice that the final number will only be congruent to \[b^2(b+1)\equiv0\pmod{3}\] if either $b\equiv2\pmod{3}, or

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

See Also

2021 AMC 10A (ProblemsAnswer KeyResources)
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

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