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

Revision as of 12:03, 8 November 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 (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 (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 $b^3-(b-2)^2\equiv0\pmod{3}$ 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 definition of bases, this number is a polynomial in terms of $b$ ( $(2b^3+0b^2+2b^1+1b^0)-(2b^2+2b^1+1b^0)$ to be exact). Thus, for two values of $b$ that share the same residue modulo $3$ , the two resulting numbers will share the same residue modulo $3$ , so either both or neither will be divisible by $3$ .

Choices $\textbf{(A)} ~3,\textbf{(B)} ~4,\textbf{(C)} ~6,\textbf{(D)} ~7,\textbf{(E)} ~8$ are congruent to $0,1,0,1,2$ modulo $3$ , respectively. This means $\textbf{(A)} ~3$ and $\textbf{(C)} ~6$ are either both wrong or both right (and the latter obviously cannot be the case), and likewise with $\textbf{(B)} ~4$ and $\textbf{(D)} ~7$ . This leaves $\boxed{\textbf{(E)} ~8}$ , the only 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

@@ Line 25: / Line 25: @@
 - icecreamrolls8
-==Solution 3 (Educated Guess)==
+==Solution 3 (Residues)==
-Note that choices <math>\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},\textbf{(E)}</math> are congruent to <math>0,1,0,1,2</math> modulo <math>3,</math> respectively. Since only one of these choices is correct, we pick <math>\boxed{\textbf{(E)} ~8}</math> due to its uniqueness.
+By definition of bases, this number is a polynomial in terms of <math>b</math> (<math>(2b^3+0b^2+2b^1+1b^0)-(2b^2+2b^1+1b^0)</math> to be exact). Thus, for two values of <math>b</math> that share the same residue modulo <math>3</math>, the two resulting numbers will share the same residue modulo <math>3</math>, so either both or neither will be divisible by <math>3</math>.
-~MRENTHUSIASM
+Choices <math>\textbf{(A)} ~3,\textbf{(B)} ~4,\textbf{(C)} ~6,\textbf{(D)} ~7,\textbf{(E)} ~8</math> are congruent to <math>0,1,0,1,2</math> modulo <math>3</math>, respectively. This means <math>\textbf{(A)} ~3</math> and <math>\textbf{(C)} ~6</math> are either both wrong or both right (and the latter obviously cannot be the case), and likewise with <math>\textbf{(B)} ~4</math> and <math>\textbf{(D)} ~7</math>. This leaves <math>\boxed{\textbf{(E)} ~8}</math>, the only choice with a unique residue.
+~MRENTHUSIASM (revised by [[User:emerald_block|emerald_block]])
 ==Video Solution (Simple and Quick)==

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