Difference between revisions of "2021 Fall AMC 12A Problems/Problem 10"

Revision as of 19:16, 26 September 2023

The following problem is from both the 2021 Fall AMC 10A #12 and 2021 Fall AMC 12A #10, so both problems redirect to this page.

Problem

The base-nine representation of the number $N$ is $27{,}006{,}000{,}052_{\text{nine}}.$ What is the remainder when $N$ is divided by $5?$

$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) }4$

Solution 1 (Modular Arithmetic)

Recall that $9\equiv-1\pmod{5}.$ We expand $N$ by the definition of bases: $\begin{align*} N&=27{,}006{,}000{,}052_9 \\ &= 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2 \\ &\equiv 2\cdot(-1)^{10} + 7\cdot(-1)^9 + 6\cdot(-1)^6 + 5\cdot(-1) + 2 &&\pmod{5} \\ &\equiv 2-7+6-5+2 &&\pmod{5} \\ &\equiv -2 &&\pmod{5} \\ &\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}. \end{align*}$

~Aidensharp ~Kante314 ~MRENTHUSIASM

(Note: by the long division algorithm, you can work from left to right accumulating the answer, and don't need to count the digits to get the power. You only need to take the current leading digit mod 5, negate it, and add it to the next digit, and repeat the process until you you reach the units digit.(~oinava)

Solution 2 (Powers of 9)

We need to first convert $N$ into a regular base- $10$ number: $N = 27{,}006{,}000{,}052_9 = 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2.$

Now, consider how the last digit of $9$ changes with changes of the power of $9:$ $\begin{align*} 9^0&=1 \\ 9^1&=9 \\ 9^2&=81 \\ 9^3&=\dots 9 \\ 9^4&= \dots 1 \\ & \ \vdots \end{align*}$ Note that if $x$ is odd, then $9^x \equiv 4\pmod{5}.$ On the other hand, if $x$ is even, then $9^x \equiv 1\pmod{5}.$

Therefore, we have $\begin{align*} N&\equiv 2\cdot(1) + 7\cdot(4) + 6\cdot(1) + 5\cdot(4) + 2\cdot(1) &&\pmod{5} \\ &\equiv 2+28+6+20+2 &&\pmod{5} \\ &\equiv 58 &&\pmod{5} \\ &\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}. \\ \end{align*}$ Note that for the odd case, $9^x \equiv -1\pmod{5}$ may simplify the process further, as given by Solution 1.

~Wilhelm Z

Video Solution

https://youtu.be/Mv38a8oanFk

~Education, the Study of Everything

Video Solution

https://youtu.be/SCGzEOOICr4?t=101

Video Solution

https://youtu.be/zq3UPu4nwsE?t=358

https://youtu.be/wlDlByKI7A8?t=885

Video Solution by WhyMath

https://youtu.be/4faUhsTmS5A

~savannahsolver

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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 12 Problems and Solutions

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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.

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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 10"

Revision as of 19:16, 26 September 2023

Contents

Problem

Solution 1 (Modular Arithmetic)

Solution 2 (Powers of 9)

Video Solution

Video Solution

Video Solution

Video Solution by WhyMath

See Also

@@ Line 16: / Line 16: @@
 &\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}.
 \end{align*}</cmath>
 ~Aidensharp ~Kante314 ~MRENTHUSIASM
+(Note: by the long division algorithm, you can work from left to right accumulating the answer, and don't need to count the digits to get the power. You only need to take the current leading digit mod 5,  negate it, and add it to the next digit, and repeat the process until you you reach the units digit.(~oinava)
 ==Solution 2 (Powers of 9)==