2021 Fall AMC 12A Problems/Problem 10

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

Recall that $9\equiv-1\pmod{5}.$ We expand $N$ by the definition of bases: $\begin{align*} N&=27{,}006{,}000{,}052_9 \\ &= 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2 \\ &\equiv 2\cdot(-1)^{10} + 7\cdot(-1)^9 + 6\cdot(-1)^6 + 5\cdot(-1) + 2 &&\pmod{5} \\ &= 2-7+6-5+2 \\ &= -2 \\ &\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}. \end{align*}$ ~Aidensharp ~kante314 ~MRENTHUSIASM

Solution 2

In the base-10 representation, modulo 5, we have $\begin{align*} N & = 2 + 5 \cdot 9 + 6 \cdot 9^6 + 7 \cdot 9^9 + 2 \cdot 9^{10} \\ & \equiv 2 + 0 \cdot \left( -1 \right) + 1 \cdot \left( - 1 \right)^6 + 2 \cdot \left( - 1 \right)^9 + 2 \cdot \left( -1 \right)^{10} \\ & \equiv 2 + 0 + 1 - 2 + 2 \\ & \equiv 3 . \end{align*}$

Therefore, the answer is $\boxed{\textbf{(D) }3}$ .

~Steven Chen (www.professorchenedu.com)

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.

Page

Toolbox

Search

2021 Fall AMC 12A Problems/Problem 10

Contents

Problem

Solution 1

Solution 2

See Also