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 (9's Identity)

We need to first convert $N$ into a regular base-10 integer:

$\begin{align*} N&=27{,}006{,}000{,}052_9 \\ &= 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2 \\ \end{align*}$

Now, consider how the last digit of $9$ changes with changes of the power of $9$ :

$\begin{align*} 9^0=1 \\ 9^1=9 \\ 9^2=81 \\ 9^3=729 \\ 9^4=6561 \\ ...... \end{align*}$

Note that if $x$ is odd:

$\begin{align*} 9^x &\equiv 4\pmod{5} \\ \end{align*}$

If $x$ is even:

$\begin{align*} 9^x &\equiv 1\pmod{5} \\ \end{align*}$

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

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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2021 Fall AMC 12A Problems/Problem 10

Contents

Problem

Solution 1

Solution 2 (9's Identity)

See Also