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

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==Solution 2 (Powers of 9)==
 
==Solution 2 (Powers of 9)==
  
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We need to first convert <math>N</math> into a regular base-<math>10</math> integer: <cmath>N = 27{,}006{,}000{,}052_9 = 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2.</cmath>
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We need to first convert <math>N</math> into a regular base-<math>10</math> number: <cmath>N = 27{,}006{,}000{,}052_9 = 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2.</cmath>
  
 
Now, consider how the last digit of <math>9</math> changes with changes of the power of <math>9:</math>
 
Now, consider how the last digit of <math>9</math> changes with changes of the power of <math>9:</math>

Revision as of 20:09, 26 November 2021

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

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&=729 \\ 9^4&=6561 \\ & \ \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

See Also

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

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