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

(Created page with "{{duplicate|2021 Fall AMC 10A #12 and 2021 Fall AMC 12A #10}} ==Problem== The base-nine re...")
 
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<math>\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) }4</math>
 
<math>\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) }4</math>
  
−
==Solution==
+
==Solution 1==
 
Recall that <math>9\equiv-1\pmod{5}.</math> We expand <math>N</math> by the definition of bases:
 
Recall that <math>9\equiv-1\pmod{5}.</math> We expand <math>N</math> by the definition of bases:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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\end{align*}</cmath>
 
\end{align*}</cmath>
 
~Aidensharp ~kante314 ~MRENTHUSIASM
 
~Aidensharp ~kante314 ~MRENTHUSIASM
 +
 +
== Solution 2 ==
 +
In the base-10 representation, modulo 5, we have
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(D) }3}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)
  
 
==See Also==
 
==See Also==

Revision as of 20:08, 25 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

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)

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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