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

(Solution 1)
(Solution 2: Removed repetitive solution. Prof. Chen agreed to this through PM ...)
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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 00:53, 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

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

See Also

2021 Fall AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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