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

(Corrected the sign error, and reformatted. Made Sol 1 more concise.)
m (The solutions in this page are very similar. So, I combined them and give credits to everyone ...)
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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 1==
+
==Solution==
−
Recall that <math>9\equiv-1\pmod{5}.</math> We have
+
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*}
−
27{,}006{,}000{,}052_9 &= 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2 \\
+
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\cdot(-1)^{10} + 7\cdot(-1)^9 + 6\cdot(-1)^6 + 5\cdot(-1) + 2 &&\pmod{5} \\
 
&= 2-7+6-5+2 \\
 
&= 2-7+6-5+2 \\
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&\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}.
 
&\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
-Aidensharp ~MRENTHUSIASM
+
-Aidensharp ~kante314 ~MRENTHUSIASM
−
 
 
−
==Solution 2==
 
−
We convert this into base <math>10,</math> so
 
−
<cmath>2 \cdot 9^{10}+7 \cdot 9^9+6 \cdot 9^6+5 \cdot 9+2</cmath>
 
−
Notice that <math>9 \equiv -1 \mod 5,</math>
 
−
<cmath>2 \cdot (-1)^10+7 \cdot (-1)^9+6 \cdot (-1)^6+5 \cdot (-1)+2=2-7+6-5+2</cmath>
 
−
Simplifying, <math>-2 \mod 5 \implies 3 \mod 5.</math> So, the answer is <math>\boxed{3}.</math>
 
−
 
 
−
-kante314
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=11|num-a=13}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:32, 22 November 2021

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 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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