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

m (→‎Problem)
(Corrected the sign error, and reformatted. Made Sol 1 more concise.)
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==Problem==
 
==Problem==
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The base-nine representation of the number <math>N</math> is <math>27,006,000,052_{\text{nine}}.</math> What is the remainder when <math>N</math> is divided by <math>5?</math>
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The base-nine representation of the number <math>N</math> is <math>27{,}006{,}000{,}052_{\text{nine}}.</math> What is the remainder when <math>N</math> is divided by <math>5?</math>
  
 
<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>
  
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==Solution==
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==Solution 1==
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Using module rules, we can find the remainder:
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Recall that <math>9\equiv-1\pmod{5}.</math> We have
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<cmath>\begin{align*}
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<math>27,006,000,052_9 = 2(9^{10})+7(9^9)+6(9^6)+5(9^1)+2</math>
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27{,}006{,}000{,}052_9 &= 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2 \\
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&\equiv 2\cdot(-1)^{10} + 7\cdot(-1)^9 + 6\cdot(-1)^6 + 5\cdot(-1) + 2 &&\pmod{5} \\
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<math>2(9^{10})+7(9^9)+6(9^6)+5(9^1)+2\equiv 2({-}1^{10})+7({-}1^9)+6({-}1^6)+5({-}1^1)+2 (\text{mod }5)</math>
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&= 2-7+6-5+2 \\
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&= -2 \\
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<math>2({-}1^{10})+7({-}1^9)+6({-}1^6)+5({-}1^1)+2\equiv 2-7+6-5+2(\text{mod }5)</math>
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&\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}.
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\end{align*}</cmath>
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<math>2-7+6-5+2\equiv -2(\text{mod }5)</math>
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-Aidensharp ~MRENTHUSIASM
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<math>-2\equiv 3(\text{mod }5)</math>
 
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Thus, the answer is <math>\boxed{\textbf{(D)}\ 3}</math>.
 
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-Aidensharp
 
  
 
==Solution 2==
 
==Solution 2==
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Simplifying, <math>-2 \mod 5 \implies 3 \mod 5.</math> So, the answer is <math>\boxed{3}.</math>
 
Simplifying, <math>-2 \mod 5 \implies 3 \mod 5.</math> So, the answer is <math>\boxed{3}.</math>
  
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- kante314
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-kante314
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==See Also==
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{{AMC10 box|year=2021 Fall|ab=A|num-b=11|num-a=13}}
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{{MAA Notice}}

Revision as of 20:26, 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 1

Recall that $9\equiv-1\pmod{5}.$ We have \begin{align*} 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 ~MRENTHUSIASM

Solution 2

We convert this into base $10,$ so \[2 \cdot 9^{10}+7 \cdot 9^9+6 \cdot 9^6+5 \cdot 9+2\] Notice that $9 \equiv -1 \mod 5,$ \[2 \cdot (-1)^10+7 \cdot (-1)^9+6 \cdot (-1)^6+5 \cdot (-1)+2=2-7+6-5+2\] Simplifying, $-2 \mod 5 \implies 3 \mod 5.$ So, the answer is $\boxed{3}.$

-kante314

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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