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| − | ==Problem==
| + | #REDIRECT [[2021_Fall_AMC_12A_Problems/Problem_10]] |
| − | 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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| − | <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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| − | Using module rules, we can find the remainder:
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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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| − | <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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| − | <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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| − | <math>2-7+6-5+2\equiv -2(\text{mod }5)</math>
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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
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| − | ==Solution 2==
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| − | We convert this into base <math>10,</math> so
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| − | <cmath>2 \cdot 9^{10}+7 \cdot 9^9+6 \cdot 9^6+5 \cdot 9+2</cmath>
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| − | Notice that <math>9 \equiv -1 \mod 5,</math>
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| − | <cmath>2 \cdot (-1)^10+7 \cdot (-1)^9+6 \cdot (-1)^6+5 \cdot (-1)+2=2-7+6-5+2</cmath>
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| − | 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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