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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 1==
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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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| − | 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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| − | &= 2-7+6-5+2 \\
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| − | &= -2 \\
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| − | &\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}.
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| − | \end{align*}</cmath>
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| − | -Aidensharp ~MRENTHUSIASM
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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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| − | ==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}}
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