2021 Fall AMC 10A Problems/Problem 12

Revision as of 20:26, 22 November 2021 by MRENTHUSIASM (talk | contribs) (Corrected the sign error, and reformatted. Made Sol 1 more concise.)

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AMC 10 Problems and Solutions

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