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

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-Aidensharp
 
-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

Revision as of 20:08, 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

Using module rules, we can find the remainder:

$27,006,000,052_9 = 2(9^{10})+7(9^9)+6(9^6)+5(9^1)+2$

$2(9^{10})+7(9^9)+6(9^6)+5(9^1)+2\equiv 2({-}1^{10})+7({-}1^9)+6({-}1^6)+5({-}1^1)+2 (\text{mod }5)$

$2({-}1^{10})+7({-}1^9)+6({-}1^6)+5({-}1^1)+2\equiv 2-7+6-5+2(\text{mod }5)$

$2-7+6-5+2\equiv -2(\text{mod }5)$

$-2\equiv 3(\text{mod }5)$

Thus, the answer is $\boxed{\textbf{(D)}\ 3}$.

-Aidensharp

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

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