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

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==Problem==
 
==Problem==
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)}\
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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>
3 \qquad\textbf{(E)}\ 4</math>
 
  
 
==Solution==
 
==Solution==

Revision as of 21:15, 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