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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 10"
m (You don't have to do that here, you can just add all the numbers to get 3.)
 
(19 intermediate revisions by 7 users not shown)
Line 11: Line 11:
 
N&=27{,}006{,}000{,}052_9 \\
 
N&=27{,}006{,}000{,}052_9 \\
 
&= 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2 \\
 
&= 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} \\
+
&\equiv 2\cdot(-1)^{10} + 7\cdot(-1)^9 + 6\cdot(-1)^6 + 2 &&\pmod{5} \\
−
&\equiv 2-7+6-5+2 &&\pmod{5} \\
+
&\equiv 2-7+6+2 &&\pmod{5} \\
−
&\equiv -2 &&\pmod{5} \\
 
 
&\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}.
 
&\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
~Aidensharp ~kante314 ~MRENTHUSIASM
+
~Aidensharp ~Kante314 ~MRENTHUSIASM ~anabel.disher
  
 
==Solution 2 (Powers of 9)==
 
==Solution 2 (Powers of 9)==
Line 24: Line 23:
 
Now, consider how the last digit of <math>9</math> changes with changes of the power of <math>9:</math>
 
Now, consider how the last digit of <math>9</math> changes with changes of the power of <math>9:</math>
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
9^0&=1 \\
+
9^0&=1, \\
−
9^1&=9 \\
+
9^1&=9, \\
−
9^2&=81 \\
+
9^2&=\ldots 1, \\
−
9^3&=729 \\
+
9^3&=\ldots 9, \\
−
9^4&=6561 \\
+
9^4&=\ldots 1, \\
 
& \ \vdots
 
& \ \vdots
 
\end{align*}</cmath>
 
\end{align*}</cmath>
Line 43: Line 42:
  
 
~Wilhelm Z
 
~Wilhelm Z
−
==Video Solution by TheBeautyofMath==
 
−
for AMC 10: https://youtu.be/zq3UPu4nwsE?t=358
 
  
−
for AMC 12: https://youtu.be/wlDlByKI7A8?t=885
+
==Remark==
 +
By the long division algorithm, you can work from left to right accumulating the answer, and don't need to count the digits to get the power. You only need to take the current leading digit modulo <math>5</math>, negate it, and add it to the next digit, and repeat the process until you reach the units digit.
  
−
~IceMatrix
+
~oinava
 +
 
 +
==Video Solution ==
 +
https://youtu.be/Mv38a8oanFk
 +
 
 +
~Education, the Study of Everything
 +
 
 +
== Video Solution==
 +
https://youtu.be/SCGzEOOICr4?t=101
 +
 
 +
==Video Solution ==
 +
https://youtu.be/zq3UPu4nwsE?t=358
 +
 
 +
https://youtu.be/wlDlByKI7A8?t=885
  
 
==Video Solution by WhyMath==
 
==Video Solution by WhyMath==
−
https://youtu.be/4faUhsTmS5A ~savannahsolver
+
https://youtu.be/4faUhsTmS5A
 +
 
 +
~savannahsolver
 +
 
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=9|num-a=11}}
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=9|num-a=11}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=11|num-a=13}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:38, 13 August 2024

The following problem is from both the 2021 Fall AMC 10A #12 and 2021 Fall AMC 12A #10, so both problems redirect to this page.

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 (Modular Arithmetic)

Recall that $9\equiv-1\pmod{5}.$ We expand $N$ by the definition of bases: \begin{align*} N&=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 + 2 &&\pmod{5} \\ &\equiv 2-7+6+2 &&\pmod{5} \\ &\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}. \end{align*} ~Aidensharp ~Kante314 ~MRENTHUSIASM ~anabel.disher

Solution 2 (Powers of 9)

We need to first convert $N$ into a regular base-$10$ number: \[N = 27{,}006{,}000{,}052_9 = 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2.\]

Now, consider how the last digit of $9$ changes with changes of the power of $9:$ \begin{align*} 9^0&=1, \\ 9^1&=9, \\ 9^2&=\ldots 1, \\ 9^3&=\ldots 9, \\ 9^4&=\ldots 1, \\ & \ \vdots \end{align*} Note that if $x$ is odd, then $9^x \equiv 4\pmod{5}.$ On the other hand, if $x$ is even, then $9^x \equiv 1\pmod{5}.$

Therefore, we have \begin{align*} N&\equiv 2\cdot(1) + 7\cdot(4) + 6\cdot(1) + 5\cdot(4) + 2\cdot(1) &&\pmod{5} \\ &\equiv 2+28+6+20+2 &&\pmod{5} \\ &\equiv 58 &&\pmod{5} \\ &\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}. \\ \end{align*} Note that for the odd case, $9^x \equiv -1\pmod{5}$ may simplify the process further, as given by Solution 1.

~Wilhelm Z

Remark

By the long division algorithm, you can work from left to right accumulating the answer, and don't need to count the digits to get the power. You only need to take the current leading digit modulo $5$, negate it, and add it to the next digit, and repeat the process until you reach the units digit.

~oinava

Video Solution

https://youtu.be/Mv38a8oanFk

~Education, the Study of Everything

Video Solution

https://youtu.be/SCGzEOOICr4?t=101

Video Solution

https://youtu.be/zq3UPu4nwsE?t=358

https://youtu.be/wlDlByKI7A8?t=885

Video Solution by WhyMath

https://youtu.be/4faUhsTmS5A

~savannahsolver

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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