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Difference between revisions of "2017 AMC 10B Problems/Problem 14"
m (Made aesthetic changes to Solution 3 and made it slightly more complete.)
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The pattern for <math>0</math> is <math>0</math>, no matter what power, so <math>0</math> doesn't work. Likewise, the pattern for <math>5</math> is always <math>5</math>. Doing the same for the rest of the digits, we find that the units digits of <math>1^{16}</math>, <math>2^{16}</math> ,<math>3^{16}</math>, <math>4^{16}</math> ,<math>6^{16}</math>, <math>7^{16}</math> ,<math>8^{16}</math> and <math>9^{16}</math> all have the remainder of <math>1</math> when divided by <math>5</math>, so <math>\boxed{\textbf{(D) } \frac 45}</math>.
 
The pattern for <math>0</math> is <math>0</math>, no matter what power, so <math>0</math> doesn't work. Likewise, the pattern for <math>5</math> is always <math>5</math>. Doing the same for the rest of the digits, we find that the units digits of <math>1^{16}</math>, <math>2^{16}</math> ,<math>3^{16}</math>, <math>4^{16}</math> ,<math>6^{16}</math>, <math>7^{16}</math> ,<math>8^{16}</math> and <math>9^{16}</math> all have the remainder of <math>1</math> when divided by <math>5</math>, so <math>\boxed{\textbf{(D) } \frac 45}</math>.
  
−
==Solution 3 (casework)==
+
==Solution 3 (Casework)==
−
We can use modular arithmetic for each case of <math>n(\text{mod }5)</math>
+
We can use modular arithmetic for each residue of <math>n \pmod 5</math>
  
  
  
−
If <math>n \equiv 0(\text{mod }5)</math>, then <math>n^{16} \equiv 0^{16} \equiv 0 (\text{mod }5)</math>
+
If <math>n \equiv 0 \pmod 5</math>, then <math>n^{16} \equiv 0^{16} \equiv 0 \pmod 5</math>
  
  
−
If <math>n \equiv 1(\text{mod }5)</math>, then <math>n^{16} \equiv 1^{16} \equiv 1 (\text{mod }5)</math>
+
If <math>n \equiv 1 \pmod 5</math>, then <math>n^{16} \equiv 1^{16} \equiv 1 \pmod 5</math>
  
  
−
If <math>n \equiv 2(\text{mod }5)</math>, then <math>n^{16} \equiv (n^2)^8 \equiv 4^8 \equiv (-1)^8 \equiv 1 (\text{mod }5)</math>
+
If <math>n \equiv 2 \pmod 5</math>, then <math>n^{16} \equiv (n^2)^8 \equiv (2^2)^8 \equiv 4^8 \equiv (-1)^8 \equiv 1 \pmod 5</math>
  
  
−
If <math>n \equiv 3(\text{mod }5)</math>, then <math>n^{16} \equiv (n^4)^4 \equiv 81^4 \equiv (1)^4 \equiv 1 (\text{mod }5)</math>
+
If <math>n \equiv 3 \pmod 5</math>, then <math>n^{16} \equiv (n^2)^8 \equiv (3^2)^8 \equiv 9^8 \equiv (-1)^8 \equiv 1 \pmod 5</math>
  
  
−
If <math>n \equiv 4(\text{mod }5)</math>, then <math>n^{16} \equiv (-1)^{16} \equiv 1 (\text{mod }5)</math>
+
If <math>n \equiv 4 \pmod 5</math>, then <math>n^{16} \equiv 4^{16} \equiv (-1)^{16} \equiv 1 \pmod 5</math>
  
  
  
−
In <math>4</math> out of the <math>5</math> cases, the result was <math>1(\text{mod }5)</math>, and since each case occurs equally, the answer is <math>\boxed{\textbf{(D) }\frac{4}{5}}</math>
+
In <math>4</math> out of the <math>5</math> cases, the result was <math>1 \pmod 5</math>, and since each case occurs equally as <math>2020 \equiv 0 \pmod 5</math>, the answer is <math>\boxed{\textbf{(D) }\frac{4}{5}}</math>
  
  
 
{{AMC10 box|year=2017|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2017|ab=B|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:25, 26 December 2017

Problem

An integer $N$ is selected at random in the range $1\leq N \leq 2020$ . What is the probability that the remainder when $N^{16}$ is divided by $5$ is $1$?

$\textbf{(A)}\ \frac{1}{5}\qquad\textbf{(B)}\ \frac{2}{5}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{4}{5}\qquad\textbf{(E)}\ 1$

Solution 1

By Fermat's Little Theorem, $N^{16} = (N^4)^4 \equiv 1 \text{ (mod 5)}$ when N is relatively prime to 5. However, this happens with probability $\boxed{\textbf{(D) } \frac 45}$.

Solution 2

Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits $0-9$ . The pattern for $0$ is $0$, no matter what power, so $0$ doesn't work. Likewise, the pattern for $5$ is always $5$. Doing the same for the rest of the digits, we find that the units digits of $1^{16}$, $2^{16}$ ,$3^{16}$, $4^{16}$ ,$6^{16}$, $7^{16}$ ,$8^{16}$ and $9^{16}$ all have the remainder of $1$ when divided by $5$, so $\boxed{\textbf{(D) } \frac 45}$.

Solution 3 (Casework)

We can use modular arithmetic for each residue of $n \pmod 5$


If $n \equiv 0 \pmod 5$, then $n^{16} \equiv 0^{16} \equiv 0 \pmod 5$


If $n \equiv 1 \pmod 5$, then $n^{16} \equiv 1^{16} \equiv 1 \pmod 5$


If $n \equiv 2 \pmod 5$, then $n^{16} \equiv (n^2)^8 \equiv (2^2)^8 \equiv 4^8 \equiv (-1)^8 \equiv 1 \pmod 5$


If $n \equiv 3 \pmod 5$, then $n^{16} \equiv (n^2)^8 \equiv (3^2)^8 \equiv 9^8 \equiv (-1)^8 \equiv 1 \pmod 5$


If $n \equiv 4 \pmod 5$, then $n^{16} \equiv 4^{16} \equiv (-1)^{16} \equiv 1 \pmod 5$


In $4$ out of the $5$ cases, the result was $1 \pmod 5$, and since each case occurs equally as $2020 \equiv 0 \pmod 5$, the answer is $\boxed{\textbf{(D) }\frac{4}{5}}$


2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AMC 10 Problems and Solutions

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