Difference between revisions of "2017 AMC 10B Problems/Problem 14"

(Solution 1)
m (Video Solution)
 
(9 intermediate revisions by 4 users not shown)
Line 5: Line 5:
  
 
==Solution 1==
 
==Solution 1==
By Fermat's Little Theorem, <math>N^{16} = (N^4)^4 \equiv 1 \text{ (mod 5)}</math> when N is relatively prime to 5. Hence, this happens with probability <math>\boxed{\textbf{(D) } \frac 45}</math>.
+
Notice that we can rewrite <math>N^{16}</math> as <math>(N^{4})^4</math>. By [[Fermat's Little Theorem]], we know that <math>N^{(5-1)} \equiv 1 \pmod {5}</math>  if <math>N \not \equiv 0 \pmod {5}</math>. Therefore for all <math>N \not \equiv 0 \pmod {5}</math> we have <math>N^{16} \equiv (N^{4})^4 \equiv 1^4 \equiv 1 \pmod 5</math>. Since  <math>1\leq N \leq 2020</math>, and <math>2020</math> is divisible by <math>5</math>, <math>\frac{1}{5}</math> of the possible <math>N</math> are divisible by <math>5</math>. Therefore, <math>N^{16} \equiv 1 \pmod {5}</math> with probability <math>1-\frac{1}{5},</math> or <math>\boxed{\textbf{(D) } \frac 45}</math>.
  
 
==Solution 2==
 
==Solution 2==
Line 34: Line 34:
 
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>
 
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>
  
 +
== Video Solution 1==
 +
https://youtu.be/zfChnbMGLVQ?t=2410
 +
 +
~ pi_is_3.14
 +
 +
==Video Solution 2==
 +
https://youtu.be/Oj3Z1JhvoiE
 +
 +
~savannahsolver
  
 
{{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}}

Latest revision as of 19:00, 17 January 2021

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

Notice that we can rewrite $N^{16}$ as $(N^{4})^4$. By Fermat's Little Theorem, we know that $N^{(5-1)} \equiv 1 \pmod {5}$ if $N \not \equiv 0 \pmod {5}$. Therefore for all $N \not \equiv 0 \pmod {5}$ we have $N^{16} \equiv (N^{4})^4 \equiv 1^4 \equiv 1 \pmod 5$. Since $1\leq N \leq 2020$, and $2020$ is divisible by $5$, $\frac{1}{5}$ of the possible $N$ are divisible by $5$. Therefore, $N^{16} \equiv 1 \pmod {5}$ with probability $1-\frac{1}{5},$ or $\boxed{\textbf{(D) } \frac 45}$.

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 $\boxed{\textbf{(D) }\frac{4}{5}}$

Video Solution 1

https://youtu.be/zfChnbMGLVQ?t=2410

~ pi_is_3.14

Video Solution 2

https://youtu.be/Oj3Z1JhvoiE

~savannahsolver

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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

Invalid username
Login to AoPS