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

Revision as of 11:55, 31 December 2019

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

2017 AMC 10B (Problems • Answer Key • Resources)
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.

@@ Line 5: / Line 5: @@
 ==Solution 1==
-Notice that we can rewrite <math>N^{16}</math> as <math>N^{4^2}</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 \cdot N^4 \equiv 1 \cdot 1 \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>.
+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==

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