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Difference between revisions of "2017 AMC 10B Problems/Problem 14"
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==Solution 2==
 
==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 <math>0-9</math> .
 
Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits <math>0-9</math> .
−
The pattern for <math>0</math> is <math>0</math>, no matter what power, so <math>0</math> doesn't work. 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> <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> <math>\boxed{\textbf{(D) } \frac 45}</math>.
 +
 
 
==See Also==
 
==See Also==
 
{{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 12:43, 16 February 2017

Problem

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

Solution

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 $5$ $\boxed{\textbf{(D) } \frac 45}$.

See Also

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

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