Difference between revisions of "2017 AMC 10B Problems/Problem 14"
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==Problem== | ==Problem== | ||
An integer <math>N</math> is selected at random in the range <math>1\leq N \leq 2020</math> . What is the probability that the remainder when <math>N^{16}</math> is divided by <math>5</math> is <math>1</math>? | An integer <math>N</math> is selected at random in the range <math>1\leq N \leq 2020</math> . What is the probability that the remainder when <math>N^{16}</math> is divided by <math>5</math> is <math>1</math>? | ||
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| + | <math>\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</math> | ||
==Solution 1== | ==Solution 1== | ||
Revision as of 16:42, 18 February 2017
Contents
Problem
An integer 
is selected at random in the range 
. What is the probability that the remainder when 
is divided by 
is 
?
Solution 1
By Fermat's Little Theorem, 
when N is relatively prime to 5. However, this happens with probability 
.
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 
.
The pattern for 
is , no matter what power, so doesn't work. Likewise, the pattern for is always . Doing the same for the rest of the digits, we find that the units digits of 
, 
,
, 
,
, 
,
and 
all have the remainder of when divided by , so .
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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