2017 AMC 10B Problems/Problem 14: Difference between revisions
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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 probablilty 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 probablilty that the remainder when <math>N^{16}</math> is divided by <math>5</math> is <math>1</math>? | ||
==Solution== | ==Solution== | ||
By Fermat's Little Theorem, <math>N^{16} = (N^4)^4 \equiv 1 \text{ (mod 5)}</math> when N is relatively prime to 5. However, this happens with probability <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 12:29, 16 February 2017
Problem
An integer 
is selected at random in the range 
. What is the probablilty that the remainder when 
is divided by 
is 
?
Solution
By Fermat's Little Theorem, 
when N is relatively prime to 5. However, this happens with probability 
.
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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