Difference between revisions of "2017 AMC 10B Problems/Problem 14"
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==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>. | 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>. | ||
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| + | ==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> . | ||
| + | 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> | ||
==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 12:33, 16 February 2017
Contents
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 
.
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. 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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
