Difference between revisions of "2017 AMC 10B Problems/Problem 14"
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==Solution 1== | ==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>. Hence, this happens with probability <math>\boxed{\textbf{(D) } \frac 45}</math>. | |
==Solution 2== | ==Solution 2== |
Revision as of 11:46, 31 December 2019
Contents
[hide]Problem
An integer is selected at random in the range . What is the probability that the remainder when is divided by is ?
Solution 1
Notice that we can rewrite as . By Fermat's Little Theorem, we know that if . Therefore for all we have . Hence, 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 .
Solution 3 (Casework)
We can use modular arithmetic for each residue of
If , then
If , then
If , then
If , then
If , then
In out of the cases, the result was , and since each case occurs equally as , the answer is
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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All AMC 10 Problems and Solutions |
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