Difference between revisions of "2017 AMC 10B Problems/Problem 14"
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In <math>4</math> out of the <math>5</math> cases, the result was <math>1 \pmod 5</math>, and since each case occurs equally as <math>2020 \equiv 0 \pmod 5</math>, the answer is <math>\boxed{\textbf{(D) }\frac{4}{5}}</math> | In <math>4</math> out of the <math>5</math> cases, the result was <math>1 \pmod 5</math>, and since each case occurs equally as <math>2020 \equiv 0 \pmod 5</math>, the answer is <math>\boxed{\textbf{(D) }\frac{4}{5}}</math> | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/zfChnbMGLVQ?t=2410 | ||
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+ | ~ pi_is_3.14 | ||
==Video Solution== | ==Video Solution== |
Revision as of 01:17, 17 January 2021
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
Notice that we can rewrite as . By Fermat's Little Theorem, we know that if . Therefore for all we have . Since , and is divisible by , of the possible are divisible by . Therefore, with probability or .
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
Video Solution
https://youtu.be/zfChnbMGLVQ?t=2410
~ pi_is_3.14
Video Solution
~savannahsolver
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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