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 | + | 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>? |
− | ==Solution== | + | |
− | By Fermat's Little Theorem, <math>N^{16} | + | <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== | ||
+ | Notice that we can rewrite <math>N^{16}</math> as <math>(N^{4})^4</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})^4 \equiv 1^4 \equiv 1 \pmod 5</math>. Since <math>1\leq N \leq 2020</math>, and <math>2020</math> is divisible by <math>5</math>, <math>\frac{1}{5}</math> of the possible <math>N</math> are divisible by <math>5</math>. Therefore, <math>N^{16} \equiv 1 \pmod {5}</math> with probability <math>1-\frac{1}{5},</math> or <math>\boxed{\textbf{(D) } \frac 45}</math>. | ||
==Solution 2== | ==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> . | 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> <math>\boxed{\textbf{(D) } \frac 45}</math>. | + | The pattern for <math>0</math> is <math>0</math>, no matter what power, so <math>0</math> doesn't work. Likewise, the pattern for <math>5</math> is always <math>5</math>. 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>, so <math>\boxed{\textbf{(D) } \frac 45}</math>. |
+ | |||
+ | ==Solution 3 (Casework)== | ||
+ | We can use modular arithmetic for each residue of <math>n \pmod 5</math> | ||
+ | |||
+ | |||
+ | |||
+ | If <math>n \equiv 0 \pmod 5</math>, then <math>n^{16} \equiv 0^{16} \equiv 0 \pmod 5</math> | ||
+ | |||
+ | |||
+ | If <math>n \equiv 1 \pmod 5</math>, then <math>n^{16} \equiv 1^{16} \equiv 1 \pmod 5</math> | ||
+ | |||
+ | |||
+ | If <math>n \equiv 2 \pmod 5</math>, then <math>n^{16} \equiv (n^2)^8 \equiv (2^2)^8 \equiv 4^8 \equiv (-1)^8 \equiv 1 \pmod 5</math> | ||
+ | |||
+ | |||
+ | If <math>n \equiv 3 \pmod 5</math>, then <math>n^{16} \equiv (n^2)^8 \equiv (3^2)^8 \equiv 9^8 \equiv (-1)^8 \equiv 1 \pmod 5</math> | ||
+ | |||
+ | |||
+ | If <math>n \equiv 4 \pmod 5</math>, then <math>n^{16} \equiv 4^{16} \equiv (-1)^{16} \equiv 1 \pmod 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> | ||
+ | |||
+ | == Video Solution 1== | ||
+ | https://youtu.be/zfChnbMGLVQ?t=2410 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/Oj3Z1JhvoiE | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
==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}} |
Latest revision as of 20:05, 1 May 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 1
https://youtu.be/zfChnbMGLVQ?t=2410
~ pi_is_3.14
Video Solution 2
~savannahsolver
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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