Difference between revisions of "2018 AMC 10A Problems/Problem 19"

(Solution)
(Solution)
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<cmath>7*3=1</cmath>
 
<cmath>7*3=1</cmath>
 
<cmath>1*3=3</cmath>
 
<cmath>1*3=3</cmath>
We see that the unit digit of <math>3^x</math> for some integer <math>x</math> will only be <math>1</math> when <math>x</math> is a multiple of <math>4</math>. Now, let's count how many numbers in <math>B</math> are divisible by <math>4</math>. This can be done by simply listing:
+
We see that the unit digit of <math>3^x</math>, for some integer <math>x</math>, will only be <math>1</math> when <math>x</math> is a multiple of <math>4</math>. Now, let's count how many numbers in <math>B</math> are divisible by <math>4</math>. This can be done by simply listing:
 
<cmath>2000,2004,2008,2012,2016.</cmath>
 
<cmath>2000,2004,2008,2012,2016.</cmath>
 
There are <math>5</math> numbers in <math>B</math> divisible by <math>4</math> out of the <math>2018-1999+1=20</math> total numbers. Therefore, the probability that <math>3</math> is picked from <math>A</math> and a number divisible by <math>4</math> is picked from <math>B</math> is <math>\frac{1}{5}*\frac{5}{20}=\frac{1}{20}</math>.  
 
There are <math>5</math> numbers in <math>B</math> divisible by <math>4</math> out of the <math>2018-1999+1=20</math> total numbers. Therefore, the probability that <math>3</math> is picked from <math>A</math> and a number divisible by <math>4</math> is picked from <math>B</math> is <math>\frac{1}{5}*\frac{5}{20}=\frac{1}{20}</math>.  
Line 17: Line 17:
 
<cmath>3*7=1</cmath>
 
<cmath>3*7=1</cmath>
 
<cmath>1*7=7</cmath>
 
<cmath>1*7=7</cmath>
We see that the unit digit of <math>7^y</math> for some integer <math>y</math> will only be <math>1</math> when <math>y</math> is a multiple of <math>4</math>. This is exactly the same conditions as our last case with <math>3</math> so the probability of this case is also <math>\frac{1}{20}</math>.  
+
We see that the unit digit of <math>7^y</math>, for some integer <math>y</math>, will only be <math>1</math> when <math>y</math> is a multiple of <math>4</math>. This is exactly the same conditions as our last case with <math>3</math> so the probability of this case is also <math>\frac{1}{20}</math>.  
Since <math>5*5=25</math> and <math>25</math> ends in <math>5</math>, the units digit of <math>5^w</math> for some integer <math>w</math> will always be <math>5</math>. Thus, the probability in this case is <math>0</math>.
+
Since <math>5*5=25</math> and <math>25</math> ends in <math>5</math>, the units digit of <math>5^w</math>, for some integer, <math>w</math> will always be <math>5</math>. Thus, the probability in this case is <math>0</math>.
 
The last case we need to consider is when the number <math>9</math> is chosen from <math>A</math>. This happens with probability <math>\frac{1}{5}</math>. We list out the repeading units digit for <math>9</math> as we have done for <math>3</math> and <math>7</math>:
 
The last case we need to consider is when the number <math>9</math> is chosen from <math>A</math>. This happens with probability <math>\frac{1}{5}</math>. We list out the repeading units digit for <math>9</math> as we have done for <math>3</math> and <math>7</math>:
 
<cmath>9*9=1</cmath>
 
<cmath>9*9=1</cmath>
 
<cmath>1*9=9</cmath>
 
<cmath>1*9=9</cmath>
We see that the units digit of <math>9^z</math> is <math>1</math> when <math>z</math> is an even number. From the <math>20</math> numbers in <math>B</math>, we see that exactly half of them are even. The probability in this case is <math>\frac{1}{5}*\frac{1}{2}=\frac{1}{10}.</math>
+
We see that the units digit of <math>9^z</math>, for some integer <math>z</math>, is <math>1</math> only when <math>z</math> is an even number. From the <math>20</math> numbers in <math>B</math>, we see that exactly half of them are even. The probability in this case is <math>\frac{1}{5}*\frac{1}{2}=\frac{1}{10}.</math>
 
Finally, we can ad all of our probabilities together to get  
 
Finally, we can ad all of our probabilities together to get  
 
<cmath>\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.</cmath>
 
<cmath>\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.</cmath>
  
 
~Nivek
 
~Nivek

Revision as of 15:37, 8 February 2018

A number $m$ is randomly selected from the set $\{11,13,15,17,19\}$, and a number $n$ is randomly selected from $\{1999,2000,2001,\ldots,2018\}$. What is the probability that $m^n$ has a units digit of $1$?

$\textbf{(A) }   \frac{1}{5}   \qquad        \textbf{(B) }   \frac{1}{4}   \qquad    \textbf{(C) }   \frac{3}{10}   \qquad   \textbf{(D) } \frac{7}{20} \qquad  \textbf{(E) }   \frac{2}{5}$

Solution

Since we only care about the unit digit, our set $\{11,13,15,17,19 \}$ can be turned into $\{1,3,5,7,9 \}$. Call this set $A$ and call $\{1999, 2000, 2001, \cdots , 2018 \}$ set $B$. Let's do casework on the element of $A$ that we choose. Since $1*1=1$, any number from $B$ can be paired with $1$ to make $1^n$ have a units digit of $1$. Therefore, the probability of this case happening is $\frac{1}{5}$ since there is a $\frac{1}{5}$ chance that the number $1$ is selected from $A$. Let us consider the case where the number $3$ is selected from $A$. Let's look at the unit digit when we repeatedly multiply the number $3$ by itself: \[3*3=9\] \[9*3=7\] \[7*3=1\] \[1*3=3\] We see that the unit digit of $3^x$, for some integer $x$, will only be $1$ when $x$ is a multiple of $4$. Now, let's count how many numbers in $B$ are divisible by $4$. This can be done by simply listing: \[2000,2004,2008,2012,2016.\] There are $5$ numbers in $B$ divisible by $4$ out of the $2018-1999+1=20$ total numbers. Therefore, the probability that $3$ is picked from $A$ and a number divisible by $4$ is picked from $B$ is $\frac{1}{5}*\frac{5}{20}=\frac{1}{20}$. Similarly, we can look at the repeating units digit for $7$: \[7*7=9\] \[9*7=3\] \[3*7=1\] \[1*7=7\] We see that the unit digit of $7^y$, for some integer $y$, will only be $1$ when $y$ is a multiple of $4$. This is exactly the same conditions as our last case with $3$ so the probability of this case is also $\frac{1}{20}$. Since $5*5=25$ and $25$ ends in $5$, the units digit of $5^w$, for some integer, $w$ will always be $5$. Thus, the probability in this case is $0$. The last case we need to consider is when the number $9$ is chosen from $A$. This happens with probability $\frac{1}{5}$. We list out the repeading units digit for $9$ as we have done for $3$ and $7$: \[9*9=1\] \[1*9=9\] We see that the units digit of $9^z$, for some integer $z$, is $1$ only when $z$ is an even number. From the $20$ numbers in $B$, we see that exactly half of them are even. The probability in this case is $\frac{1}{5}*\frac{1}{2}=\frac{1}{10}.$ Finally, we can ad all of our probabilities together to get \[\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.\]

~Nivek

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