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Difference between revisions of "2018 AMC 10A Problems/Problem 19"
(→‎Solution 1)
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== Solution 1 ==
 
== Solution 1 ==
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Since we only care about the unit digit, our set <math>\{11,13,15,17,19 \}</math> can be turned into <math>\{1,3,5,7,9 \}</math>. Call this set <math>A</math> and call <math>\{1999, 2000, 2001, \cdots , 2018 \}</math> set <math>B</math>. Let's do casework on the element of <math>A</math> that we choose. Since <math>1*1=1</math>, any number from <math>B</math> can be paired with <math>1</math> to make <math>1^n</math> have a units digit of <math>1</math>. Therefore, the probability of this case happening is <math>\frac{1}{5}</math> since there is a <math>\frac{1}{5}</math> chance that the number <math>1</math> is selected from <math>A</math>. Let us consider the case where the number <math>3</math> is selected from <math>A</math>. Let's look at the unit digit when we repeatedly multiply the number <math>3</math> by itself:
+
Since we only care about the unit digit, our set <math>\{11,13,15,17,19 \}</math> can be turned into <math>\{1,3,5,7,9 \}</math>. Call this set <math>A</math> and call <math>\{1999, 2000, 2001, \cdots , 2018 \}</math> set <math>B</math>. Let's do casework on the element of <math>A</math> that we choose. Since <math>1\cdot 1=1</math>, any number from <math>B</math> can be paired with <math>1</math> to make <math>1^n</math> have a units digit of <math>1</math>. Therefore, the probability of this case happening is <math>\frac{1}{5}</math> since there is a <math>\frac{1}{5}</math> chance that the number <math>1</math> is selected from <math>A</math>. Let us consider the case where the number <math>3</math> is selected from <math>A</math>. Let's look at the unit digit when we repeatedly multiply the number <math>3</math> by itself:
−
<cmath>3*3=9</cmath>
+
<cmath>3\cdot 3=9</cmath>
−
<cmath>9*3=7</cmath>  
+
<cmath>9\cdot 3=7</cmath>  
−
<cmath>7*3=1</cmath>
+
<cmath>7\cdot 3=1</cmath>
−
<cmath>1*3=3</cmath>
+
<cmath>1\cdot 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>.  
 
Similarly, we can look at the repeating units digit for <math>7</math>:
 
Similarly, we can look at the repeating units digit for <math>7</math>:
−
<cmath>7*7=9</cmath>
+
<cmath>7\cdot 7=9</cmath>
−
<cmath>9*7=3</cmath>
+
<cmath>9\cdot 7=3</cmath>
−
<cmath>3*7=1</cmath>
+
<cmath>3\cdot 7=1</cmath>
−
<cmath>1*7=7</cmath>
+
<cmath>1\cdot 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\cdot 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 repeating 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 repeating 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\cdot 9=1</cmath>
−
<cmath>1*9=9</cmath>
+
<cmath>1\cdot 9=9</cmath>
−
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>
+
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}\cdot \frac{1}{2}=\frac{1}{10}.</math>
 
Finally, we can add all of our probabilities together to get  
 
Finally, we can add 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>

Revision as of 22:45, 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 1

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 $1\cdot 1=1$, 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: \[3\cdot 3=9\] \[9\cdot 3=7\] \[7\cdot 3=1\] \[1\cdot 3=3\] 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 $7$: \[7\cdot 7=9\] \[9\cdot 7=3\] \[3\cdot 7=1\] \[1\cdot 7=7\] 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 $5\cdot 5=25$ 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 repeating units digit for $9$ as we have done for $3$ and $7$: \[9\cdot 9=1\] \[1\cdot 9=9\] 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 $\frac{1}{5}\cdot \frac{1}{2}=\frac{1}{10}.$ Finally, we can add all of our probabilities together to get \[\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.\]

~Nivek

Solution 2

Since only the units digit is relevant, we can turn the first set into $\{1,3,5,7,9\}$. Note that $x^4 \equiv 1 \mod 10$ for all odd digits $x$, except for 5. Looking at the second set, we see that it is a set of all integers between 1999 and 2018. There are 20 members of this set, which means that, $\mod 4$, this set has 5 values which correspond to $\{0,1,2,3\}$, making the probability equal for all of them. Next, check the values for which it is equal to $1 \mod 10$. There are $4+1+0+1+2=8$ values for which it is equal to 1, remembering that $5^{4n} \equiv 1 \mod 10$ only if $n=0$, which it is not. There are 20 values in total, and simplifying $\frac{8}{20}$ gives us $\boxed{\frac{2}{5}}$ or $\boxed{E}$.

$QED\blacksquare$

See Also

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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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