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

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== Solution 1 ==
 
== Solution 1 ==
−
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:
+
Since we only care about the units 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\cdot 3=9</cmath>
 
<cmath>3\cdot 3=9</cmath>
 
<cmath>9\cdot 3=7</cmath>  
 
<cmath>9\cdot 3=7</cmath>  
Line 13: Line 13:
 
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}\cdot \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}\cdot \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\cdot 7=9</cmath>
 
<cmath>7\cdot 7=9</cmath>
Line 21: Line 21:
 
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\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>.
 
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\cdot 9=1</cmath>
 
<cmath>9\cdot 9=1</cmath>
 
<cmath>1\cdot 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}\cdot \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{\textbf{(E)} ~\frac{2}{5}}.</cmath>
  
 
~Nivek
 
~Nivek
 +
 +
~very minor edits by virjoy2001
  
 
== Solution 2 ==
 
== Solution 2 ==
−
Since only the units digit is relevant, we can turn the first set into <math>\{1,3,5,7,9\}</math>. Note that <math>x^4 \equiv 1 \mod 10</math> for all odd digits <math>x</math>, 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, <math>\mod 4</math>, this set has 5 values which correspond to <math>\{0,1,2,3\}</math>, making the probability equal for all of them. Next, check the values for which it is equal to <math>1 \mod 10</math>. There are <math>4+1+0+1+2=8</math> values for which it is equal to 1, remembering that <math>5^{4n} \equiv 1 \mod 10</math> only if <math>n=0</math>, which it is not. There are 20 values in total, and simplifying <math>\frac{8}{20}</math> gives us <math>\boxed{\frac{2}{5}}</math> or <math>\boxed{E}</math>.
+
Since only the units digit is relevant, we can turn the first set into <math>\{1,3,5,7,9\}</math>. Note that <math>x^4 \equiv 1 \mod 10</math> for all odd digits <math>x</math>, 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, <math>\mod 4</math>, this set has 5 values which correspond to <math>\{0,1,2,3\}</math>, making the probability equal for all of them. Next, check the values for which it is equal to <math>1 \mod 10</math>. There are <math>4+1+0+1+2=8</math> values for which it is equal to 1, remembering that <math>5^{4n} \equiv 1 \mod 10</math> only if <math>n=0</math>, which it is not. There are 20 values in total, and simplifying <math>\frac{8}{20}</math> gives us <math>\boxed{\textbf{(E)} ~\frac{2}{5}}</math>.
  
 
<math>QED\blacksquare</math>
 
<math>QED\blacksquare</math>
 +
 
==Solution 3==
 
==Solution 3==
−
By Euler's Theorem, we have that <math>a^{4}=1(\mod 10)</math>, iff <math>\gcd(a,10)=1</math>. Hence <math>m=11,13,17,19</math>, <math>n=2000,2004,2008,2012,2016</math> work. Also note that <math>11^{\text{any positive integer}}\equiv 1(\mod 10)</math> because <math>11^b=(10+1)^b=10^b+10^{b-1}1+...+10(1)+1</math>, and the latter mod 10 is clearly 1. So <math>m=11</math>, <math>n=1999,2001,2002,2003,2005,...,2018</math> work(not counting multiples of 4 as we would be double counting if we did). We can also note that <math>19^{2a}\equiv 1(\mod 10)</math> because <math>19^{2a}=361^{a}</math>, and by the same logic as why <math>11^{\text{any positive integer}}\equiv 1(\mod 10)</math>, we are done. Hence <math>m=19</math>, and <math>n=2002, 2006, 2010, 2014, 2018</math> work(not counting any of the aforementioned cases as that would be double counting). We cannot make anymore observations that add more <math>m^n</math> with units digit <math>1</math>, hence the number of <math>m^n</math> that have units digit one is <math>4\cdot 5+1\cdot 15+1\cdot 5=40</math>. And the total number of combinations of an element of the set of all <math>m</math> and an element of the set of all <math>n</math> is <math>5\cdot 20=100</math>. Hence the desired probability is <math>\frac{40}{100}=\frac{2}{5}</math>, which is answer choice <math>\textbf{(E)}</math>.  
+
By Euler's Theorem, we have that <math>a^{4} \equiv 1 \pmod {10}</math>, if <math>\gcd(a,10)=1</math>.  
 +
Hence <math>m=11,13,17,19</math>, <math>n=2000,2004,2008,2012,2016</math> work.  
 +
 
 +
Also note that <math>11^{\text{any positive integer}}\equiv 1 \pmod {10}</math> because <math>11^b=(10+1)^b=10^b+10^{b-1}1+...+10(1)+1</math>, and the latter <math>\pmod {10}</math> is clearly <math>1</math>. So <math>m=11</math>, <math>n=1999,2001,2002,2003,2005,...,2018</math> work (not counting multiples of 4 as we would be double counting if we did).  
 +
 
 +
We can also note that <math>19^{2a}\equiv 1 \pmod {10}</math> because <math>19^{2a}=361^{a}</math>, and by the same logic as why <math>11^{\text{any positive integer}}\equiv 1 \pmod {10}</math>, we are done. Hence <math>m=19</math>, and <math>n=2002, 2006, 2010, 2014, 2018</math> work (not counting any of the aforementioned cases as that would be double counting).  
 +
 
 +
We cannot make any more observations that add more <math>m^n</math> with unit digit <math>1</math>, hence the number of <math>m^n</math> that have units digit one is <math>4\cdot 5+1\cdot 15+1\cdot 5=40</math>. And the total number of combinations of an element of the set of all <math>m</math> and an element of the set of all <math>n</math> is <math>5\cdot 20=100</math>. Hence the desired probability is <math>\frac{40}{100}=\frac{2}{5}</math>, which is answer choice <math>\boxed{\textbf{(E)} ~\frac{2}{5}}</math>.  
 
~vsamc
 
~vsamc
  
−
==Video Solution==
+
==Solution 4 (Easy)==
−
https://youtu.be/M22S82Am2zM
+
When a number's unit's digit is <math>1</math>, then any power to this number will also end in <math>1</math> (since <math>1^{n}</math> for any <math>n</math> is always <math>1</math>), so we have <math>20</math> choices for <math>11</math>.
  
 +
When a number's unit's digit is <math>3</math>, then <math>3^{4n}</math> for any <math>n</math> will produce a number ending with 1. So, <math>20 \div 4 = 5</math> choices for <math>13</math>.
 +
 +
<math>5^{n}</math> always ends in 5, so there are <math>0</math> possibilities for <math>15</math>.
 +
 +
When a number's unit's digit is <math>7</math>, then this is also the same thing with <math>3</math>, so we have <math>5</math> choices.
 +
 +
When a number's unit's digit is <math>9</math>, then <math>9^{2n}</math> will produce a number ending in <math>1</math>, so we have <math>20 \div 2 = 10</math> possibilities.
 +
 +
Hence, we have a total of <math>5 \cdot 20 = 100</math> ways, so the probability is <math>\frac{20+5+0+5+10}{100} = \frac {40}{100} = \boxed{\textbf{(E)} ~\frac{2}{5}}</math>.
 +
 +
~MrThinker
 +
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/M22S82Am2zM?t=630
 
~IceMatrix
 
~IceMatrix
 +
 +
==Video Solution 2==
 +
https://youtu.be/njyn611TJi0
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 20:13, 6 August 2024

Problem

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 units 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 $\frac{1}{5}\cdot \frac{5}{20}=\frac{1}{20}.$ 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 \[\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\textbf{(E)} ~\frac{2}{5}}.\]

~Nivek

~very minor edits by virjoy2001

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 $\boxed{\textbf{(E)} ~\frac{2}{5}}$.

$QED\blacksquare$

Solution 3

By Euler's Theorem, we have that $a^{4} \equiv 1 \pmod {10}$, if $\gcd(a,10)=1$. Hence $m=11,13,17,19$, $n=2000,2004,2008,2012,2016$ work.

Also note that $11^{\text{any positive integer}}\equiv 1 \pmod {10}$ because $11^b=(10+1)^b=10^b+10^{b-1}1+...+10(1)+1$, and the latter $\pmod {10}$ is clearly $1$. So $m=11$, $n=1999,2001,2002,2003,2005,...,2018$ work (not counting multiples of 4 as we would be double counting if we did).

We can also note that $19^{2a}\equiv 1 \pmod {10}$ because $19^{2a}=361^{a}$, and by the same logic as why $11^{\text{any positive integer}}\equiv 1 \pmod {10}$, we are done. Hence $m=19$, and $n=2002, 2006, 2010, 2014, 2018$ work (not counting any of the aforementioned cases as that would be double counting).

We cannot make any more observations that add more $m^n$ with unit digit $1$, hence the number of $m^n$ that have units digit one is $4\cdot 5+1\cdot 15+1\cdot 5=40$. And the total number of combinations of an element of the set of all $m$ and an element of the set of all $n$ is $5\cdot 20=100$. Hence the desired probability is $\frac{40}{100}=\frac{2}{5}$, which is answer choice $\boxed{\textbf{(E)} ~\frac{2}{5}}$. ~vsamc

Solution 4 (Easy)

When a number's unit's digit is $1$, then any power to this number will also end in $1$ (since $1^{n}$ for any $n$ is always $1$), so we have $20$ choices for $11$.

When a number's unit's digit is $3$, then $3^{4n}$ for any $n$ will produce a number ending with 1. So, $20 \div 4 = 5$ choices for $13$.

$5^{n}$ always ends in 5, so there are $0$ possibilities for $15$.

When a number's unit's digit is $7$, then this is also the same thing with $3$, so we have $5$ choices.

When a number's unit's digit is $9$, then $9^{2n}$ will produce a number ending in $1$, so we have $20 \div 2 = 10$ possibilities.

Hence, we have a total of $5 \cdot 20 = 100$ ways, so the probability is $\frac{20+5+0+5+10}{100} = \frac {40}{100} = \boxed{\textbf{(E)} ~\frac{2}{5}}$.

~MrThinker

Video Solution by TheBeautyofMath

https://youtu.be/M22S82Am2zM?t=630 ~IceMatrix

Video Solution 2

https://youtu.be/njyn611TJi0

~savannahsolver

See Also

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