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

m
m ((Phrased stuff better))
Line 35: Line 35:
 
<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 because we already did that). 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}=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>.  
  
 
==Video Solution==
 
==Video Solution==

Revision as of 17:35, 3 February 2020

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 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 $\frac{1}{5}\cdot \frac{5}{20}=\frac{1}{20}$. 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 \[\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.\]

~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 $\boxed{E}$.

$QED\blacksquare$

Solution 3

By Euler's Theorem, we have that $a^{4}=1(\mod 10)$, iff $\gcd(a,10)=1$. Hence $m=11,13,17,19$, $n=2000,2004,2008,2012,2016$ work. Also note that $11^{\text{any positive integer}}\equiv 1(\mod 10)$ because $11^b=(10+1)^b=10^b+10^{b-1}1+...+10(1)+1$, and the latter mod 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 $19^{2a}\equiv 1(\mod 10)$ because $19^{2a}=361^{a}$, and by the same logic as why $11^{\text{any positive integer}}\equiv 1(\mod 10)$, 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 anymore observations that add more $m^n$ with units 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 $\textbf{(E)}$.

Video Solution

https://youtu.be/M22S82Am2zM

~IceMatrix

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS