2018 AMC 10A Problems/Problem 19

Revision as of 16:35, 8 February 2018 by Nivek (talk | contribs) (Solution)

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 care about the unit digit only, 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 from $B$ can be paired with $1$ to make $m^n$ 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 $9^z$ is $1$ 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

Invalid username
Login to AoPS