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

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

~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 $\boxed{\frac{2}{5}}$ or $\boxed{E}$.

$QED\blacksquare$

Solution 3

By Euler's Theorem, we have that $a^{4}=1 \pmod {10}$, 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 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 $\textbf{(E)}$. ~vsamc

Video Solution

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

See Also

 2018 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 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.

Invalid username
Login to AoPS