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

Revision as of 15:38, 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

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 $1*1=1$ , 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: $3*3=9$ $9*3=7$ $7*3=1$ $1*3=3$ 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 repeating 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 $9^z$ , 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 $\frac{1}{5}*\frac{1}{2}=\frac{1}{10}.$ 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

@@ Line 19: / Line 19: @@
 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>.
-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 repeading 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>1*9=9</cmath>

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

Revision as of 15:38, 8 February 2018

Solution