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

Revision as of 18:07, 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 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 $1*1=1$ , 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 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 $\boxed{E}$ .

$QED\blacksquare$

@@ Line 3: / Line 3: @@
 <math>\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} </math>
-==Solution==
+== 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*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*3=9</cmath>
@@ Line 27: / Line 27: @@
 ~Nivek
+== 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>.
+<math>QED\blacksquare</math>
 ==See Also==
 {{AMC10 box|year=2018|ab=A|num-b=18|num-a=20}}
+{{MAA Notice}}

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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Difference between revisions of "2018 AMC 10A Problems/Problem 19"

Revision as of 18:07, 8 February 2018

Solution 1

Solution 2

See Also