Difference between revisions of "2018 AMC 10A Problems/Problem 19"
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+ | ==Problem== | ||
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A number <math>m</math> is randomly selected from the set <math>\{11,13,15,17,19\}</math>, and a number <math>n</math> is randomly selected from <math>\{1999,2000,2001,\ldots,2018\}</math>. What is the probability that <math>m^n</math> has a units digit of <math>1</math>? | A number <math>m</math> is randomly selected from the set <math>\{11,13,15,17,19\}</math>, and a number <math>n</math> is randomly selected from <math>\{1999,2000,2001,\ldots,2018\}</math>. What is the probability that <math>m^n</math> has a units digit of <math>1</math>? | ||
<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> | <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 | + | Since we only care about the units 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\cdot 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 | + | <cmath>3\cdot 3=9</cmath> |
− | <cmath>9 | + | <cmath>9\cdot 3=7</cmath> |
− | <cmath>7 | + | <cmath>7\cdot 3=1</cmath> |
− | <cmath>1 | + | <cmath>1\cdot 3=3</cmath> |
We see that the unit digit of <math>3^x</math>, for some integer <math>x</math>, will only be <math>1</math> when <math>x</math> is a multiple of <math>4</math>. Now, let's count how many numbers in <math>B</math> are divisible by <math>4</math>. This can be done by simply listing: | We see that the unit digit of <math>3^x</math>, for some integer <math>x</math>, will only be <math>1</math> when <math>x</math> is a multiple of <math>4</math>. Now, let's count how many numbers in <math>B</math> are divisible by <math>4</math>. This can be done by simply listing: | ||
<cmath>2000,2004,2008,2012,2016.</cmath> | <cmath>2000,2004,2008,2012,2016.</cmath> | ||
− | There are <math>5</math> numbers in <math>B</math> divisible by <math>4</math> out of the <math>2018-1999+1=20</math> total numbers. Therefore, the probability that <math>3</math> is picked from <math>A</math> and a number divisible by <math>4</math> is picked from <math>B</math> is <math>\frac{1}{5} | + | There are <math>5</math> numbers in <math>B</math> divisible by <math>4</math> out of the <math>2018-1999+1=20</math> total numbers. Therefore, the probability that <math>3</math> is picked from <math>A</math> and a number divisible by <math>4</math> is picked from <math>B</math> is <math>\frac{1}{5}\cdot \frac{5}{20}=\frac{1}{20}.</math> |
Similarly, we can look at the repeating units digit for <math>7</math>: | Similarly, we can look at the repeating units digit for <math>7</math>: | ||
− | <cmath>7 | + | <cmath>7\cdot 7=9</cmath> |
− | <cmath>9 | + | <cmath>9\cdot 7=3</cmath> |
− | <cmath>3 | + | <cmath>3\cdot 7=1</cmath> |
− | <cmath>1 | + | <cmath>1\cdot 7=7</cmath> |
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>. | 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 | + | Since <math>5\cdot 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> | + | 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 | + | <cmath>9\cdot 9=1</cmath> |
− | <cmath>1 | + | <cmath>1\cdot 9=9</cmath> |
− | We see that the units digit of <math>9^z</math>, for some integer <math>z</math>, is <math>1</math> only when <math>z</math> is an even number. From the <math>20</math> numbers in <math>B</math>, we see that exactly half of them are even. The probability in this case is <math>\frac{1}{5} | + | We see that the units digit of <math>9^z</math>, for some integer <math>z</math>, is <math>1</math> only when <math>z</math> is an even number. From the <math>20</math> numbers in <math>B</math>, we see that exactly half of them are even. The probability in this case is <math>\frac{1}{5}\cdot \frac{1}{2}=\frac{1}{10}.</math> |
− | Finally, we can | + | Finally, we can add all of our probabilities together to get |
<cmath>\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.</cmath> | <cmath>\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.</cmath> | ||
~Nivek | ~Nivek | ||
+ | |||
+ | ~very minor edits by virjoy2001 | ||
+ | |||
+ | == 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> | ||
+ | ==Solution 3== | ||
+ | By Euler's Theorem, we have that <math>a^{4}=1 \pmod {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 \pmod {10}</math> because <math>11^b=(10+1)^b=10^b+10^{b-1}1+...+10(1)+1</math>, and the latter <math>\pmod {10}</math> is clearly <math>1</math>. 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 \pmod {10}</math> because <math>19^{2a}=361^{a}</math>, and by the same logic as why <math>11^{\text{any positive integer}}\equiv 1 \pmod {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 any more 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>. | ||
+ | ~vsamc | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/M22S82Am2zM?t=630 | ||
+ | ~IceMatrix | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2018|ab=A|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Number Theory Problems]] | ||
+ | [[Category:Introductory Probability Problems]] |
Latest revision as of 17:59, 7 January 2021
Problem
A number is randomly selected from the set , and a number is randomly selected from . What is the probability that has a units digit of ?
Solution 1
Since we only care about the units digit, our set can be turned into . Call this set and call set . Let's do casework on the element of that we choose. Since , any number from can be paired with to make have a units digit of . Therefore, the probability of this case happening is since there is a chance that the number is selected from . Let us consider the case where the number is selected from . Let's look at the unit digit when we repeatedly multiply the number by itself: We see that the unit digit of , for some integer , will only be when is a multiple of . Now, let's count how many numbers in are divisible by . This can be done by simply listing: There are numbers in divisible by out of the total numbers. Therefore, the probability that is picked from and a number divisible by is picked from is Similarly, we can look at the repeating units digit for : We see that the unit digit of , for some integer , will only be when is a multiple of . This is exactly the same conditions as our last case with so the probability of this case is also . Since and ends in , the units digit of , for some integer, will always be . Thus, the probability in this case is . The last case we need to consider is when the number is chosen from . This happens with probability We list out the repeating units digit for as we have done for and : We see that the units digit of , for some integer , is only when is an even number. From the numbers in , we see that exactly half of them are even. The probability in this case is Finally, we can add all of our probabilities together to get
~Nivek
~very minor edits by virjoy2001
Solution 2
Since only the units digit is relevant, we can turn the first set into . Note that for all odd digits , 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, , this set has 5 values which correspond to , making the probability equal for all of them. Next, check the values for which it is equal to . There are values for which it is equal to 1, remembering that only if , which it is not. There are 20 values in total, and simplifying gives us or .
Solution 3
By Euler's Theorem, we have that , iff . Hence , work.
Also note that because , and the latter is clearly . So , work (not counting multiples of 4 as we would be double counting if we did).
We can also note that because , and by the same logic as why , we are done. Hence , and work (not counting any of the aforementioned cases as that would be double counting).
We cannot make any more observations that add more with units digit , hence the number of that have units digit one is . And the total number of combinations of an element of the set of all and an element of the set of all is . Hence the desired probability is , which is answer choice . ~vsamc
Video Solution
https://youtu.be/M22S82Am2zM?t=630 ~IceMatrix
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
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 |
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