Difference between revisions of "2018 AMC 10A Problems/Problem 19"
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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> | 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 add all of our probabilities together to get | 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{\textbf{(E)} ~\frac{2}{5}}.</cmath> |
~Nivek | ~Nivek | ||
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== Solution 2 == | == 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} | + | 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{\textbf{(E)} ~\frac{2}{5}}</math>. |
<math>QED\blacksquare</math> | <math>QED\blacksquare</math> | ||
+ | |||
==Solution 3== | ==Solution 3== | ||
− | By Euler's Theorem, we have that <math>a^{4} | + | By Euler's Theorem, we have that <math>a^{4} \equiv 1 \pmod {10}</math>, if <math>\gcd(a,10)=1</math>. |
Hence <math>m=11,13,17,19</math>, <math>n=2000,2004,2008,2012,2016</math> work. | Hence <math>m=11,13,17,19</math>, <math>n=2000,2004,2008,2012,2016</math> work. | ||
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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 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 | + | We cannot make any more observations that add more <math>m^n</math> with unit 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>\boxed{\textbf{(E)} ~\frac{2}{5}}</math>. |
~vsamc | ~vsamc | ||
− | ==Video Solution== | + | ==Solution 4 (Easy)== |
− | https://youtu.be/M22S82Am2zM?t= | + | When a number's unit's digit is <math>1</math>, then any power to this number will also end in <math>1</math> (since <math>1^{n}</math> for any <math>n</math> is always <math>1</math>), so we have <math>20</math> choices for <math>11</math>. |
+ | |||
+ | When a number's unit's digit is <math>3</math>, then <math>3^{4n}</math> for any <math>n</math> will produce a number ending with 1. So, <math>20 \div 4 = 5</math> choices for <math>13</math>. | ||
+ | |||
+ | <math>5^{n}</math> always ends in 5, so there are <math>0</math> possibilities for <math>15</math>. | ||
+ | |||
+ | When a number's unit's digit is <math>7</math>, then this is also the same thing with <math>3</math>, so we have <math>5</math> choices. | ||
+ | |||
+ | When a number's unit's digit is <math>9</math>, then <math>9^{2n}</math> will produce a number ending in <math>1</math>, so we have <math>20 \div 2 = 10</math> possibilities. | ||
+ | |||
+ | Hence, we have a total of <math>5 \cdot 20 = 100</math> ways, so the probability is <math>\frac{20+5+0+5+10}{100} = \frac {40}{100} = \boxed{\textbf{(E)} ~\frac{2}{5}}</math>. | ||
+ | |||
+ | ~MrThinker | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/M22S82Am2zM?t=630 | ||
~IceMatrix | ~IceMatrix | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/njyn611TJi0 | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 20:13, 6 August 2024
Contents
[hide]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 .
Solution 3
By Euler's Theorem, we have that , if . 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 unit 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
Solution 4 (Easy)
When a number's unit's digit is , then any power to this number will also end in (since for any is always ), so we have choices for .
When a number's unit's digit is , then for any will produce a number ending with 1. So, choices for .
always ends in 5, so there are possibilities for .
When a number's unit's digit is , then this is also the same thing with , so we have choices.
When a number's unit's digit is , then will produce a number ending in , so we have possibilities.
Hence, we have a total of ways, so the probability is .
~MrThinker
Video Solution by TheBeautyofMath
https://youtu.be/M22S82Am2zM?t=630 ~IceMatrix
Video Solution 2
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.