2018 AMC 10A Problems/Problem 19
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 ?
Since we only care about the unit 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
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 .
By Euler's Theorem, we have that , iff . Hence , work. Also note that because , and the latter mod 10 is clearly 1. 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 anymore 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
|2018 AMC 10A (Problems • Answer Key • Resources)|
|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|