2018 AMC 10 A
Complete problem set with solutions and individual problem pages
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 ? (2018 AMC 10A Problem, Question#19)
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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 . Looking at the second set, we see that it is a set of all integers between and . There are members of this set, which means that, , this set has 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 , remembering that only if , which it is not. There are values in total, and simplifying gives us or .
