2018 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 19 Hard

A number m is randomly selected from the set \left\{11,13,15,17,19 \right\}, and a number n is randomly selected from \left\{1999,2000,2001,\cdots ,2018 \right\}. What is the probability that m^n has a units digit of 1? (2018 AMC 10A Problem, Question#19)

  • A.

    \frac{1}{5}

  • B.

    \frac{1}{4}

  • C.

    \frac{3}{10}

  • D.

    \frac{7}{20}

  • E.

    \frac{2}{5}

Answer:E

Since we only care about the unit digit, our set \left\{11,13,15,17,19 \right\} can be turned into \left\{1,3,5,7,9 \right\}. Call this set A and call \left\{1999,2000,2001,\cdots ,2018 \right\} set B. Let's do casework on the element of A that we choose. Since 1\cdot 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\cdot 3=99\cdot 3=77\cdot 3=11\cdot 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}\cdot \frac{5}{20}=\frac{1}{20}. Similarly, we can look at the repeating units digit for 7: 7\cdot 7=99\cdot 7=33\cdot 7=11\cdot 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\cdot 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\cdot 9=11\cdot 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}\cdot \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}}.

Since only the units digit is relevant, we can turn the first set into \left\{1,3,5,7,9 \right\}. Note that x^{4}\equiv\rm 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, \rm mod~4, this set has 5 values which correspond to \left\{0,1,2,3 \right\}, making the probability equal for all of them. Next, check the values for which it is equal to 1~\rm mod~10. There are 4+1+0+1+2=8 values for which it is equal to 1, remembering that 5^{4n}\equiv\rm 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{\rm E}.