Problem:
A number m is randomly selected from the set {11,13,15,17,19}, and a number n is randomly selected from {1999,2000,2001,…,2018}. What is the probability that mn has a units digit of 1?
Answer Choices:
A. 51
B. 41
C. 103
D. 207
E. 52
Solution:
For m∈{11,13,15,17,19}, let p(m) denote the probability that mn has units digit 1 , where n is chosen at random from the set S={1999,2000,2001,…,2018}. Then the desired probability is equal to 51(p(11)+p(13)+p(15)+p(17)+p(19)). Because any positive integral power of 11 always has units digit 1,p(11)=1, and because any positive integral power of 15 always has units digit 5, p(15)=0. Note that S has 20 elements, exactly 5 of which are congruent to jmod4 for each of j=0,1,2,3. The units digits of powers of 13 and 17 cycle in groups of 4. More precisely,
(13kmod10)k=19992018=(7,1,3,9,7,1,…,3,9)
and
(17kmod10)k=19992018=(3,1,7,9,3,1,…,7,9)
Thus p(13)=p(17)=205=41. Finally, note that the units digit of 19k is 1 or 9, according to whether k is even or odd, respectively. Thus p(19)=21. Hence the requested probability is
51(1+41+0+41+21)=(E)52
The problems on this page are the property of the MAA's American Mathematics Competitions