![]() And when D was 0, that left 8 points remaining. The only way to account for this odd number of points was with two safeties and a PAT. When D was 1, that left 5 points remaining. When D was 2, that again left 2 points remaining, either due to a 2-point conversion or a safety. Here, you could look at subcases when D (the number of field goals) was 2, 1 or 0. When A was one, you needed to account for 8 remaining points (with at most one conversion). There were three ways to do this: with two PATs, with one 2-point conversion and with one safety. When A was 2, you somehow needed to account for 2 remaining points (with at most two conversions). There were three cases to consider: when A was 2, 1 or 0. ![]() But for those counting by hand, like organizing your thinking based on the number of touchdowns ( A) was a good strategy, since touchdowns were worth the most points. ![]() Then the problem became finding the number of distinct ordered quintuples ( A, B, C, D, E) such that 6 A + B + 2 C + 3 D + 2 E = 14, with the constraints that A, B, C, D, E ≥ 0 and B + C ≤ A (i.e., a team couldn’t have more conversions than touchdowns).Ī few solvers wrote computer code to count up the number of such quintuples. Suppose a team scored A touchdowns, B PATs, C 2-point conversions, D field goals and E safeties. So scoring a field goal and then a safety was the same as a safety and then a field goal (i.e., there was only one distinct way a team could score 5 points). Using the aforementioned methods of scoring, how many distinct ways could a team have scored 14 points? Note that the sequence in which a team scores these points didn’t matter. But that’s not necessarily how those 14 points were scored. So when a team has scored 14 points, it’s safe to assume that they scored two touchdowns and two PATs. Some methods of scoring points are more common than others. Based purely on this angle, you want to determine whether the point of light is more likely to be a star or a plane. Suddenly, you spot a point of light at a particular angle of elevation above the horizon. ![]() One cloudless night, the sky above you contains an equal number of visible stars and planes. Riddler Expressįind all pairs of integers a and b that are solutions to the following equation: a Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter or send me an email. Submit a correct answer for either, 1 and you may get a shoutout in the next column. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. ![]()
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