![]() Suddenly, you spotted a point of light at a particular angle of elevation above the horizon. Last week, on a cloudless night, the sky above you contained an equal number of visible stars and planes. Sometimes the answer … is that there is no answer.Ĭongratulations to □ Michael Seifert □ of Quaker Hill, Connecticut, winner of last week’s Riddler Classic. Therefore, this equation had no integer solutions. Meanwhile, the right side of the equation couldn’t be a multiple of 3. ![]() Sure enough, it’s never a multiple of 3!Īt this point, you knew the left side of the equation had to be a multiple of 3. If you’re not convinced, check the values of b 2+4 for different integer values of b. In other words, b 2+4 was never congruent to 0 (mod 3). Finally, if b was congruent to 2 (mod 3), then b 2 was again congruent to 1 (mod 3), which meant b 2+4 was congruent to 2 (mod 3). If b was congruent to 1 (mod 3), then so was b 2, which meant b 2+4 was congruent to 2 (mod 3). If b was a multiple of 3, then so was b 2, which meant b 2+4 was congruent to 1 (mod 3). ![]() ( a+2) was guaranteed to be even, that meant any solution had to have an even value for b.Conversely, if b was even, then so were b 2 and b 2+4. If b was odd, then b 2 was also odd, as was b 2+4. Now what about the right side? At first, it wasn’t immediately clear how b 2+4 might have related to factors of 2, 3 or 6. In terms of factors, that was everything you could say for sure about the left side of the equation. ( a+2) was a multiple of both 2 and 3, that meant it was also a multiple of 6.In number theory parlance, either a was a multiple of 3, a was congruent to 1 (mod 3), 2 in which case a+2 was a multiple of 3, or a was congruent to 2 (mod 3), in which case a+1 was a multiple of 3.įinally, since a That was because exactly one of the three factors had to be a multiple of 3. How would you arrange four squares to get as many rectangles as possible? And what is this number of rectangles? With this arrangement, it’s possible to trace three rectangles: the square on the left, the square on the right and the larger rectangle around both squares. For example, if you had two squares instead of four, you could place the squares side by side, as shown below: Your goal is to position the squares so that you can trace as many rectangles as possible using the edges of the squares. You can place the squares so that their edges align, but their interiors cannot overlap. You have four squares that you can place on a large, flat table. 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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