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).
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