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[ Information | Background | The CyberTeaser | Quantum News | Miscellaneous Stuff ] Solution to CyberTeaser B166If Ciliegia works at the slower rate, then he finishes 6 stools between Friday and Sunday, and his total number of stools is a multiple of 5 as well as 3. Any such number is a multiple of 15, and in fact 15 itself works, if we start Ciliegia working on Wednesday. And there are no other solutions possible: if he started working n days before Friday, then 5(n + 1) = 3[(n + 1) + 2], so n = 2. But there is another possible interpretation of the problem. Perhaps Ciliegia, working at his fast pace, will finish on Friday; and at his slow pace, on a Sunday more than one week later. If n is the number of days before Friday that he started and m is the number of weeks between the Friday and the Sunday, then Ciliegia works n + 1 days at the fast rate and 7m + (n + 1) + 2 days at the slow rate, so we have 5(n + 1) = 3(7m + n + 3), or 21m = 2n – 4. We need a solution of this equation in integers. There are general and standard ways to do this, which the reader is invited to consult in any book on number theory. Meanwhile, we will note that since 2n – 4 is even, m must also be even. Letting m = 2k, we have 42k = 2n – 4, or 21k = n – 2, or n = 21k + 2. But 21k, being a multiple of 7, represents an integer number of weeks. So n days before Friday is still a Wednesday, no matter how many weeks intervene. Note: the explanation here differs slightly from that printed in the magazine, due to the intervention of alert Quantum reader Rick Dilling. (The answer, however, is the same: Wednesday.) CyberTeaser Archive
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