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Anytime each of three consecutive months has exactly four Fridays, Jack's birthday will fall in one of those three months. Which month is that?

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    Is it Ramadan??2012-02-13

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The three months with $4$ Fridays each have $12$ Fridays together, so they can have at most $12\cdot7+6=90$ days. Conversely, if three consecutive months have at most $90$ days, they sometimes contain only $12$ Fridays, and since every month contains at least $4$ Fridays they then contain $4$ Fridays each. In summary, three consecutive months may contain $4$ Fridays each iff they have at most $90$ days. Three consecutive months can have at most $90$ days iff one of them is February. Thus Jack's birthday is in February.

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    You were just a bit faster :)2012-02-13
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    The birthday is not Feb 29th :-). Also, just for completion (assuming no leap year and assuming I have got it right), if Jan 7th is Friday, then Jan,Feb,Mar have 4 fridays each and if Feb 7th is friday, Feb,Mar,Apr have four fridays and Jan has 5.2012-02-13
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    @Aryabhata: There are two more possibilities: February 6th also gives four Fridays each in February, March and April; and December 7 gives four Fridays each in December, January and February; in fact that last constellation will occur this/next year.2012-02-13
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    @joriki: I was just trying to prove that there are at least two sets of months, with only Feb being common among them. Basically eliminating the case that it will always be jan,feb,mar or something like that (in which case, the answer could be any one of those).2012-02-13
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    @Aryabhata: I thought I'd covered that by saying that they have at most 90 days *iff* one of them is February. That implies that there are examples for each combination with February, so it can only be February. Also your examples both include February and March, so it seems they don't serve the intended purpose?2012-02-13
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    @joriki: Feb _has_ to be one of them, sure. Question is, is there any other month with that property. You are right, Mar is common to both and the completion is incomplete :-) Considering Dec,Jan,Feb completes it, like you pointed out. Apologies.2012-02-13
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    @Aryabhata: I don't understand -- how can there be any other month with that property if the months have at most $90$ days iff one of them is February? That other month would have to be in any consecutive triple of months that contains February, but clearly February is the only month with that property.2012-02-13
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    @joriki: You have shown that "12 fridays" implies "$\le 90$ days", the latter statement being equivalent to "having Feb". So only the implication "12 fridays" implies "having Feb" has actually been shown. This does not mean Feb is the _unique_ month with that property. I was just trying to confirm that Feb is indeed unique. What am I missing?2012-02-13
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    @joriki: "Three consecutive months have at most 90 days iff one of them is February". January-March 2012 together have 91 days. So your quick solution via "iff" falls apart. Also one needs to show not only that December-Febrauri and Februari-April (sometimes) have at most $90$ days, but also that this allows them to sometimes have 4 Fridays each (in order to rigorously exclude March and Januari as possibilities). The answer by Beni Bogosel does that (almost, the missing details are easy to supply).2012-02-13
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    @Aryabhata: Ah, sorry, I see what you mean now. I was using the first implication also as an equivalence but hadn't stated or justified it as such. I've completed the argument now.2012-02-13
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    @Marc: On your first point: You're right, I ignored leap days, but the solution doesn't "fall apart"; I just had to add "can" :-) Your second point, if I understand correctly, is essentially Aryabhata's point, which I just addressed before seeing your comment. About Beni's answer: it seems to me that it has a similar gap; it argues that a day of the week repeats exactly $12$ times, but not that each month must then contain $4$.2012-02-13
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    To both of you: Your criticisms rest on a certain interpretation of the question. They're valid if you regard the answer as a proof that there is exactly one month with this property. However, one could also interpret the question to imply that there is exactly one month, and to ask only which one it is. I think I was half treating it that way, half also trying to show why there can be only one such month, but not applying full rigour to that part of the answer.2012-02-13
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    @joriki: It wasn't crtitcism :-) I was in fact implicitly trying to clarify that the question is well-formed (If you notice, my very first comment was not directed towards you). FWIW, I think I was the first one to upvote your answer :-)2012-02-13
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    @Aryabhata: OK, sorry -- I was probably too much under the impression of Marc's "your quick solution falls apart" :-) I see what you mean.2012-02-13
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    @joriki: I admit that my first point was a bit overly severe, since actually the fact that in rare occasions a given three consecutive months fail to have${}\leq90$ days makes no difference for the problem, given that even when they do they usually fail to have $4$ Fridays each. I'll make amends by supplying the last missing details: December 2012 and January, Febrauri 2013 will each have 4 Fridays, as will Februari, March, and April 2014; this excludes March and January as solutions (providing Jack lives that long;-).2012-02-14
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Three consecutive months which do not contain February have at least $91$ days. Because $91=7 \times 13$, each day of the week will repeat itself at least 13 times, so Jack's birthday cannot fall in any of these months.

But if we consider February (in years in which it has $28$ days) then any three consecutive months containing February have at most 90 days, so there is a day of the week which repeats itself exactly 12 times in these three months. As time goes by, this day will be a friday. So Jack's birthday will fall in a month out of any three which contain february. Therefore Jack's birthday comes in February.