bonanova Posted October 24, 2010 Report Share Posted October 24, 2010 Most puzzle aficionados know how to solve the "equation" Halloween = Christmas. Let's extend the puzzle a bit. In 2008, and on average every seven years, one can write Halloween = Thanksgiving = Christmas. Explain. . If we concatenate the month and day numbers we get 1031 = 1225. How could this be possible? . If we assign distinct, nonzero numbers to letters and let period be multiplication we get Oct.31 = Dec.25. What value must "c" have? . Quote Link to comment Share on other sites More sharing options...
0 araver Posted October 24, 2010 Report Share Posted October 24, 2010 Thanksgiving is the fourth Tuesday in November. In 2008 this was November 27th (NOV 27). Christmas is always December 25th (DEC 25). Halloween is always October 31th (OCT 31). We can interpret DEC, NOV and OCT as base 10 (DECimal), base 9 (NOVenary) and base 8 (OCTal). (Actually this is not very surprising since both base names and month names come from Latin and early Roman calendar did not have January and February, so December was actually the tenth month.) So: NOV 27 = 27 (base 9) = 2x9+7 = 25 (base 10) = DEC 25 OCT 31 = 31 (base 8) = 3x8+1= 25 (base 10) 27th would fall as the fourth Tuesday in November once every 7 years, if no leap years existed. However, because of leap years - 27th actually jumps a day of the month every 4 years. Which means the normal interval of 6 (not 7) years is not actually accurate E.g. 2008, 2014, 2025, 2031 etc - it jumps over 2020 as 2020 is a leap year and 27th Nov jumps from Wednesday in 2019 to Friday in 2020. Last digit of left hand is T*1 mod 10=T and last digit on right hand is C*5 mod 10. So T=0 or T=5. The next left-hand digit is (C+3*T) mod 10, while the next right-hand digit is (2*C+5*E+(5*C)div10) mod 10. (where 5*C div 10 represents the carry-on from the left-most digit multiplication). So, (C+3*T) mod 10 = ((2*C+(5*C)div10)mod10 + 5*E mod10)mod 10. Note that 5*E mod 10 is either 0 or 5 so there are only 2 possibilities to check for each possible value for C. If T=0 then C*5 mod 10 =0 so C is even and C is not 0 (C and T are distinct): If C=2 then C+3*T mod 10=2 and (2*C+5*E+(5*C)div10) mod 10=(3+5*E mod10)mod 10 which is either 3 or 8 so the equality is impossible. If C=4 then C+3*T mod 10=4 and (2*C+5*E+(5*C)div10) mod 10=(8+2+5*E mod10)mod 10 either 0 or 5 - impossible. If C=6 then C+3*T mod 10=6 and (2*C+5*E+(5*C)div10) mod 10=(12+3+5*E mod10)mod 10 either 0 or 5 - impossible. If C=8 then C+3*T mod 10=8 and (2*C+5*E+(5*C)div10) mod 10=(16+4+5*E mod10)mod 10 either 0 or 5 - impossible. If T=5 then C*5 mod 10 =5 so C is odd and C is not 5 (C and T are distinct): If C=1 then C+3*T mod 10=6 and (2*C+5*E+(5*C)div10) mod 10=(2+0+5*E mod10)mod 10 either 2 or 7 - impossible. If C=3 then C+3*T mod 10=8 and (2*C+5*E+(5*C)div10) mod 10=(6+1+5*E mod10)mod 10 either 7 or 2 - impossible. If C=9 then C+3*T mod 10=4 and (2*C+5*E+(5*C)div10) mod 10=(18+4+5*E mod10)mod 10 either 2 or 7 - impossible. So C=7 and T=5. We get C+3*T mod 10=2 and (2*C+5*E+(5*C)div10) mod 10=(14+3+5*E mod10) mod 10 is 2 if and only if E is odd (either 1,3 or 9). Moving on for E, D,C the solution 675*31 = 837*25 is found. Not sure about Other than 10 base10 = 12 base8 and 31 base8 = 25 base10, don't have a clue. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted October 24, 2010 Author Report Share Posted October 24, 2010 Thanksgiving is the fourth Tuesday in November. In 2008 this was November 27th (NOV 27). Christmas is always December 25th (DEC 25). Halloween is always October 31th (OCT 31). We can interpret DEC, NOV and OCT as base 10 (DECimal), base 9 (NOVenary) and base 8 (OCTal). (Actually this is not very surprising since both base names and month names come from Latin and early Roman calendar did not have January and February, so December was actually the tenth month.) So: NOV 27 = 27 (base 9) = 2x9+7 = 25 (base 10) = DEC 25 OCT 31 = 31 (base 8) = 3x8+1= 25 (base 10) 27th would fall as the fourth Tuesday in November once every 7 years, if no leap years existed. However, because of leap years - 27th actually jumps a day of the month every 4 years. Which means the normal interval of 6 (not 7) years is not actually accurate E.g. 2008, 2014, 2025, 2031 etc - it jumps over 2020 as 2020 is a leap year and 27th Nov jumps from Wednesday in 2019 to Friday in 2020. Last digit of left hand is T*1 mod 10=T and last digit on right hand is C*5 mod 10. So T=0 or T=5. The next left-hand digit is (C+3*T) mod 10, while the next right-hand digit is (2*C+5*E+(5*C)div10) mod 10. (where 5*C div 10 represents the carry-on from the left-most digit multiplication). So, (C+3*T) mod 10 = ((2*C+(5*C)div10)mod10 + 5*E mod10)mod 10. Note that 5*E mod 10 is either 0 or 5 so there are only 2 possibilities to check for each possible value for C. If T=0 then C*5 mod 10 =0 so C is even and C is not 0 (C and T are distinct): If C=2 then C+3*T mod 10=2 and (2*C+5*E+(5*C)div10) mod 10=(3+5*E mod10)mod 10 which is either 3 or 8 so the equality is impossible. If C=4 then C+3*T mod 10=4 and (2*C+5*E+(5*C)div10) mod 10=(8+2+5*E mod10)mod 10 either 0 or 5 - impossible. If C=6 then C+3*T mod 10=6 and (2*C+5*E+(5*C)div10) mod 10=(12+3+5*E mod10)mod 10 either 0 or 5 - impossible. If C=8 then C+3*T mod 10=8 and (2*C+5*E+(5*C)div10) mod 10=(16+4+5*E mod10)mod 10 either 0 or 5 - impossible. If T=5 then C*5 mod 10 =5 so C is odd and C is not 5 (C and T are distinct): If C=1 then C+3*T mod 10=6 and (2*C+5*E+(5*C)div10) mod 10=(2+0+5*E mod10)mod 10 either 2 or 7 - impossible. If C=3 then C+3*T mod 10=8 and (2*C+5*E+(5*C)div10) mod 10=(6+1+5*E mod10)mod 10 either 7 or 2 - impossible. If C=9 then C+3*T mod 10=4 and (2*C+5*E+(5*C)div10) mod 10=(18+4+5*E mod10)mod 10 either 2 or 7 - impossible. So C=7 and T=5. We get C+3*T mod 10=2 and (2*C+5*E+(5*C)div10) mod 10=(14+3+5*E mod10) mod 10 is 2 if and only if E is odd (either 1,3 or 9). Moving on for E, D,C the solution 675*31 = 837*25 is found. Not sure about Other than 10 base10 = 12 base8 and 31 base8 = 25 base10, don't have a clue. araver, nicely done on [1] and [3]. I'm wondering whether you're saying that Thanksgiving, because of leap year, does not occur on the 27th every seven years on average? On different years it's Nov 22 thru Nov 28. In a given year is one date more likely? Hint: [2] is not a "base" relationship; it's a rather ad hoc equivalence included to make a set of three. Quote Link to comment Share on other sites More sharing options...
0 araver Posted October 24, 2010 Report Share Posted October 24, 2010 araver, nicely done on [1] and [3]. I'm wondering whether you're saying that Thanksgiving, because of leap year, does not occur on the 27th every seven years on average? On different years it's Nov 22 thru Nov 28. In a given year is one date more likely? No, you're right. On average no day of the week is more "expected" to be 27th of November. So it is Thursday 27th in 1/7 years for an infinite amount of years. However, for humans (let's say a 100 years life expectancy) what you actually see is a different "rule": Thanksgiving is 27th November once every 6 years with a couple of big gaps as exceptions e.g. between 2014 - 2025. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 24, 2010 Report Share Posted October 24, 2010 No, you're right. On average no day of the week is more "expected" to be 27th of November. So it is Thursday 27th in 1/7 years for an infinite amount of years. However, for humans (let's say a 100 years life expectancy) what you actually see is a different "rule": Thanksgiving is 27th November once every 6 years with a couple of big gaps as exceptions e.g. between 2014 - 2025. Not really. Is Thursday in 57/400 years, as Sunday, Wednesday or Saturday. Is Monday or Tuesday in 58/400 years and Friday in 56/400 years. The Gregorian calendar has a cycle of 400 years. Quote Link to comment Share on other sites More sharing options...
0 araver Posted October 24, 2010 Report Share Posted October 24, 2010 Not really. Is Thursday in 57/400 years, as Sunday, Wednesday or Saturday. Is Monday or Tuesday in 58/400 years and Friday in 56/400 years. The Gregorian calendar has a cycle of 400 years. Indeed, in this cyclic calendar system (which we as a race seem to choose to acknowledge for now ) it is not as I suggested - as the cycle length(400) is not divisible by the number of weekdays(7). I did not see that . I got a different distribution of weekday possibilities for 27th Nov after checking (Monday 56,58,57,57,58,56,58) but the fact remains. Quote Link to comment Share on other sites More sharing options...
0 araver Posted October 25, 2010 Report Share Posted October 25, 2010 Hint: [2] is not a "base" relationship; it's a rather ad hoc equivalence included to make a set of three. 1031=97*10+61 1225=97*12+61 so they're equal modulo 97 and also share an interesting property - their quotients are the first two digits concatenated. Am I getting too far here? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted October 26, 2010 Author Report Share Posted October 26, 2010 1031=97*10+61 1225=97*12+61 so they're equal modulo 97 and also share an interesting property - their quotients are the first two digits concatenated. Am I getting too far here? That's it. Better, even, than the stock answer of equal mod 194. Quote Link to comment Share on other sites More sharing options...
Question
bonanova
Most puzzle aficionados know how to solve the "equation" Halloween = Christmas.
Let's extend the puzzle a bit.
Explain.
.
How could this be possible?
.
What value must "c" have?
.
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