The twelve months January, February, March, .........., November and December (in that order) of a year are respectively assigned the values 0,1,2,.....,10 and 11.
A given month corresponding to a specific year is defined as
succinct when the remainder obtained by dividing a given year by 12 precisely corresponds to the value assigned to a month containing
five Sundays. For example, 1990 when divided by 12 yields a remainder of 10; but November, 1990 did not contain five Sundays. Consequently November, 1990 was not a
succinct month.
Considering all the succinct months occurring between 1945 A.D. and 2176 A.D. inclusively in accordance with the
Gregorian calendar system, when does the
50th succinct month occur?
Determine the
maximum number of years between two consecutive succinct months in the stated period.
The succinct months are listed below, numbered:
# Mo Year # Mo Year
1 2 1946 41 9 2061
2 4 1948 42 11 2063
3 6 1950 43 5 2069
4 0 1956 44 7 2071
5 2 1958 45 9 2073
6 4 1960 46 11 2075
7 6 1962 47 3 2079
8 9 1965 48 5 2081
9 11 1967 49 7 2083
10 2 1970 50 3 2091
11 7 1975 51 8 2096
12 9 1977 52 10 2098
13 11 1979 53 0 2100
14 5 1985 54 3 2103
15 7 1987 55 8 2108
16 9 1989 56 10 2110
17 11 1991 57 0 2112
18 3 1995 58 4 2116
19 5 1997 59 6 2118
20 7 1999 60 8 2120
21 3 2007 61 10 2122
22 8 2012 62 0 2124
23 10 2014 63 2 2126
24 0 2016 64 4 2128
25 4 2020 65 6 2130
26 6 2022 66 0 2136
27 8 2024 67 2 2138
28 10 2026 68 4 2140
29 0 2028 69 6 2142
30 2 2030 70 9 2145
31 4 2032 71 11 2147
32 6 2034 72 2 2150
33 0 2040 73 7 2155
34 2 2042 74 9 2157
35 4 2044 75 11 2159
36 6 2046 76 5 2165
37 9 2049 77 7 2167
38 11 2051 78 9 2169
39 2 2054 79 11 2171
40 7 2059 80 3 2175
The 50th is April 2091.
The puzzles dates seem rather arbitrary; I don't know the significance of stopping at 2176.
The maximum years between such is 8, between August 1999 and April 2007, and between the same months of 2083 and 2091. This amounts to 2800 days between the starting days of the two months involved.
The longest number of days between the beginnings of succinct months lies outside the range given in the puzzle, between the starts of January 2196 and September 2204: 3165 days. This was done by extending the check to 1200 years beyond 1945, to 3145, as 1200 years is both a multiple of 12 and a multiple of the 400-year cycle of the Gregorian calendar.
The program below uses the Julian Day Number, a sequential numbering of days starting in 4713 BC.
DEFDBL A-Z
CLS
FOR y = 1945 TO 2176
mo = y MOD 12 + 1
ye = y
da = 1
GOSUB greg.to.jd
jd1 = jd
mo = mo + 1: IF mo > 12 THEN mo = 1: ye = y + 1
GOSUB greg.to.jd
jd2 = jd
dow = (jd1 + 1) MOD 7 ' sunday = 0
toGo = (7 - dow) MOD 7
effLength = jd2 - jd1 - toGo
sundays = -INT(-effLength / 7) ' ceiling
IF sundays = 5 THEN
ct = ct + 1
row = (ct - 1) MOD 40: col = ((ct - 1) 40) * 40
LOCATE row + 2, col + 2
PRINT USING "### ## ####"; ct; y MOD 12; y
'IF ct = 50 THEN STOP
END IF
NEXT y
END
greg.to.jd:
10100 REM :greg mo/da/ye --> jd at noon
10110 GOSUB jul.to.jd
10120 jd = jd + 2 - INT(cw(1) / 100) + INT(cw(1) / 400)
10130 RETURN
jul.to.jd:
10150 REM :jul mo/da/ye --> jd at noon
10160 cw(0) = mo: cw(1) = ye: IF mo < 3 THEN cw(0) = mo + 12: cw(1) = ye - 1
10170 jd = INT(365.25 * cw(1)) + INT(30.61 * (cw(0) + 1)) + da + 1720995!
10180 RETURN
Edited on December 6, 2008, 7:00 pm
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Posted by Charlie
on 2008-12-06 18:58:44 |
computer solution
