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Succinct months (Posted on 2008-12-06) Difficulty: 2 of 5
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.

See The Solution Submitted by K Sengupta    
Rating: 2.0000 (1 votes)

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Solution computer solution Comment 2 of 2 |

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
  Posted by Charlie on 2008-12-06 18:58:44

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