The days of the week (d.o.w) which are Sunday, Monday,
Tuesday,Wednesday,Thursday,Friday and Saturday (in that order) are respectively denoted by the numbers 0,1,2,3,4,5 and 6 . Any given year commencing with a particular d.o.w is assigned that value corresponding to that d.o.w. For example, the value ‘0’ would be assigned to a year commencing with a Sunday.
# A year is defined as ‘Matched’ if the remainder obtained, when the year is divided by 7, corresponds precisely with the value assigned to that particular year. For example, 2003 A.D. is NOT A Matched year since 2003 leaves a remainder of 1 upon division by 7 but January 1,2003 occurred on a Wednesday which is denoted by 3.
Determine the total number of ‘Matched’ years between 1960 A.D. and 2560 A.D.(both years inclusive) in accordance with the Western (Gregorian) Calendar System.
With the given that the Gregorian Calendar is made up of 365 days and adjusted every 4 years with the inclusion of a leap day it can be noted that each year begins with a d.o.w. sequentially following the prior one, skipping a day for the year immediately following a leapyear. This cycle repeats every 7 leapyear cycles (4 years), adjusted again for centennials non-divisible by 400.
3560134561234601245602345012 ß 7 leapyear cycles
1234560123456012345601234560 ß 4 d.o.w. cycles
As can be seen above, the first year of the ‘matched’ d.o.w. and mod leapyear cycle occurs 11 years before and 17 years after the current year of the cycle.
Since (2003 – 1960) = 43 => 1 d.o.w cycle (28 years) + 15 years, and as 15 years is greater than the 11 years before, it includes a complete ‘matched’ d.o.w.-mod leapyear cycle. Therefore, there are 8 ‘matched’ years (2 ‘matched’ d.o.w.-mod leapyear cycles) between 1960 and 2003.
And (2560 – 2003 (+1)) = 558 => 558 - 12 (3 leapyear cycles for non-leapyear centennials) => 19 d.o.w. cycles + 5 years. As 5 years is less 17 years, no additional ‘matched’ d.o.w.-mod leapyear cycles are included. Thus there are 76 ‘matched’ years between 2003 and 2560 (inclusive of both years).
In conclusion, the number of ‘matched’ years is 84 (= 8 + 76).
Posted by Dej Mar
on 2006-01-23 18:50:57