If on monday, January, 1968 is the first day of the year. Which are the other 3 years which have the same day as monday for their first months irst day?
7 days later it is the same day again, so it is convenient to use a number system in which 7 = 0. It turns out that such a system exists. You can do addition and multiplication in the usual way and put 7 = 0 without any contradictions.
A year later it is 365 days later if the year is not a leap year and 366 days later otherwise.
You can compute which day it is 365 days later as follows by by identifying 7 with zero:
10 = 10 - 0 = 10 - 7 = 3
square both sides of 10 = 3:
100 = 9 = 9 - 0 = 9 - 7 = 2
multiply both sides by 3:
300 = 6 = -1
now 63 = 9*7 = 0, so:
363 = -1
add 2 to both sides:
365 = 1
So, 365 days later is the same day of the week as it is 1 day later. In case of a leap year it will be 2 days later.
1968 was a leap year, so Januari 1 1969 was a wednesday. If you go another year forward you move one day forward, etc. until you arrive at 1972. When you go to 1973 you have to add 2 days because 1972 was a leap year.
So, you just add the total number of days you move if you move forward from one Newyearsday to the next until the sum becomes 7:
2 + 1 + 1 + 1 + 2 = 7 -->
1973 is a year on which Januari 1 falls on a monday.
Why should there only be three other years starting with Monday, besides 1968? Do you mean since then?
Three such years are 1973, 1979 and 1990. You can find others using
http://www.timeanddate.com/calendar/index.html?year=1990&country=1
1968 is a leap year so 366 days
366/7 = 52 weeks and 2 days
so
1969 Wed Jan 1
365/7 = 52 weeks and one day
so
1970 Thurs Jan 1
1971 Fri Jan 1
1972 Sat Jan 1 Leap!
1973 Mon Jan 1 Remarkable <---
1974 Tues
1975 Wed
1976 Thurs Leap!
1977 Sat
1978 Sun
1979 Mon <-------
1980 Tues leap!
1981 Thurs
1982 Fri
1983 Sat
1984 Sun leap
1985 Tues (skip mon because of leap)
1986 Wed
1987 Thurs
1988 Fri leap!
1989 Sun
1990 Mon <----------
Check that carefully !
Of course there are a lot more than 3 of them
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To find the other three years in which Monday is the first day of the year, we can use a combination of manual calculation and some basic knowledge about calendars.
Here's how you can find the answer step by step:
Step 1: Determine the day of the week for January 1, 1968.
To calculate this, we can use Zeller's Congruence formula, which is a mathematical formula used to find the day of the week for any given date.
Zeller's Congruence formula:
h = (q + [(13(m + 1)) / 5] + K + [(K / 4)] + [(J / 4)] - 2J) % 7
Where:
h = Day of the week (0 = Saturday, 1 = Sunday, 2 = Monday, ..., 6 = Friday)
q = Day of the month
m = Month (3 = March, 4 = April, ..., 12 = December, 13 = January, 14 = February)
K = Year of the century (year % 100)
J = Zero-based century (actually year / 100)
Using this formula for January 1, 1968:
q = 1
m = 13 (January corresponds to month 13 of the previous year)
K = 68
J = 19 (20th century)
Plugging these values into the formula:
h = (1 + [(13(13 + 1)) / 5] + 68 + [(68 / 4)] + [(19 / 4)] - 2(19)) % 7
Simplifying:
h = (1 + [(169) / 5] + 68 + [17] + [4] - 38) % 7
h = (1 + 33.8 + 68 + 17 + 4 - 38) % 7
h = (85.8) % 7
h ≈ 1
Based on the calculation, January 1, 1968, was a Monday.
Step 2: Find the three other years in which Monday is also the first day of the year.
To determine this, we need to consider that the days of the week shift by one day in common years and two days in leap years.
Knowing that January 1, 1968, was a Monday, we can follow these steps to find the other three years:
1. Add 1 year to 1968:
1968 + 1 = 1969
2. Calculate the day of the week for January 1, 1969:
Using Zeller's Congruence formula as we did before:
h = (1 + [(13(14 + 1)) / 5] + 69 + [(69 / 4)] + [(19 / 4)] - 2(19)) % 7
Simplifying:
h = (1 + [(273) / 5] + 69 + [17] + [4] - 38) % 7
h = (1 + 54.6 + 69 + 17 + 4 - 38) % 7
h = (103.6) % 7
h ≈ 4
January 1, 1969, was a Wednesday.
3. Repeat steps 1 and 2 two more times, adding 1 year each time:
1969 + 1 = 1970
1970 + 1 = 1971
4. Calculate the days of the week for January 1, 1970, and January 1, 1971, using the same process as before.
For January 1, 1970:
h = (1 + [(13(15 + 1)) / 5] + 70 + [(70 / 4)] + [(19 / 4)] - 2(19)) % 7
Simplifying:
h = (1 + [(195) / 5] + 70 + [17] + [4] - 38) % 7
h = (1 + 39 + 70 + 17 + 4 - 38) % 7
h = (93) % 7
h = 2
January 1, 1970, was a Thursday.
For January 1, 1971:
h = (1 + [(13(16 + 1)) / 5] + 71 + [(71 / 4)] + [(19 / 4)] - 2(19)) % 7
Simplifying:
h = (1 + [(208) / 5] + 71 + [17] + [4] - 38) % 7
h = (1 + 41.6 + 71 + 17 + 4 - 38) % 7
h = (96.6) % 7
h ≈ 5
January 1, 1971, was a Friday.
So, the three other years in which Monday is the first day of the year are: 1969 (Wednesday), 1970 (Thursday), and 1971 (Friday).