Step 1: Understanding the Question:
The question asks us to determine the exact day of the week for a specific historical date: 26 January 1950.
This is a standard calendar problem that requires calculating the total number of "odd days" from the beginning of the Gregorian calendar to this specific date.
Step 2: Key Formula or Approach:
We use the odd days counting method.
An ordinary year has 365 days, which equals 52 weeks and 1 odd day.
A leap year has 366 days, which equals 52 weeks and 2 odd days.
400 years (and multiples like 800, 1200, 1600, 2000) have exactly 0 odd days.
100 years have 5 odd days, 200 years have 3 odd days, and 300 years have 1 odd day.
The final remainder of odd days maps to the week: 0 = Sunday, 1 = Monday, ..., 4 = Thursday.
Step 3: Detailed Explanation:
The period up to the date can be split as: 1600 years + 300 years + 49 completed years + 26 days of January 1950.
The first 1600 years contribute exactly 0 odd days.
The next 300 years (1601 to 1900) contribute exactly 1 odd day.
Now we analyze the 49 completed years from 1901 to 1949.
To find the number of leap years in this 49-year period, we divide 49 by 4, which gives 12 leap years.
The remaining 37 years are ordinary years.
The total odd days for these 49 years is $(12 \times 2) + (37 \times 1)$.
This equals $24 + 37 = 61$ odd days.
We find the remainder of 61 divided by 7 to simplify: $61 \equiv 5 \pmod 7$.
Finally, we look at the year 1950. The date is 26 January, meaning 26 days have passed.
We find the remainder of 26 divided by 7: $26 \equiv 5 \pmod 7$.
Now, we sum all the odd days calculated: $0 \text{ (1600)} + 1 \text{ (300)} + 5 \text{ (49 years)} + 5 \text{ (Jan 1950)}$.
Total odd days = $1 + 5 + 5 = 11$.
We divide 11 by 7 to find the final remainder: $11 \equiv 4 \pmod 7$.
Mapping the remainder 4 to the days of the week (0=Sun, 1=Mon, 2=Tue, 3=Wed, 4=Thu), we get Thursday.
Step 4: Final Answer:
The day of the week on 26 January 1950 was Thursday.