What day of the week was 26 January 2003?

Date: 26 January 2003

Let's assume 1 January 2024 is a monday

From 26 January 2003 to 1 January 2024, calculate the total days difference.

From 26 January 2003 to 26 January 2023 is 20 years.

In 20 years:

Leap years: 2004, 2008, 2012, 2016, 2020 = 5 leap years.

Regular years: 20 - 5 = 15

Total days =  (15×365) + (5×366)=7300

From 26 January 2023 to 1 January 2024:

Days = 340.

Total days = 7300+340=7640

Find the remainder:

7640 mod 7 = 0

0 mod 7 lands on the same day

Thus, 26 January 2003 was a Sunday.

Given: Date: 26 January 2003

For the Gregorian calendar:

​$$h = \left( q + \left\lfloor \frac{13(m + 1)}{5} \right\rfloor + K + \left\lfloor \frac{K}{4} \right\rfloor + \left\lfloor \frac{J}{4} \right\rfloor - 2J \right) \mod 7$$

Where:

h is the day of the week (0 = Saturday, 1 = Sunday, 2 = Monday, 3 = Tuesday, 4 = Wednesday, 5 = Thursday, 6 = Friday).

q is the day of the month.

m is the month (3 = March, 4 = April, ..., 12 = December, January and February are counted as months 13 and 14 of the previous year).

k is the year of the century (year mod 100).

j is the zero-based century (actually floor(year/100)).

26 January 2003

Adjusted month: January (13) and year: 2002 (since January is considered the 13th month of the previous year)

Day = 26 

month = 13 

year 02

century (j) = 20

​$$h = \left( 26 + \left\lfloor \frac{13(13 + 1)}{5} \right\rfloor + 02 + \left\lfloor \frac{02}{4} \right\rfloor + \left\lfloor \frac{20}{4} \right\rfloor - 2\times20 \right) \mod 7$$

​​$$h = \left( 26 + \left\lfloor \frac{182}{5} \right\rfloor + 02 + 0 + 5 - 40 \right) \mod 7$$

​​h= (26+36+02+0+5−40)  mod 7

h=  29  mod   7

h = 1

Therefore, 26 January 2003 was a Sunday.

Thus, correct option is (b)​​