Historical story

What day of the week was 11th April 1948?

To find the day of the week for a given date, you can use Zeller's congruence:

$$h = (q + (26 * m +1) /10 + k + k/4 + j/4 + 5j) \ mod \ 7$$

where:

- h is the day of the week (0=Sunday, 1=Monday, ..., 6=Saturday).

- q is the day of the month.

- m is the month (1=January, 2=February, ..., 12=December).

- k is the year of the century (the last two digits of the year).

- j is the century (the first two digits of the year).

For 11th April 1948:

q = 11, m = 4, k = 48, j = 19

Substituting these values into Zeller's congruence:

$$h = (11 + (26*4+1)/10 + 48 + 48/4 + 19/4 + 5*19) \ mod \ 7$$

$$h = (11 + 105/10 + 48 + 12 + 4+ 95) \ mod \ 7$$

$$h = (11+10+48+12+4+95) \ mod \ 7$$

$$h = 180 \ mod \ 7$$

$$h = 1$$

Therefore, 11th April 1948 was a Sunday.

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