Concept:
A month has 5 Fridays if it has enough days such that one of the Fridays falls within the first three days of a 31-day month, the first two days of a 30-day month, or on the first day of a 29-day month. Specifically, in a month with 31 days, if it starts on a Friday, Thursday, or Wednesday, it will contain 5 Fridays.
Calculation:
First, we determine if 1926 is a leap year. Since 1926 is not divisible by 4, it is an ordinary year with 365 days.
Now, let us find the day of the week on January 1, 1926 by calculating the number of odd days:
Odd days until the year 1600 = 0
Odd days from 1601 to 1900 = 1
Odd days from 1901 to 1925 (25 years):
Number of leap years in 25 years = 25 ÷ 4 = 6 leap years
Total odd days in 25 years = 25 (one for each year) + 6 (extra for leap years) = 31 days
31 days ≡ 3 odd days (since 31 ÷ 7 leaves a remainder of 3)
Total odd days until December 31, 1925 = 1 + 3 = 4 odd days.
The 4th odd day corresponds to Thursday. Therefore, January 1, 1926 was a Friday.
January has 31 days. Since it started on a Friday, the Fridays in January 1926 were:
January 1, 8, 15, 22, and 29.
This gives a total of 5 Fridays.
Checking the other options for confirmation:
Hence, January is the month that had 5 Fridays.
Alternate Method
Using the day determination formula for January 1, 1926:
Day = (Date + Month Code + Century Code + Year + Leap Year Code) mod 7
Total = (1 + 0 + 0 + 26 + 6) = 33
33 ÷ 7 = 4 weeks and a remainder of 5.
In the day code system (0=Sun, 1=Mon, 2=Tue, 3=Wed, 4=Thu, 5=Fri, 6=Sat), the remainder 5 represents Friday.
Since January 1 is a Friday and the month has 31 days, it contains 5 Fridays.
Hence, the correct answer is January.