Problem

You are given the following information,
but you may prefer to do some research for yourself.

1 Jan 1900 was a Monday.
Thirty days has September,
April, June and November.
All the rest have thirty-one,
Saving February alone,
Which has twenty-eight, rain or shine.
And on leap years, twenty-nine.
A leap year occurs on any year evenly divisible by 4, 
but not on a century unless it is divisible by 400.

How many Sundays fell on the first of the month during
the twentieth century (1 Jan 1901 to 31 Dec 2000)?

Answer

171

---

Python

#!/usr/bin/env python
from calendar import monthrange; from itertools import product
print(len([(year, month) for year, month in product(list(range(1901, 2001)), list(range(1, 13))) if monthrange(year, month)[0] == 6]))

#!/usr/bin/env python
months = [
    31,
    28,
    31,
    30,
    31,
    30,
    31,
    31,
    30,
    31,
    30,
    31
]

week = ['sun',
        'mon',
        'tue',
        'wed',
        'thu',
        'fri',
        'sat'
       ]

leapyears = [year for year in range(1, 101) if year % 4 == 0]

day = 0 # 1 Jan 1901
month = 0 # January
year = 1 # 1901
weekday = 2 # Tuesday

days = {day: week[weekday]} 

since_last_month = 0
since_last_year = 0

first_of_months = {}
while year <= 100:
    #print "%s/%s/%s - %s" % (month+1, since_last_month+1, year+1900, week[weekday])
    # if it's the first of the month, make a note 
    # of the weekday.
    if since_last_month == 0:
        first_of_months["%s/%s/%s" % (month+1, since_last_month+1, year+1900)] = week[weekday]

    # increment the counters..
    day += 1
    since_last_month += 1
    since_last_year += 1
    
    # check what year it is
    days_of_year = 365
    if year in leapyears:
        days_of_year += 1
    if since_last_year >= days_of_year:
        year += 1
        since_last_year = 0

    # check what month it is
    days_of_month = months[month]
    if month == 1 and year in leapyears: # february
        days_of_month += 1
    if since_last_month >= days_of_month:
        month += 1
        since_last_month = 0
        if month > len(months)-1:
            month = 0

    # check what day of the week it is
    weekday += 1
    if weekday > len(week)-1:
        weekday = 0
    days[day] = week[weekday]

def date_sort(date):
    date_str = date[0]
    month,day,year = date_str.split('/')
    return int(month)*int(day)*int(year)

print(len([weekday for weekday in list(first_of_months.values()) if weekday == 'sun']))

---

Ruby

#!/usr/bin/env ruby
require 'date'
puts Date.new(1901,1,1).upto(Date.new(2000,12,31)).find_all { |d| d.mday == 1 && d.wday == 0 }.count

---

Java

public final class p019 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p019().run());
	}
	
	
	/* 
	 * We use Zeller's congruence to compute the day of week when given the year, month, and day.
	 * Then we simply check the first day of all the months in the given range by brute force.
	 * 
	 * Zeller's congruence is well-known and a bit long to explain.
	 * See: https://en.wikipedia.org/wiki/Zeller%27s_congruence
	 */
	
	public String run() {
		int count = 0;
		for (int y = 1901; y <= 2000; y++) {
			for (int m = 1; m <= 12; m++) {
				if (dayOfWeek(y, m, 1) == 0)  // Sunday
					count++;
			}
		}
		return Integer.toString(count);
	}
	
	
	// Return value: 0 = Sunday, 1 = Monday, ..., 6 = Saturday.
	private static int dayOfWeek(int year, int month, int day) {
		if (year < 0 || year > 10000 || month < 1 || month > 12 || day < 1 || day > 31)
			throw new IllegalArgumentException();
		
		// Zeller's congruence algorithm
		int m = (month - 3 + 4800) % 4800;
		int y = (year + m / 12) % 400;
		m %= 12;
		return (y + y/4 - y/100 + (13 * m + 2) / 5 + day + 2) % 7;
	}
	
}

---

Mathematica

(* 
 * We simply use Mathematica's built-in date library to compute the answer by brute force.
 *)
<< Miscellaneous`Calendar`
Sum[Boole[DayOfWeek[{y, m, 1}] === Sunday], {y, 1901, 2000}, {m, 1, 12}]