Project Euler Q19

Question 19, here's the 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)?

nothing much I'll just brute force, counting from first day to the last.
first create a function for get the how many day in that month.
then just create 3 for loops eh? lol
_________________________here's the code__________
function getMonthDate(month,year)
if (month == 2) then
if (year % 4) == 0 and year ~= 1900 then
return 29
else
return 28
end
elseif (month == 1) or
(month == 3) or
(month == 5) or
(month == 7) or
(month == 8) or
(month == 10) or
(month == 12) then
return 31
else
return 30
end
end
function getDay(day)
if day == 1 then
return "sunday"
elseif day == 2 then
return "monday"
elseif day == 3 then
return "tuesday"
elseif day == 4 then
return "wednesday"
elseif day == 5 then
return "thursday"
elseif day == 6 then
return "friday"
elseif day == 7 then
return "saturday"
else
return "error"
end
end


d = { num = 1, day = 2, month = 1, year = 1900}
match = 0

for i = d.year, 2000 do
for j = d.month, 12 do
--print("_____________________________")
--print("the month is "..d.month)
--print ("this month has "..getMonthDate(d.month, d.year).." days")
for k = d.num, getMonthDate(d.month, d.year) do
--print("the day is "..d.num.." on "..getDay(d.day))
if (d.num == 1) and (d.day == 1) and (d.year > 1900) then
match = match + 1
end
if d.day == 7 then d.day = 0 end
if d.num == getMonthDate(d.month, d.year) then d.num = 0 end
d.day = d.day + 1
d.num = d.num + 1

end
if d.month == 12 then d.month = 0 end
d.month = d.month + 1
end
d.year = d.year + 1
end

print (match)