Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

When will Ramadan fall across 3 months?

Date: 5 Mar 1995 07:50:34 -0500
From: D. Albert Szilak
Subject: Calendar problem

   Here's the problem & answer. How is the answer obtained? 
What's the proof?  This year the Muslim holy month of Ramadan 
fell across 3 months of the western calendar. It began 31 Jan & 
ended 1 Mar. Each month in the Islamic calendar has exactly 30 
days. The beginning of Ramadan each year moves back 11 days 
in the western calendar. So next year it will begin 20 Jan. 1996 
is a Leap year. Question is, when will Ramadan again fall across 
3 months?  In other words, when again will it begin on 31 Jan in 
a non-leap year? Answer I have is after 719 years, in 2714? 
What's the calculation look like?


Date: 7 Mar 1995 13:40:02 -0500
From: Dr. Ken
Subject: Re: Calendar problem

Hello there!

Do you know how modulo arithmetic works?  It consists of 
looking at the remainders upon division by the base.  So if 
we're doing arithmetic modulo 5, for example, all the 
following act exactly the same: 2,7,12,17,22,..., and all the 
following are the same: 16,11,6,1,-4,-9,-14,....  If we add 4
and 3 modulo 5, we get 7, which is congruent to 2, so 
4+3==2 (Mod 5).  I'll use the symbol == to mean "is 
congruent to."  Modulo arithmetic is a pretty useful 
technique to use in certain problems (like this one!).  

How does it apply here?  Well, we want to find out when the 
first day of Ramadan will fall on the 31st day of the Gregorian 
calendar.  So we want to add 365-11 a bunch of times (to see 
where Ramadan will fall the next year), and see when the 
remainder upon division by 365 is 31.  So we'd want to solve 
the congruence equation 31 + n(365 - 11)==31 (Mod 365) for 
n, i.e. 31 + 354n ==31 (Mod 365).

That will give us n, the number of years from now (in the 
Islamic calendar) until Ramadan falls on the 31st of January.

Actually, that's the simplified version.  See, leap year gets in 
there and muddles it up.  For the sake of this problem, let's 
assume leap year occurs every 4 years, with complete 
regularity, because I don't know offhand when the 
irregularities are.  So our regular cycle, starting with this year, 
has 1 year with 365 days, then 1 with 366, then 2 with 365, 
for a total of 1461 days in a complete cycle.  Then we want to 
find out which days in this cycle are desirable for the 
beginning of Ramadan to fall on, (31, 762, 1127) and solve 
the new congruences:  
               31 + 354n ==  31 (Mod 1461)
               31 + 354n == 762 (Mod 1461)
               31 + 354n ==1127 (Mod 1461)

That's pretty much how you would solve it.  Of course, we 
didn't take into account the irregularities in the leap year 
schedule, so we're going to be off.  But this is the technique, 

-Ken "Dr." Math
Associated Topics:
High School Puzzles

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994-2013 The Math Forum