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One Solar Cycle = Three 365 day years + One 366 day year.
One Lunar Cycle = Lunar Cycle


Within the lunar year, there are 364 days, so in order to not fall behind, ever third year we add an intercalary month that consists of 30 days. That is the yellow segment you see in the picture. This little cycle happens again - 3 years, at the end of the third year adding an intercalary month. But then, in order to completely align ourselves to the solar cycle, what we do is: after two years, add an intercalary month. (FYI, FTR, that little yellow segment, the intercalary month, is part of the third year, not the fourth)

enter image description here


Now, in the picture above, you can see the little cycle. For reference, the lunar year has 12 months. Month One will have 30 days, month two 29, month three 30, etc. The third year, however, will have 13 months (because of the intercalary month).


Now let's say that the first day ($x$, shown in the picture) of the first 3 year cycle is Sunday. After the intercalary month, year four, $x$ = Sunday, also for the beginning of the 6th year. However, once the two year cycle ends, 9th year begins (a new lunar cycle), the first day, $x$, is now Wednesday.


From this, I assumed that the next 3 year cycle, and the beginning of the 4th year, will be Wednesday.
From this you can also assume, that the first day of the third lunar cycle will be $x + 3 = Saturday$

How can I use this information to derive a formula that will let me calculate $x$ for the a given lunar cycle (i.e, the day of the week that the lunar cycle will begin on? (e.g. Lunar cycle 189 will start on Saturday....)

I am appealing to help on this site, because it is the only place I know of (online) where a bunch of people are willing to give up their time to answer questions - so, thank you a ton!


I haven't got the slightest idea of how to solve this, or even if it's possible (however, I think it is highly probable that it is possible).

Thanks a ton, guys!

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  • $\begingroup$ 8 solar years has 8*365 + 2 = 2922 days. 8 unaugmented lunar years has 8* 364 = 2912 days, or only 10 days fewer. So why are you adding 90 extra days per 8 year cycle? That puts you much farther ahead than you would be behind without the intercalary months? And what are the "etc" of the month lengths? Do they alternate between 30 and 29? Or is only the 2nd month 29 days and all the rest 30? or are there 31 day months too? $\endgroup$ – Paul Sinclair Mar 21 '17 at 16:52
  • $\begingroup$ "Within the lunar year, there are 364 days, so in order to not fall behind, ever third year we add an intercalary month that consists of 30 days" This doesn't quite compute. $\endgroup$ – n.m. Mar 21 '17 at 16:56
  • $\begingroup$ I was simplifying a solar year - One solar year = 365.25 days * 4 = 1461 days. 1461 days * 2 = 2292 = 8 years. 8 lunar years = 364 days * 3 = 1062 + 30 = 1092. 1092 * 2 = 2184 days (6 years) + 364(2) + 30 = 2922 days = 8 Lunar years $\endgroup$ – Carlos Carlsen Mar 21 '17 at 16:56
  • $\begingroup$ I am talking about reconciling 1 lunar cycle to 2 solar cycles. $\endgroup$ – Carlos Carlsen Mar 21 '17 at 16:57
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    $\begingroup$ 6*30 + 6*29 = 180 + 174 = 354, not 364, so you fall short by 11.25 days each year. So 8 unaugmented lunar years will be 90 days shorter than 8 solar years. Now it adds up. $\endgroup$ – Paul Sinclair Mar 21 '17 at 23:23
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Now that the error in your numbers has been spotted, the question you ask is fairly easy. The 8 year cycle lasts $2922$ days. A week has $7$ days, so we just divide: $2922 = 417 \times 7 + 3$. Thus as you've figured out, each cycle starts 3 days later in the week than the previous cycle. To calculate the beginning day for cycle $n$, you need $$x - 1 \equiv 3(n-1) \equiv 3n - 3 \mod 7$$

The subtraction of $1$ from $x$ is so that your answer will be $1$ to $7$ instead of $0$ to $6$. Subtracting $1$ from $n$ is needed because you are counting that cycle that begins on Sunday as $1$ rather than $0$.

If you have a computer, you can use x = mod(3*n -3, 7) + 1 If you are doing this by hand, I recommend

  1. Subtract $1$ fron $n$.
  2. Divide the result by $7$, discarding the quotient and keeping the remainder.
  3. Multiply the remainder by $3$.
  4. Divide that by $7$ again, discarding the quotient and keeping the remainder.
  5. Add $1$.

The reason for dividing by $7$ twice is that it makes the numbers you are dealing with much smaller, which makes the subsequent calculations easier, more than making up for the added difficulty of doing the second division.

For example, you mentioned the starting day for cycle 189:

  1. Subtract 1: 188
  2. Divide by 7:$$\require{enclose} \begin{array}{r} 26 \\[-3pt] 7 \enclose{longdiv}{188} \\[-3pt] \underline{14}\phantom{8} \\[-3pt] 48 \\[-3pt] \underline{42}\\[-3pt] 6\end{array} $$We don't care about the $26$, but the remainder is $6$.
  3. Multiply by $3$: $18$
  4. Divide by $7$ again: $18 = 2\times 7 + 4$, so $4$.
  5. Add 1: $5$.

Thus cycle 189 starts on day $5$ = Thursday.

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  • $\begingroup$ This is great information! Thank you so much!! $\endgroup$ – Carlos Carlsen Mar 22 '17 at 18:05
  • $\begingroup$ You don't know how much you have helped @Paul Sinclair! $\endgroup$ – Carlos Carlsen Mar 22 '17 at 22:13
  • $\begingroup$ Could you help me with another problem??? $\endgroup$ – Carlos Carlsen Mar 24 '17 at 0:52
  • $\begingroup$ Only if I know what the problem is. $\endgroup$ – Paul Sinclair Mar 24 '17 at 14:55
  • $\begingroup$ :D Thanks - math.stackexchange.com/questions/2200651/… $\endgroup$ – Carlos Carlsen Mar 24 '17 at 16:57

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