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WQ Archive 211 - 220

Question 211

Which 2 reasons make 357-day years impossible in the currently fixed Hebrew calendar?

Answer

The first answer to this question is found in Properties of Hebrew Year Periods - Part 1 (10. The 357-Day Year Paradox).

This part of the answer is explained as follows, where

d  = 24h   0p; d' = 23h 1079p; d" = 24h   1p
c  =  8h 876p; c' =  8h  875p; c" =  8h 877p
g  = 15h 204p; g' = 15h  203p; g" = 15h 205p
b  = 21h 589p; b' = 21h  588p; b" = 21h 590p
f  = any fractional part of a day such that 0 <= f <= d'
p' = the number of days that Tishrei 1 is postponed at the start of the year
p" = the number of days that Tishrei 1 is postponed at the end   of the year
T' = the molad of Tishrei of some year H'
T" = the molad of Tishrei of some subsequent year H"
D' = the day of 1 Tishrei H'
D" = the day of 1 Tishrei H"
[f+m] = the integer value of f+m

If the rule to eliminate 356 day years is applied when
     
     M = 354d, [f+m] = 1d, p" = 2d and p' = 0d
     
then 
     
     L = 354d + 1d + 2d - 0d = 357d .
     
The 357-day year does not occur in the fixed Hebrew calendar. 

The paradox can be explained as follows.

To eliminate the 356-day years, a postponement to Thursday (5d)
is made whenever the fractional part of the molad of Tishrei on a
Tuesday (3d) is greater than d'-m.

The fractional part of the molad is given by f + m - [f+m].

Since f + m - [f+m] > d'- m  and [f+m] = d, (by hypothesis)
          f + m - d > d'- m
Hence     f + m     > d + (d'- m)
and           m     > d + (d'- m) - f

The inequality shows that the minimum value for m
will be when f is at its maximum value
which is d' (by definition).

Hence, the minimum value of m required to develop
a 357 day year is given by
          
          m > d + (d'- m) - d'
          m > d - m
          m > d / 2 
          

In the fixed Hebrew calendar, the actual value of m for the 
12 month year is 8h 876p, which is less than d/2.

That is why the 357-day year cannot be generated in the
fixed Hebrew calendar.

The second reason for which the 357-day year cannot be generated in the fixed Hebrew calendar is found in a Talmudic ruling.


Question 212

Which rule in the Babylonian Talmud eliminates the possibility of a 357-day year?

Answer

The following Mishnah appears on page 8b of tractate Arakin in the Babylonian Talmud.

There are never less than four full months in the year, nor did it seem right to have more than eight. The two loaves were consumed never earlier than the second, nor later than the third day. The shewbread was consumed never earlier than the ninth nor later than the eleventh day. An infant may never be circumcised earlier than the eighth nor later than the the twelfth day.

The first line of the Mishnah effectively limits the Hebrew years to no less than 4 29-day months in the 12-month years, and no less than 5 29-day months in the 13-month years. Hence, the 12-month years can have at most 356 days, while the 13-month years are limited to at most 385-days.

Hence, the ruling on page 8b of tractate Arakin in the Babylonian Talmud eliminates the possibility of a 357-day year.


Question 213

What impact does the prohibition of the 357-day years have on any of the arithmetic rules of the Hebrew calendar?


Answer

The following Mishnah appears on page 8b of tractate Arakin in the Babylonian Talmud.

There are never less than four full months in the year, nor did it seem right to have more than eight. The two loaves were consumed never earlier than the second, nor later than the third day. The shewbread was consumed never earlier than the ninth nor later than the eleventh day. An infant may never be circumcised earlier than the eighth nor later than the the twelfth day.

The first line of the Mishnah effectively limits the Hebrew years to no less than 4 29-day months in the 12-month years, and no less than 5 29-day months in the 13-month years. Hence, the 12-month years can have at most 356 days, while the 13-month years are limited to at most 385-days.

Hence, the ruling on page 8b of tractate Arakin in the Babylonian Talmud eliminates the possibility of a 357-day year.

Not apparent in the reference literature is the unknown fact that this Mishnah has its greatest impact on the form of the very first Dehiyyah (postponement rule) which has been set as Lo ADU Rosh. This rule prevents the first day of Tishrei from occurring either on a Sunday, Wednesday, or Friday.

What is of significance here is not the days eliminated for 1 Tishrei, but rather, the fact that the Mishnah prohibits two consecutive weekdays from being named as postponement days. In other words, Sunday and Monday could not be both named as postponement days for Rosh Hashannah, for if that had been the case, then a 357-day year would have been possible.

Referring to Properties of Hebrew Year Periods - Part 1, the length L of any period of Hebrew years is given by

L = D" - D' = M + INT(f+m) + p" - p'
where

M  = the integer    portion of the molad period for the given period of Hebrew years
m  = the fractional portion of the molad period for the given period of Hebrew years
f  = any fractional part of a day such that 0 <= f <= d'
p' = the number of days that Tishrei 1 is postponed at the start of the year
p" = the number of days that Tishrei 1 is postponed at the end   of the year
D' = the day of 1 Tishrei H'
D" = the day of 1 Tishrei H"

Assuming any 2 consecutive days as forbidden days for 1 Tishrei, then, whenever the day of the molad of the subsequent year is the first of the proscribed 2 days, p" = 2d.

When p' = 0, INT(f+m) = 1, and p" = 2, then the single year length becomes L = 354 + 1 + 2 = 357 days.

Thus, the 357-day single Hebrew year would be created were 2 consecutive weekdays allowed for the postponement of 1 Tishrei.

Consequently, it is seen that the Mishnah at Arakin 8b actually limits the number of postponement days possible to a maximum of 3, each of which is separated from the other by at least one day.

There is a side issue as to which 3 days can be selected for such purposes since there are exactly 7 ways of making that selection. These selections quite obviously are, {1d,3d,5d}, {2d,4d,6d}, {3d,5d,7d}, {1d,4,6d}, {2d,5d,7d}, {1d,3d,6d}, and {2d,4d,7d}. It is interesting to note that, by coincidence, ADU lies at the very centre of these 7 possibilities.


Question 214

Assuming Tishrei 1 postponements of not more than 6 days, what is the length of the smallest Hebrew year that could then be generated?


Answer

Referring to Properties of Hebrew Year Periods - Part 1, the length L of any period of Hebrew years is given by

L = D" - D' = M + INT(f+m) + p" - p'
where

M  = the integer    portion of the molad period for the given period of Hebrew years
m  = the fractional portion of the molad period for the given period of Hebrew years
f  = any fractional part of a day such that 0 <= f <= d'
p' = the number of days that Tishrei 1 is postponed at the start of the year
p" = the number of days that Tishrei 1 is postponed at the end   of the year
D' = the day of 1 Tishrei H'
D" = the day of 1 Tishrei H"

Were the Tishrei 1 postponement up to 6 days, then it would appear that, that if p" = 0d while p' = 6d, then the smallest length of the single year possible, would be L = 354d - 6d + 0d = 348d.

However, for reasons that are peculiarly delightful, the smallest length that would be generated is 350d. The following table shows the simulation of the required conditions.

Qeviyyot when DAYS 1 2 3 4 5 6 are OUT
YEAR LENGTH IN DAYS
DAY 350 357 378 385 TOTALS
Sat 163785 271671 40000 214016 689472
TOTALS 163785 271671 40000 214016 689472

It is also interesting to note that the largest single Hebrew year length remains locked at 385d and not the anticipated 390d.

The reasons for these year length limits will be left to anyone willing to tell the Weekly Question why.


Question 215

Assuming the postponements of 1 Tishrei, what are the minimum number of days of postponement required in order to generate 350-day years?

Answer

Referring to Properties of Hebrew Year Periods - Part 1, the length L of any period of Hebrew years is given by

L = D" - D' = M + INT(f+m) + p" - p'
where

M  = the integer    portion of the molad period for the given period of Hebrew years
m  = the fractional portion of the molad period for the given period of Hebrew years
f  = any fractional part of a day such that 0 <= f <= d'
p' = the number of days that Tishrei 1 is postponed at the start of the year
p" = the number of days that Tishrei 1 is postponed at the end   of the year
D' = the day of 1 Tishrei H'
D" = the day of 1 Tishrei H"

Were the year length 350 days, then it would appear that the minimum Tishrei 1 postponement needed would be when p" = 0d thereby making L = 354d - p' + 0d = 350d.

From that equality, it becomes clear that p' = 354d - 350d = 4d.

Hence, the smallest number of week days of postponement needed to achieve a 350-day year would be 4 consecutive week days.

The following table, showing the first 4 week days as forbidden to 1 Tishrei, illustrates the answer.

Qeviyyot - WEEK DAYS 1 2 3 4 Forbidden
YEAR LENGTH IN DAYS
DAY 350 351 352 355 356 357 379 380 383 384 385 TOTALS
Thu 39369 62208 62208 0 0 147255 3712 36288 0 0 141440 492480
Fri 0 0 0 0 62208 0 0 0 0 36288 0 98496
Sat 0 0 0 62208 0 0 0 0 3712 32576 0 98496
TOTALS 39369 62208 62208 62208 62208 147255 3712 36288 3712 68864 141440 689472


Question 216

Would any combination of at least 4 week days forbidden to 1 Tishrei lead to 350-day years?


Answer

NO! The 350-day year is only possible when at least 4 consecutive week-days are forbidden for 1 Tishrei. These forbidden days may include any 4 consecutive week-days.

Referring to Properties of Hebrew Year Periods - Part 1, the length L of any period of Hebrew years is given by

L = D" - D' = M + INT(f+m) + p" - p'
where

M  = the integer    portion of the molad period for the given period of Hebrew years
m  = the fractional portion of the molad period for the given period of Hebrew years
f  = any fractional part of a day such that 0 <= f <= d'
p' = the number of days that Tishrei 1 is postponed at the start of the year
p" = the number of days that Tishrei 1 is postponed at the end   of the year
D' = the day of 1 Tishrei H'
D" = the day of 1 Tishrei H"

Were the year length 350 days, then it would appear that the minimum Tishrei 1 postponement needed would be when p" = 0d thereby making L = 354d - p' + 0d = 350d.

From that equality, it becomes clear that p' = 354d - 350d = 4d.

Hence, the smallest number of week days of postponement needed to achieve a 350-day year would be 4 consecutive week days.

The following tables, showing the first 4, 5, and 6 week days as forbidden to 1 Tishrei, illustrate the answer.

Keviyyot - DAYS 1 2 3 4 OUT
YEAR LENGTH IN DAYS
DAY 350 351 352 355 356 357 379 380 383 384 385 TOTALS
Thu 39369 62208 62208 0 0 147255 3712 36288 0 0 141440 492480
Fri 0 0 0 0 62208 0 0 0 0 36288 0 98496
Sat 0 0 0 62208 0 0 0 0 3712 32576 0 98496
TOTALS 39369 62208 62208 62208 62208 147255 3712 36288 3712 68864 141440 689472

Keviyyot - DAYS 1 2 3 4 5 OUT
YEAR LENGTH IN DAYS
DAY 350 351 356 357 378 379 384 385 TOTALS
Fri 101577 62208 0 209463 3712 36288 0 177728 590976
Sat 0 0 62208 0 0 0 36288 0 98496
TOTALS 101577 62208 62208 209463 3712 36288 36288 177728 689472

Keviyyot - DAYS 1 2 3 4 6 OUT
YEAR LENGTH IN DAYS
DAY 350 352 355 357 380 383 385 TOTALS
Thu 39369 124416 0 147255 40000 0 141440 492480
Sat 0 0 124416 0 0 40000 32576 196992
TOTALS 39369 124416 124416 147255 40000 40000 174016 689472

Keviyyot - DAYS 1 2 3 4 5 6 OUT
YEAR LENGTH IN DAYS
DAY 350 357 378 385 TOTALS
Sat 163785 271671 40000 214016 689472
TOTALS 163785 271671 40000 214016 689472


The Hebrew-Gregorian Calendar Correspondence Cycle indicates that

The Hebrew calendar fully repeats itself in a cycle of 689,472 Hebrew years.
The Gregorian calendar fully repeats itself in a cycle of 400 Gregorian years.

Due to the fact that both the Hebrew years and the Gregorian years are of different lengths, the correspondence between the Hebrew and Gregorian calendars repeats itself in a cycle of
14,389,970,112 Hebrew years which is also 14,390,140,400 Gregorian years!


Question 217

What is the length of the Hebrew to Julian calendar correspondence cycle?


Answer

The Hebrew-Gregorian Calendar Correspondence Cycle indicates that

The Hebrew calendar fully repeats itself in a cycle of 689,472 Hebrew years.
The Gregorian calendar fully repeats itself in a cycle of 400 Gregorian years.

Due to the fact that both the Hebrew years and the Gregorian years are of different lengths, the correspondence between the Hebrew and Gregorian calendars repeats itself in a cycle of
14,389,970,112 Hebrew years which is also 14,390,140,400 Gregorian years!

The length of these two cycles is derived as follows:

The Hebrew calendar cycles perfectly over 689,472 Hebrew years, and is exactly 251,827,457 days long.

The Gregorian calendar cycles prefectly over 400 Gregorian years, and is exactly 146,097 days long.


Let HC = the Hebrew calendar cycle = 689,472 Hebrew years = 251,827,457 days
Let GC = the  Gregorian calendar cycle = 400 Gregorian years = 146,097 days
Let JC = the Julian calendar cycle = 28 Julian years = 10227 days

In order that the Hebrew and Gregorian calendars repeat their corresponding dates in the same order for all of their years, it is necessary that, in this cycle, an integral number of Hebrew calendar cycles be equal in length to an integral number of Gregorian calendar cycles.

Since the Hebrew years are of different length than the Gregorian years, the cycles must be first calculated in days.

Hence, for the Hebrew-Gregorian mapping cycle it is necessary that

HC * x = GC * y where x and y are integers, so that 251,827,457 days * x = 146,097 days * y

Now, 251,827,457 days = 7 * 47 * 131 * 5843 days and 146,097 days = 7 * 27 * 773 days

reducing the above equation to

47 * 131 * 5843 * x = 27 * 773 * y

From that, it may be seen that the smallest integers x and y satisfying the equation are
x = 27 * 773 and y = 47 * 131 * 5843.

Consequently, in complete years, it takes

27 *773 * 689,472 Hebrew years = 14,389,970,112 Hebrew years = 5,255,890,855,047 days

to equal

47 * 131 * 5843 * 400 Gregorian years = 14,390,140,400 Gregorian years = 5,255,890,855,047 days

We similarly develop the Hebrew-Julian calendar mapping cycle.

HC * x = JC * y

7 * 47 * 131 * 5843 * x = 7 * 3 * 487 * y

Consequently, in complete years, it takes

3 * 487 * 689,472 Hebrew years = 1,007,318,592 Hebrew years = 367,919,914,677 days to equal

47 * 131 * 5843 * 28 Julian years = 1,007,309,828 Julian years = 367,919,914,677 days

Hence, the length of the Hebrew to Julian correspondence (mapping) cycle is 367,919,914,677 days.


Question 218

Why is Pesach so late in April 2005?


Answer

The difference de between the lengths of 13 months (13 * m) and the solar year is

de = 13 * m - 235 * m / 19 = 12 * m / 19

For HY Hebrew years

13 * m * HY = HY * 235 * m / 19 + HY * de

Whenever HY * de > m, m is subtracted so that 0 <= HY * de - m < m

0 <= HY * 12 * m / 19 - m < m 0 <= HY * 12 * m / 19 - 19 * m / 19 < m 0 <= HY * 12 - 19 < 19

That result is the remainder of HY * 12 / 19.

Let R(12 * HY, 19) represent that remainder.

Then, the number of months in HY years can be given as

NUMMONTHS(HY) = HY * 235 / 19 + R(12 * HY, 19) / 19

In the expression, the cycle as defined in the R() function needs to be slightly modified so as to line up with the Hebrew calendar 13-month distribution. This is done by adding 5 to the expression HY * 12 + Y.

Hence,


NUMMONTHS(HY) =  HY * 235 / 19 + R(12 * HY + 5, 19) / 19
= [HY * 235 + R(12 * HY + 5, 19)] / 19
NUMMONTHS will always evaluate to an integer.

When HY = 9, R(12 * HY + 5, 19) = 18, which is the maximum possible value for that expression. In turn, this causes the greatest possible differences between the lunar and solar years in the mahzor qatan, and as a result, 1 Tishrei appears at the latest possible times in the solar calendars.

In the fixed Hebrew calendar, Pesach always precedes 1 Tishrei by exactly 163 days. Consequently, whenever 1 Tishrei is very late in the solar calendar, so will be the occurrence of Pesach.

As an example, Pesach 5727H began on Tuesday 25 April 1967g, while Rosh Hashannah 5728H began on Thursday 5 October 1967g. These dates are currently the latest possible days in the solar calendar for both of these Holidays. As well, the Hebrew year 5728H is also the 9th year in the fixed Hebrew calendar's mahzor qatan.

Since the Hebrew year 5766H is the 9th year of the mahzor qatan, Pesach 5765H takes place relatively late in the solar year, beginning on Sunday 24 April 2005g.

Correspondents Tzvi D. Goldman, Larry Padwa, and Ari Brodsky all sent ingenious answers whose observational deductions appear to be correct. Correspondent Tzvi D. Goldman's solution was in Hebrew and will be excerpted once an English translation has been made available for such purpose.

Correspondent Larry Padwa observed:


First, note the obvious fact that Hebrew dates (after Adar) will be
later in leap years than in common years, due to the extra month.

Leap years occur in years 3,6,8,11,14,17,19  of the nineteen year cycle.

Note that two of these years (8 and 19) follow the previous leap year
by only two years. All of the others follow the previous leap year by
three years. Thus years 8 and 19 will have Hebrew dates (after Adar I)
that are later (relative to the Gregorian calendar) than dates in the other
five leap years of the cycle because they have been subject to two leap 
years out of the past three years. That is, of years 6,7, and 8, two of them
are leap years. The same thing happens with years 17, 18, and 19.

Now Pesach in 2005G is in 5765H (post Adar I) which is year 8 of the
cycle. Hence Pesach (and all dates in 2005 after Adar I) are relatively late.
It may be noted that when HY = 1, R(HY * 12 + 5, 19) = 17.

Correspondent Ari Brodsky deduced that:-


The 8th year of the 19-year cycle concludes a span of 3 
consecutive years containing 2 leap years (the 6th and 8th years 
of the cycle are both leap years).  There is only one other year 
in the cycle, namely the 19th, that shares this property.  
Any other 3-year span, concluding in a year other than the 8th or 19th,
contains only one leap year.  

This means that holidays in the 8th or 19th year of the cycle will be 
later than in other years, because there have been 2 leap years in a 
shorter period of time than usual.  

Furthermore, the 8th year of the cycle concludes a span of 11 consecutive 
years containing 5 leap years (the 17th, 19th, 3rd, 6th, and 8th years of 
the cycle are leap years).  This is the only year in the cycle with this 
property.  

Any other 11-year span contains only 4 leap years, even the span ending 
with the 19th year of the cycle.  This means that holidays in the 8th year 
of the cycle will be later than in all other years, because there have 
been 5 leap years in a shorter period of time than usual.

The result of all this is that all Jewish calendar dates and holidays 
from Adar II of the 8th year of the cycle (in our case, this year 5765) 
until Shevat of the following year (5766) will fall later in the solar year
(i.e. relative to the civil calendar) than they do in all other years 
of the 19-year cycle.  So thats why everything seems so late this year.  
The holidays havent been this late since 19 years ago, the year 5746 
(1986).
Thank you very much correspondents Tzvi D. Goldman, Larry Padwa, and Ari Brodsky for sharing your remarkable observations.


Question 219

Which year, or years, of the mahzor qatan (19 year cycle) would have the earliest starting dates in a solar calendar?


Answer

The difference dc between the lengths of the solar year and 12 months (12 * m) is

dc = (235 * m / 19) - 12 * m = 7 * m / 19

For HY Hebrew years

12 * m * HY = HY * 235 * m / 19 - HY * dc

Whenever HY * dc > m, m is added so that 0 <= HY * de + m < m

0 <= HY * 7 * m / 19 - m < m 0 <= HY * 7 * m / 19 - 19 * m / 19 < m 0 <= HY * 7 - 19 < 19

That result is the remainder of HY * 7 / 19.

Let R(7 * HY, 19) represent that remainder.

Then, the number of months in HY years can be given as

NUMMONTHS(HY) = (HY * 235 / 19) - R(12 * HY, 19) / 19

In the expression, the cycle as defined in the R() function needs to be slightly modified so as to line up with the Hebrew calendar 13-month distribution. This is done by adding 13 to the expression HY * 7 + Y.

Hence,


NUMMONTHS(HY) =  (HY * 235 / 19) - R(7 * HY + 13, 19) / 19
= [HY * 235 - R(7 * HY + 13, 19)] / 19
NUMMONTHS will always evaluate to an integer.

When HY = 13, R(7 * HY + 13, 19) = 18, which is the minimum possible value for that expression. In turn, this causes the smallest possible differences between the lunar and solar years in the mahzor qatan. As a result, 1 Tishrei appears at the earliest possible times in the solar calendars.

Last week's formula could also have been used to answer Weekly Question 219, because
at HY = 17, R(12 * HY + 5, 19) = 0.

Since the Hebrew year 5774H is the 17th year of the mahzor qatan, Rosh Hashannah 5774H will takes place relatively early in the solar year, beginning on Thursday 5 September 2013g. Due to The Rosh Hashannah Drift, this will be the last time, for all practical purposes, that the High Holidays will begin as early as September 5.

Correspondents Larry Padwa and Ari Brodsky all sent ingenious answers whose observational deductions appear to be correct.

Correspondent Larry Padwa observed:


Year 16 has the earliest starting dates (post Adar) in a solar 
calendar.

The reasoning behind this is similar (but of course in reverse) to the
reasoning behind last week's question regarding latest dates.

Year 16 is the only year which concludes a sequence of eight 
consecutive
years which consist of only two leap years. All sequences of eight
consecutive years not ending on year 16, contain three leap years. This
"lightness" in leap year density (for lack of a better term) leading up
to year 16, causes it to have the earliest solar dates.

Runners up to year 16 are years 5 and 13, but 16 is the winner.

Correspondent Ari Brodsky deduced that:-


In answer to Question 219, Tishrei of the 17th year of the cycle will 
have the earliest starting date, as it is the only one to follow a sequence 
of 8 years containing only 2 leap years. 

(Alternatively, it is the only one to begin a sequence of 11 years 
containing 5 leap years, which demonstrates more clearly the connection 
between this question and my answer to question 218.)

I usually prefer to think of the fact that Pesach (and all holidays 
from Adar onward) are earliest in the 16th year of the cycle, which is
essentially saying the same thing as Rosh HaShana being earliest in the
17th year.
Thank you very much correspondents Larry Padwa, and Ari Brodsky for sharing your very correct observations.


Correspondent Larry Padwa sent some excellent observations based on Weekly Questions 218 and 219. Of interest were the numbers of 12-month and 13-month years that could be found within a certain number of years beginning from any year HY.


Question 220

Beginning from any year HY, what is the number of 12-month years that can be found within a given number of years ?


Answer


Let H = the number of Hebrew years.
    M = the number of months in H years
    c = the number of 12-month years in H Hebrew years
    e = the number of 13-month years in H Hebrew years

Then 
  c +   e = H
12c + 13e = M

Hence, 

c = 13 * H - M
e = M - 12 * H

When H is not a multiple of 19 years, then H can have either M or M + 1 months,
where M = INT(235 * H / 19).

For example,

120 years can have either 1484, or 1485 months.

In that case, the number of 12-month years in the 120-year period 
can be either 76 or 75.
Thank you very much correspondent Larry Padwa for sending a number of intriguing observations related to the question.


The next question, related to Weekly Question 220, first appeared as Weekly Question 184.


Question 184

Which Hebrew year or years begin the shorter periods of H Hebrew years?


Answer

Properties of Hebrew Year Periods - Part 2 explains and demonstrates the fact that all periods of Hebrew years whose year lengths are not multiples of 19 exhibit 2 different lengths in months. These two monthly lengths are one month apart.

The reason for this particular Hebrew calendar phenomenon does not come from the fact that Hebrew leap years have an extra month, or that the leap years are distributed in a particular manner. Those reasons do not explain the one month difference in the periods of time which are not multiples of 19 Hebrew years.

It is possible to demonstrate the phenomenon using simple arithmetic techniques, as evident from a reading of
Properties of Hebrew Year Periods - Part 2.

The one month differences arise from a quirk in the remaindering arithmetic that is used to calculate the Hebrew calendar. The same quirk applies also to both the Julian and Gregorian calendars. However, in those calendars, the differences in the lengths of equal numbers of years is only one day.

Weekly Question 181 noted that in the Hebrew calendar the frequency of the shorter spans of H Hebrew years occurred 12 * H MOD 19 times in any cycle of 19 Hebrew years.

The same modulus arithmetic uncovers the amazing fact that when H and Hebrew year HY are such that

(12 * H MOD 19) + [(12 * HY + 5) MOD 19] > 18

then Hebrew year HY will start the shorter period of H Hebrew years.


Example

Let H = 120 Hebrew years

Then, 12 * H MOD 19 = 12 * 120 MOD 19 = 15.

Consequently, all of the shorter periods of time will occur whenever

[(12 * HY + 5) MOD 19] > 3

This will be true whenever HY is either year 1, 2, 4, 5, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, or 19 of the mahzor qatan GUChADZaT.

By implication, the longer periods of 120 will be started whenever

[(12 * HY + 5) MOD 19] < 4

This will be true whenever HY is either year 3, 6, 14, or 17 of the mahzor qatan GUChADZaT.

It is interesting to note that Rosh Hashanah 5676H (Thu 9 Sep 1915g) began the longest possible period of 120 Hebrew years.


The next question first appeared as Weekly Question 194.


Question 194

How often is parshah Ha'azinu read on Shabbat Shuvah?


Answer

Rabbi Erwin Schild, Rabbi Emeritus of Congregation Adath Israel in Downsview, Ontario, asked this week's question.

The following terms need explaining.

A parshah or sedrah is the portion of the Mosaic scriptures (Torah) selected for reading during a particular week.

Shabbat Shuvah is the Shabbat preceding the day of Yom Kippur.

Parshah or Sedrah Ha'azinu is found in D'vorim (Deuteronomy) 32:1-52.

One of the more prevalent practices, among the Jewish people, is that of reading the entire Mosaic text over the course of one Hebrew year. At Simchat Torah, the last few verses are read, and then the entire cycle is repeated once again from Bereshit (Genesis).

The scriptural readings are divided into contiguous weekly portions, which are read in their entirety each Shabbat morning. Each division is known as a Parshah or Sedrah. These portions are arranged so as to be completely read over the course of one Hebrew year.

Each portion is given a special name. The two portions that can each be read on Shabbat Shuvah are Vayelech and Ha'azinu.

Since there are 14 ways of laying out the Hebrew years (14 keviyyot), there exist only 14 ways of dividing the annual Torah reading cycle. As a result, the 14 different divisions can be easily tabulated in very compact form. One such tabulation may be found at the back of certain editions of the Humash (Pentateuch) as translated by Alexander Harkavy, and published by the Hebrew Publishing Co. in New York (1928).

That source shows that the portion Ha'azinu is read whenever Rosh Hashannah begins on either Thursday or Shabbat.

The Keviyyot shows the statistical distribution of all the years that begin on Thursday or Shabbat in the full Hebrew calendar cycle of 689,472 years.

Keviyyot - Under ALL Postponement Rules
YEAR LENGTH IN DAYS
DAY 353 354 355 383 384 385 TOTALS
Mon 39369 0 81335 40000 0 32576 193280
Tue 0 43081 0 0 36288 0 79369
Thu 0 124416 22839 26677 0 45899 219831
Sat 29853 0 94563 40000 0 32576 196992
TOTALS 69222 167497 198737 106677 36288 111051 689472

The statistics show that parshah Ha'azinu is read on about 61% of all of the Shabbat Shuvah's.

Correspondents Ari M. Brodsky and Larry Padwa both sent correct answers.

Ari M. Brodsky suggested that:


In years when Rosh HaShana begins on either Thursday or Shabbat, there 
is no Shabbat between Yom Kippur and Sukkot.  This occurs in 8 out of the 14
possible Jewish calendar year-type. The years 5764 and 5765 are both included
in one of these types.

and Larry Padwa, doing the arithmetic, noted that:


Ha'azinu is always read on the last regular (ie non-holiday) Shabbat
before Sukkot. This will be Shabbat Shuvah if there is no Shabbat
between Yom Kippur and Sukkot. This occurs if YK falls on Monday
(196,992 times in the full cycle or 28.6% of the time) or if YK falls 
on Shabbat (219,831 times or 31.9% of the time).

Thus Ha'azinu is read on Shabbat Shuvah slightly more than 60% of the
time, or more precisely 416,823 times in the full cycle of 689,472
years.


For other Additional Notes click here.
For other Archived Weekly Questions click here.
Hebrew Calendar Science and Myths

I'd love to hear from you. Please send your thoughts to:

Remy Landau

 First  Begun 21 Jun 1998 
First  Paged  2 Feb 2005
Next Revised  2 Feb 2005