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

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 eliminate356 day yearsis applied whenM = 354d, [f+m] = 1d, p" = 2d and p' = 0dthenL = 354d + 1d + 2d - 0d = 357d .Thedoes not occur in the fixed Hebrew calendar. The357-day yearcan be explained as follows. To eliminate theparadox356-dayyears, a postponement toThursday (5d)is made whenever the fractional part of themolad of Tishreion aTuesday (3d)is greater thand'-m. The fractional part of themoladis given byf + 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) - fThe inequality shows that the minimum value formwill be whenfis at its maximum value which isd'(by definition). Hence, thevalue ofminimummrequired to develop a357 dayyear is given bym > d + (d'- m) - d' m > d - m m > d / 2In the fixed Hebrew calendar, the actual value ofmfor the12 monthyear is8h 876p, which is less thand/2. That is why the357-dayyear cannot be generated in the fixed Hebrew calendar.

The ** second** reason for which the

Which rule in the Babylonian

Talmudeliminates the possibility of a 357-day year?

The following ** Mishnah** appears on

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

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

The following ** Mishnah** appears on

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

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

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

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.

Assuming

Tishrei1 postponements of not more than 6 days, what is the length of thesmallestHebrew year that could then be generated?

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

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.

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

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

Assuming the postponements of 1

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

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

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

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 |

Would any combination of at least 4 week days forbidden to 1

Tishreilead to 350-day years?

** NO!** The

**L** of any period of Hebrew years is given by

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

Were the year length **350 days**, then it would appear that the minimum ** Tishrei 1** postponement needed would be when

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

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 |

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 |

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 |

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

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

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

The Hebrew-Gregorian Calendar Correspondence Cycle indicates that

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

The Gregorian calendar fully repeats itself in a cycle of400 Gregorian years.Due to the fact that both the

Hebrew yearsand theGregorian yearsare of different lengths, the correspondence between the Hebrew and Gregorian calendars repeats itself in a cycle of

which is also14,389,970,112 Hebrew years!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**.

Why is

Pesachso late in April 2005?

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

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,

In the fixed Hebrew calendar, ** Pesach** always precedes

As an example, ** Pesach 5727H** began on

Since the Hebrew year **5766H** is the **9th year** of the ** mahzor qatan**,

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:

It may be noted that whenFirst, 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.

Correspondent **Ari Brodsky** deduced that:-

Thank you very much correspondentsThe 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).

Which year, or years, of the

mahzor qatan(19 year cycle) would have the earliest starting dates in a solar calendar?

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

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,

Last week's formula could also have been used to answer ** Weekly Question 219**, because

at

Since the Hebrew year **5774H** is the **17th year** of the ** mahzor qatan**,

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:-

Thank you very much correspondentsIn 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.

Correspondent **Larry Padwa** sent some excellent observations based on ** Weekly Questions 218 and 219**. Of interest were the numbers of

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

Thank you very much correspondentLet 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.

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

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

**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

** Weekly Question 181** noted that in the Hebrew calendar the frequency of the shorter spans of

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

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

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

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

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

The next question first appeared as ** Weekly Question 194**.

How often is?parshah Ha'azinuread onShabbat Shuvah

**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 (

** Shabbat Shuvah** is the

** Parshah or Sedrah Ha'azinu** is found in

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

The scriptural readings are divided into contiguous weekly portions, which
are read in their entirety each ** Shabbat** morning. Each division is known as a

Each portion is given a special name. The two portions that can each
be read on ** Shabbat Shuvah** are

Since there are **14 ways** of laying out the Hebrew years
(**14 keviyyot**), there
exist only

That source shows that the portion ** Ha'azinu** is read whenever

The Keviyyot shows the statistical distribution
of all the **years** that begin on ** Thursday** or

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

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.

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