Do the longest periods of 10 Hebrew years always begin on Monday?

*YES!*

The reason why the **longest** periods of

Do theperiods of 10 Hebrew years alwaysshortestendonMonday?

*YES!*

The smallest number of ** Hebrew** years that can have the

The **shortest** period of **10** Hebrew years is **3,630 days**
and occurs **3,237 times** in the full Hebrew calendar cycle of
**689,472 years**.

The **longest** period of **10** Hebrew years is **3,664 days**
and occurs **9,473 times** in the full Hebrew calendar cycle of
**689,472 years**.

For some given period ofHHebrew years Let its lunar length be A + a where A is the integral number of days in that length and a is the outstanding fraction of a day in that length Then Ls, theshortestpossible INTEGRAL length of that period, is A - 2 days. If theHyears are not a multiple of 19 years, then the alternate length includes an additional month and the longer lunar length is A + a + 29d 12h 793p. Then Lm, themaximumpossible INTEGRAL length of that longer period, is A + 29 + 3 days = A + 32 days. The difference between the maximum length and the shortest possible length is Lm - Ls = A + 32 - (A - 2) = 34 days.The smallest number of

years that can have theHebrewday difference between itsmaximum 34durations isshortest and longest.10 yearsTherefore, for any period of

Hebrew years, the maximum possible difference of days between its shortest and longest durations isH.34 daysThe

shortestperiod of10Hebrew years is3,630 daysand occurs3,237 timesin the full Hebrew calendar cycle of689,472 years.The

longestperiod of10Hebrew years is3,664 daysand occurs9,473 timesin the full Hebrew calendar cycle of689,472 years.Let D~ = the start day of the shortest period Let D' = the start day of that period 10 years after

Then (D' - D~) Mod 7d = (A - 2) MOD 7d Since the shortest period takes place when D~ is after a 2 day postponement D~ MOD 7d = 5d. Hence, A MOD 7d = (D' - 5d + 2d) A MOD 7d = ({0d , 2d, 3d, 5d} - 3d) MOD 7d = {4d, 6d, 0d, 2d}

Since the shortest period takes place when D~ is after a 2 day postponement D~ MOD 7d = 5d. Hence, A MOD 7d = (D' - 5d + 2d) MOD 7d A MOD 7d = ({0d , 2d, 3d, 5d} - 3d) MOD 7d = {4d, 6d, 0d, 2d}

Now, Let D~ = the start day of the longest period Let D' = the start day of that period 10 years after

Then (D' - D~) Mod 7d = (A + 34d) MOD 7d Since the longest period takes place when D' is after a 2 day postponement D' MOD 7d = 5d. A MOD 7d = (5d' - D~ - 34d) MOD 7d A MOD 7d = (1d - {0d , 2d, 3d, 5d} ) MOD 7d = {1d, 6d, 5d, 3d}

For A to satisfy both the minimum and the maximum cases it is therefore necessary that A MOD 7d = 6d, since that is the only value common to both cases.

For the shortest period of 10 years (D' - D~) Mod 7d = (A - 2d) MOD 7d D' Mod 7d = (A - 2d + D~) MOD 7d = (6d - 2d + 5d) MOD 7d = 2d Similarly for the longest period of 10 years (D' - D~) Mod 7d = (A + 34d) MOD 7d D~ Mod 7d = (D' - A + 6d ) MOD 7d = (5d - 2d + 6d) MOD 7d = 2d

Therefore, the shortest periods of 10 years always end on Monday while, the longest periods of 10 years always begin on Monday.

__
__

The above conclusion holds for any period of Hebrew years that varies 34 days between its shortest and longest periods!

TheWeekly Questionwill resume after Pesach.

*YES!*

The smallest number of ** Hebrew** years that can have the

The **shortest** period of **10** Hebrew years is **3,630 days**
and occurs **3,237 times** in the full Hebrew calendar cycle of
**689,472 years**.

The **longest** period of **10** Hebrew years is **3,664 days**
and occurs **9,473 times** in the full Hebrew calendar cycle of
**689,472 years**.

For some given period ofHHebrew years Let its lunar length be A + a where A is the integral number of days in that length and a is the outstanding fraction of a day in that length Then Ls, theshortestpossible INTEGRAL length of that period, is A - 2 days. If theHyears are not a multiple of 19 years, then the alternate length includes an additional month and the longer lunar length is A + a + 29d 12h 793p. Then Lm, themaximumpossible INTEGRAL length of that longer period, is A + 29 + 3 days = A + 32 days. The difference between the maximum length and the shortest possible length is Lm - Ls = A + 32 - (A - 2) = 34 days.years that can have theHebrewday difference between itsmaximum 34durations isshortest and longest.10 yearsTherefore, for any period of

Hebrew years, the maximum possible difference of days between its shortest and longest durations isH.34 daysshortestperiod of10Hebrew years is3,630 daysand occurs3,237 timesin the full Hebrew calendar cycle of689,472 years.longestperiod of10Hebrew years is3,664 daysand occurs9,473 timesin the full Hebrew calendar cycle of689,472 years.Let D~ = the start day of the shortest period Let D' = the start day of that period 10 years after

Then (D' - D~) Mod 7d = (A - 2) MOD 7d Since the shortest period takes place when D~ is after a 2 day postponement D~ MOD 7d = 5d. Hence, A MOD 7d = (D' - 5d + 2d) A MOD 7d = ({0d , 2d, 3d, 5d} - 3d) MOD 7d = {4d, 6d, 0d, 2d}

Since the shortest period takes place when D~ is after a 2 day postponement D~ MOD 7d = 5d. Hence, A MOD 7d = (D' - 5d + 2d) MOD 7d A MOD 7d = ({0d , 2d, 3d, 5d} - 3d) MOD 7d = {4d, 6d, 0d, 2d}

Now, Let D~ = the start day of the longest period Let D' = the start day of that period 10 years after

Then (D' - D~) Mod 7d = (A + 34d) MOD 7d Since the longest period takes place when D' is after a 2 day postponement D' MOD 7d = 5d. A MOD 7d = (5d' - D~ - 34d) MOD 7d A MOD 7d = (1d - {0d , 2d, 3d, 5d} ) MOD 7d = {1d, 6d, 5d, 3d}

For A to satisfy both the minimum and the maximum cases it is therefore necessary that A MOD 7d = 6d, since that is the only value common to both cases.

For the shortest period of 10 years (D' - D~) Mod 7d = (A - 2d) MOD 7d D' Mod 7d = (A - 2d + D~) MOD 7d = (6d - 2d + 5d) MOD 7d = 2d Similarly for the longest period of 10 years (D' - D~) Mod 7d = (A + 34d) MOD 7d D~ Mod 7d = (D' - A + 6d ) MOD 7d = (5d - 2d + 6d) MOD 7d = 2d

Therefore, the shortest periods of 10 years always end on Monday while, the longest periods of 10 years always begin on Monday.

__
__

The above conclusion holds for any period of Hebrew years that varies 34 days between its shortest and longest periods!

The following first appeared as

In the foreseeable future, will there ever be a Gregorian year without a Rosh Hashannah?

** YES!** Although that future will be quite distant!

Assuming that no changes take place to either the Gregorian or the Hebrew
calendars, then *Rosh Hashannah* will **NOT** be observed during
the year ** 22,335g**.

This will be the first time that a Gregorian year will not host such an important holiday.

*Rosh Hashannah* __ 26,095H__ will fall on

The topic on **The Calendar Drift** found in the ** Additional Notes**
fully explains why the 1st day of

Correspondent **Larry Padwa** sent the following response:

I don't have calendars that go that far into the future, but I can say that the situation you describe will definitely happen, and approximately when, as follows:Currently, the latest date for Rosh Hashannah is approximately 5-Oct.

The drift is roughly four days per thousand years, so it should take about 17,000 years for Rosh Hashannah to begin on or about 12-Dec.

If the Rosh Hashannah of a leap year begins between 12-Dec and 31-Dec, then the following Gregorian year will not have a Rosh Hashannah.

Thus, in about 17,000 years there should be a Gregorian year without a Rosh Hashannah.

Without calendars, (or running of a calendar program) I cannot be more precise than that, but I think that my guess is a fair approximation.

Thank you **Larry Padwa** for sharing an excellent insight into the
dynamics which lead to this particular phenomenon of the Hebrew calendar.

In eliminating the 356 day year, what unforeseen problem would have arisen if the period of the molad had been 29d 13h 1p?

**IF** the molad period had been **29d 13h 1p** then the
**12 month** year would have been **354d 12h 12p**.

The **356 day year** could then have been eliminated by postponing
*Rosh Hashannah* to **Thursday (5d)** whenever the
*molad of Tishrei* of a **12 month year**
fell on **Tuesday (3d)** past **11h 1067p**.

This postponement would then cause the year to be **354d**.

However, suppose that in this scenario, the *molad of Tishrei* of a
**12 month year** occurred on **Thursday (5d)** at **23h 1056p**.

Then **5d 23h 1056p + 354d 12h 12p = 360d 11h 1068p**

Now **(360d 11h 1068p) MOD 7d = 3d 11h 1068p**

Since, this time of the molad is **Tuesday (3d)** past **11h 1067p**
it is necessary to postpone *Rosh Hashannah* to **Thursday (5d)**
thereby adding **2 days** to the year.

Consequently, the final sum of days is **362d**.

Since the year began at **5d**, the length of the year is
**362d - 5d = 357 days**.

Therefore, if the period of the molad had been **29d 13h 1p**
instead of **29d 12h 793p**, the unexpected problem arising
from the elimination of the **356 day** year would have been
the creation of a **357 day** year.

It appears that all of the literature that deals with the fixed Hebrew
calendar's method for eliminating the **356 day** year gives the
impression that no other Hebrew year length is created in that process.

And it appears that noone has actually demonstrated, other than by computer simulation, that this is the case.

Why is the 357 day Hebrew year impossible in the fixed Hebrew calendar?

In the fixed Hebrew calendar,
** absent Dehiyyah Molad Zakein**,
the

This postponement then causes the year to be **354d**.

However, suppose that in this scenario, the *molad of Tishrei* of a
**12 month year** occurred on **Thursday (5d)** at **23h 1079p**.

Then **5d 23h 1079p + 354d 8h 876p = 360d 8h 875p**

Now **(360d 8h 875p) MOD 7d = 3d 8h 875p**.

Since this is the ** maximum** possible time for a

Therefore, it is impossible to create a **357 day year** in the
fixed Hebrew calendar.

Correspondent **Ram Sinclair** asked this week's intriguing question
which concerns the disappearing days ** Heshvan 30 and February 29**.

CanHeshvan30 ever coincide with February 29?

*YES!*

In the fixed Hebrew calendar, ** Heshvan 30** occurs only when
the year is

In the Gregorian calendar, **February 29** occurs only when the
Gregorian year is a ** leap** year.

That simple observation made correspondent **Ram Sinclair** wonder
whether or not *Heshvan* 30 could ever coincide with February 29.

**Ram Sinclair** then researched the matter.

He correctly deduced that *Heshvan* 30 could coincide February 29
provided that *Rosh Hashannah* mapped onto January 1 of a Grgorian
leap year.

With that in mind, he noted from the topic
The Rosh Hashannah Drift

that his search
would have to start after ** Rosh Hashannah 25963H**
(

Assuming that the Hebrew and Gregorian calendar rules remain unchanged,
**Ram Sinclair** discovered that **30 Heshvan 28184H**
would coincide with

Thank you **Ram Sinclair**, not only for asking this very interesting
question, but also for sharing with us the results of your
** excellent** research.

Correspondent **Ram Sinclair**'s second discovery was an earlier
coincidence of ** Heshvan 30 and February 29**. But was this
the earliest possible such coincidence?

When will be the earliest coincidence ofHeshvan30 and February 29?

Once again, correspondent **Ram Sinclair** provided the correct answer.

I am almost sure thatFeb 29, 23680gis the first30.HeshvanI realized that in such a year,

Rosh Hashannahis onJan 1.So I looked in the table in your site and found that

22203gis the first occurrence ofJan 1 on.Rosh HashannahSo I tried every

4th yearstarting with22204g(skipping multiples of100which are not multiples of400).It was interesting because there were many close calls on the way, like

Kislev1 when there were only 29 days inHeshvan.

Thank you **Ram Sinclair**, not only for asking this very interesting
question, but also for sharing with us the results of your
** excellent** research.

Still on the idea of disappearing dates, correspondent **Dennis Kluk**
discovered that

**Thu 30 Adar 5585H** coincided with

Correspondent **Dennis Kluk** did not know if there were any other
coincidences of these dates.

So ** Weekly Question 127** will deal with this issue.

When was the most recent coincidence ofAdar30 and February 29?

Once again, correspondent **Ram Sinclair** provided the correct answer.

According to my research1824gis the most recent occurrence of29 Feb and 30.AdarSend my appreciation to

Dennis Kluk. To my understanding30is less frequent thanAdar30which makes his challenge tougher to meet.Heshvan

Thank you **Ram Sinclair**, for once again sharing with us the results of
your ** excellent** research.

And also, thank you correspondent **Dennis Kluk** for passing on your
thought provoking question.

Of course, if the most recent coincidence of
**Thu 30 Adar 5585H** was

So ** Weekly Question 128** will deal with this issue.

When will be the next coincidence ofAdar30 and the Gregorian date Feb 29?

Assuming that both the Hebrew and the Gregorian calendars rules
do not change, once again, correspondent **Ram Sinclair** provided
the correct answer.

The next occurrence will be about 80,000 years from now.I found a Feb 29 on Adar 30 in 83820g = 87579H.

I am not sure that it is the first.

Correspondent **Ram Sinclair** did find the next date for this
particular coincidence.

In order that ** Adar 30** coincide with

All of these conditions are established next in the year
**87579H (83819g/83820g)**.

Thank you **Ram Sinclair**, for once again sharing with us the results of
your ** excellent** research.

Can theDehiyyah Molad Zakeintest be bypassed when calculating forTishrei 1?

*YES!*

** Dehiyyah Molad Zakein** limits the maximum possible calculated
time of the

The * Molad Zakein* rule requires that

An unsuccesful attempt to increase that limit to **18h 642p** was made by
*Aaron Ben Meir* in 920g (4680H). For more information on this attempt,
please refer to The Ben Meir Years

When the Hebrew calendar is considered without
** Dehiyyah Molad Zakein**, and it is desired to maintain the

Those two limits then become **15h 204p and 21h 589p**.

Then, by adding **6 hours** to the *molad of Tishrei*, and also
using the limits as modified, there is no further need to include a test
for ** Dehiyyah Molad Zakein**.

In this way, the ** Dehiyyah Molad Zakein** test is entirely
bypassed.

This technique is interesting to anyone involved in the calculation of the
Hebrew calendar since it automatically applies
** Dehiyyah Molad Zakein** without having to test for it.

The technique actually was used by the 18th century
mathematical genius **Karl Friedrich Gauss** in his well known
*Pesach* formula which automatically calculates the Julian date
for *Pesach* given any Hebrew year.The formula is shown in
The Gauss Pesach Formula

The next question will deal with the effect the **6 hour** addition to
the *molad of Tishrei*.

Why does adding 6 hours to themolad of Tishreiautomatically bypass theDehiyyah Molad Zakeintest when calculating forTishrei 1?

** Dehiyyah Molad Zakein** limits the maximum possible calculated
time of the

The * Molad Zakein* rule requires that

the

By adding **6 hours** to the time of the *molad of Tishrei*, and also
to the threshold limits of *Dehiyyot GaTaRaD and BeTU'TeKaPoT*,
there is no further need to include a test
for ** Dehiyyah Molad Zakein**.

In this way, the ** Dehiyyah Molad Zakein** test is entirely
bypassed.

When the time of the *molad of Tishrei* => 18h

on either **Shabbat (0d), Monday (2d), Tuesday (3d),
or Thursday (5d)**,

*Dehiyyah Molad Zakein* requires that the first day of *Tishrei*
be postponed to

either ** 2d, 3d, 5d, or 0d**.

Adding **6 hours** to that time cause the resulting day of the week to
be either **1d, 3d, 4d, or 6d**.

*Dehiyyah LO ADU Rosh* will then force
the first day of *Tishrei* to be either ** 2d, 3d, 5d, or 0d**,
giving the same result as if the

In order to be complete, the demonstration also requires consideration
of the **6 hour** addition in the 3 cases in which the
*Molad of Tishrei* is

**1.** either on **1d, 4d, or 6d**,

**2.** or past **9h 203p on Tuesday**,

**3.** or past **15h 588p on a post-leap year Monday**.

However, the ** Weekly Question** will leave that demonstration
to its readers.

The **6 hour** addition technique is interesting to anyone involved in
the calculation of the Hebrew calendar since it automatically applies
** Dehiyyah Molad Zakein** without having to test for it.

The technique actually was used by the 18th century
mathematical genius **Karl Friedrich Gauss** in his well known
*Pesach* formula which automatically calculates the Julian date
for *Pesach* given any Hebrew year.The formula is shown in
The Gauss Pesach Formula.

The next question first appeared as ** Weekly Question 19**.

In the foreseable future can 2 consecutiveRosh Hashannah's ever begin in the same Gregorian year?

Correspondent **Larry Padwa** sent in the correct answer and said as
follows:

The answer is yes, and it follows immediately from your reply to the previous question.RH 26,096H falls on 2-Jan 22336g. Since 26096H (mod 19) = 9, 26096H is not a leap year and hence has at most 355 days. 355 days from 2-Jan falls in December of the same year, so RH 26096H and 26097H fall in the same Gregorian year.

Correspondent **Ram Sinclair** also gave the correct answer without
specific dates.

Thank you correspondents **Larry Padwa** and **Ram Sinclair** for
sharing with us the results of your excellent research.

Holding constant both the Hebrew and Gregorian calendars, it is possible to discover the occurrence of Gregorian years in which no Rosh Hashannah will occur. It is also possible to discover Gregorian years in which 2 Rosh Hashannot will be observed.

The next question first appeared as ** Weekly Question 20**.

In the foreseeable future, can 2 Gregorian years ever see the same Rosh Hashannah?

Correspondent **Larry Padwa** once again sent in a correct answer.
Thank you Larry.

The Rosh Hashannah Drift shows that the Hebrew year moves into the Gregorian calendar at the rate of about one day in every 216 years.

Left unchecked by any changes to either calendar, it is then possible to
predict the event of the first day of *Rosh Hashannah* on
**December 31**. In that instance, the second day of the festivity
will occur in the next Gregorian year.

The event of one **Rosh Hashannah** spanning 2 Gregorian years will then
first be seen to occur for **Saturday, 1 Tishrei 25716H (31 Dec 21955g)**.

Correspondents **Ram Sinclair** and **Winfried Gerum** both sent in the
dates on which the first day of *Rosh Hashannah* and the
Gregorian New Year would theoretically coincide. That is also a
correct answer because the start of *Rosh Hashanah* would be on the
eve of **December 31**, which of course, is **New Year's Eve**.

Thank you **Ram Sinclair and Winfried Gerum** for noticing and sharing
with us this very festive coincidence.

The next question first appeared as ** Weekly Question 21**.

Some traditions suggest that the ** Exodus** took place on a

either in the year

Assuming the fixed calendar rules, could theExodushave taken place on a Thursday, in either the years 2448H or 2449H?

**NO!**

According to the fixed calendar, the first day of *Pesach* would have
occurred on

**Tuesday 13 Mar -1312g** in the year **2448H**.

Similarly, the first
day of *Pesach* would have taken place on

**Sunday 31 Mar -1311g** in the year **2449H**.

By one of these very strange coincidences, the first day of *Pesach*
would have taken place on **THURSDAY 24 Mar -1313g**
in the year **2447H**.

A very quick way to derive the day of the week for the first day of
*Pesach* is to subtract **two days** from the day of the week of
the subsequent *Rosh Hashannah*.

Consequently, it may be entirely possible that the advocates of a
*Pesach* originating on a **Thursday** in the year **2448H**
may have overlooked or forgotten to realize that **Thursday** was
the first day of *Pesach* for the previous year, namely, **2447H**.

Since **17 Tammuz 5761H** occurred

First Begun 21 Jun 1998 First Paged 5 Nov 2004 Next Revised 12 Nov 2004