WQ Archive 221 - 230

Question 221

In exact fraction of year, how much longer is the Hebrew year over the Gregorian year?

In days, the difference of the average Hebrew year over the average Gregorian year is easily found to be the exact fraction (6939d + 16h + 595p) / 19 - 146097d / 400 = 10,643d / 2,462,400.

In years, the exact fraction is a bit more complicated, since it can only be given either as fraction of Hebrew year or as fraction of Gregorian year.

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!

Let H the number of Hebrew years (hy) which exactly equal G gregorian years (gy).

```
Then

H * hy = G * gy

Hence,

hy - gy
= hy - H * hy / G
= hy * (G - H) / G
= hy * 170,288 / 14,390,140,400
=       10,643 /    899,383,775 Hebrew years

= gy * G / H - gy
= gy * (G - H) / H
= gy * 170,288 / 14,389,970,112
=       10,643 /    899,373,132 Gregorian years
```
It is interesting to note that when reduced, all of the above fractions have the numerator 10,643.

Question 222

Exactly how many years are required for a one year difference between the number of Hebrew and Julian years?

In exact fraction of year, how much longer is the Hebrew year over the Gregorian year?

In days, the difference of the average Hebrew year over the average Julian year is easily found to be the exact fraction (6939d + 16h + 595p) / 19 - 1461d / 4 = -313d / 98,496.

In years, the exact fraction is a bit more complicated, since it can only be given either as fraction of Hebrew year or as fraction of Gregorian year.

Due to the fact that both the Hebrew years and the Julian years are of different lengths, the correspondence between the Hebrew and Julian calendars repeats itself in a cycle of
1,007,318,592 Hebrew years which is also 1,007,309,828 Julian years!

Let H the number of Hebrew years (hy) which exactly equal J Julian years (jy).

```
Then

H * hy = J * jy

Hence,

hy - jy
= hy - H * hy / J
= hy * (J - H) / J
= hy * -8,764 / 1,007,309,828
=        -313 /    35,975,351 Hebrew years

= jy * J / H - jy
= jy * (J - H) / H
= jy * -8,764 / 1,007,318,592
=        -313 /    35,975,664 Julian years
```
It is interesting to note that when reduced, all of the above fractions have the numerator -313.

In order that the difference between the Hebrew year count and the Julian year count dHY becomes 1 Hebrew year, it is necessary that

```
dHY =   35,975,351 / 313
= 114,937 + 70 / 313 Hebrew years have elapsed

or  the count in Julisan years dJY

dJY =   35,975,664 / 313
= 114,938 + 70 / 313 Julian years have elapsed

```
This result is consistent with the fact that J / (J - H) - H / (J - H) = 1.

Correspondents Larry Padwa and Ari Meir Brodsky sent a number of delightful observations related to Hebrew Year 5765H.

Question 223

```Hebrew Year 5765H

- began on Thursday 16 September 2004g.
- is 383 days long.
- is the 8th year of the 304th mahzor qatan.
- has molad of Tishrei of 3d 19h 287p.
- was postponed to Thursday due to Dehiyyah Molad Zaqen.
- is not affected by Dehiyyah GaTaRaD because it is a leap year.
```
Can you mention at least 3 other calendar related features of Hebrew year 5765H?

Correspondents Larry Padwa and Ari Meir Brodsky sent a number of other delightful observations related to Hebrew Year 5765H.

Here are some:-

Correspondent Larry Padwa noted that Year 5765H has the second least frequently occurring qeviyyah. Tables in The Qeviyyot indicate that the 383-day year starts on Thursday only 26,677 times in the full Hebrew calendar cycle of 689,472 years. This is about 3.87% of all of the years in the cycle. Or, as correspondent Ari Meir Brodsky noted, about 4 times in every hundred years. This qeviyyah last occurred for the year 5741H (beginning in 1980g).

The molad of Heshvan 5765H will occur on Thursday 14 October 2004g at 8h 0m 0hl. Please note that the time of this molad is at an exact number of hours.

The molad of Kislev 5765H will occur on Friday 12 November 2004g at 20h 44m 1hl. Rosh Hodesh Kislev 5765H will be observed on Sunday the 14th day of November 2004g. Therefore, Shabbat M'vorchim, during which the time of the molad of Kislev 5765H will be announced in many synagogues, will occur after the molad of Kislev 5765H. As a result, the announcement will have to note that the molad of Kislev 5765H has already occurred. This irregularity occurs approximately once in every 2 or 3 years.

Correspondents Larry Padwa and Ari Meir Brodsky both noted that when a 383-day year starts on Thursday then there are no double portions in the regular weekly Torah readings. All of the other Hebrew year layouts force at least one double parshah in the weekly Torah readings.

Correspondent Ari Meir Brodsky also found out that the Hebrew and Gregorian calendars for year 5765H will map onto the exact same dates 152 years from now. He therefore urges that we keep this year's calendars because these will be reusable 152 years from now!

Correspondent Ari Meir Brodsky drew attention to his fascinating article on the Hebrew year 5765H How Is This Year Different From All Other Years?

Yasher Koach!! to correspondents Larry Padwa and Ari Meir Brodsky for identifying and sharing with us so many of the outstanding features of Hebrew year 5765H.

Question 224

[Relative to 5765H] When next will there be a molad with a whole number of hours?

As indicated in The Moladot,

It is interesting to note that from month to month the halaqim ascend by exactly one unit, going from 0 to 17 in perfect cyclical order.

29d 12h 793p equal 765,433 parts. That is the number of parts which are added to the time of the molad from month to month.

When 765,433 is divided by 18 there is a remainder of 1, thus resulting in the 1hl difference from month to month.

Consequently, it will take 1,080 Hebrew months for all of the excess halaqim to total exactly 1 whole hour. As may inferred either through ordinary arithmetic, or referencing the tables shown in 247 Hebrew Year Periods, 1080 Hebrew months corresponds to either 87 Hebrew years and 3 months, or 87 Hebrew years and 4 months.

87 years from now, the molad of Tishrei 5852H(2091g) will be 5d 17h 355p. Hence, to add up to a whole number of hours, the number of lunar months will have to be at least a multiple of 5. The first month that would allow for this then, would have to be Adar Rishon 5852H. Its molad will be [(5d 17h 355p) + 5 * (29d 12h 793p)] MOD 7d = 6d 9h 0p.

Correspondents Larry Padwa and Dwight Blevins all provided correct answers.

Correspondent Larry Padwa solved this problem as follows:

```Ignoring whole hours, the increment in the molad from one month to the
next is 793 chalakim. There are 1080 chalakim in an hour, and since 793
and 1080 have no common factor, it is necessary to move forward 1080
months to reach the same point in the cycle of chalakim as the starting
point.

Now to calculate 1080 months from Cheshvan 5765.
1080=4*235 + 140.  Thus 1080 months comprise 4 complete 19-year cycles,
plus a remainder of 140 months. The complete cycles give 76 years, and
the 140 months beginning in year 8 of the machzor katan yield (by brute
force calculation) 11 years and 4 months, allowing for leap years at
correct intervals. This takes us 76 years + 11 years + 4 months from
Cheshvan 5765, which is Adar I 5852.  (Whew).
```

Correspondent Dwight Blevins chose this route:

```It seems that a forward movement of 1080 months would be required
(advancing 1 part per month) for the next molad to fall precisely on
the whole hour.  Based on that assumption, 1080 months advanced from
October 14, 2004, would be 1 Adar 1, February 9, 2092 (5852H).

Stated another way, the fall of 2091g, Tishrei 1 falls at 17h 355p.
Therefore, 5 more months beyond that date are required to achieve 1080
months from Heshvan, 2004g.  Thus, 793 x 5 = 3965p + 355p = 4320p/1080
= precisely 4 hours of parts, past the reference point of Heshvan, Oct.
14, 2004, thus achieving the next whole hour molad, with reference to
Heshvan, 5765H.  That whole hour molad, again, is 1 Adar 1, 5852H (Feb.
9, 2092).
```

Thank you Correspondent Larry Padwa and Dwight Blevins for providing these very good and very correct answers!

Question 225

[Relative to 5765H] When next will the molad of Tishrei be a whole number of hours?

Although, from month to month, the halaqim ascend by exactly one unit, going from 0 to 17 in perfect cyclical order, the moladot of Tishrei do not exhibit such easy relationship.

As explained in Cycles and Moladot, all periods of Hebrew years that are exact multiples of 19 years have exactly one number of months, namely, the number of 19 year periods multiplied by 235.

All other periods of Hebrew years have two numbers of months differing from each other by exactly one month.

The lower number of months, for some period of HY Hebrew years is given by INT(235 * HY / 19). The larger number of months is just one more month than that.

Consequently, the determination of some formula which will help establish the Tishrei moladot that have whole numbers of hours will be left to the mathematically skilled who just might enjoy that challenge.

Using empirical computer methods, it is easy to determine that the next molad of Tishrei to have a whole number of hours will be the molad of Tishrei 6027H (Mon 1 Oct 2266g). That molad will occur at exactly 1d 11h.

What is particularly enticing in this phenomenon is that all of the whole hour Tishrei moladot occur in an exact cyclical pattern.

Question 226

The whole hour Tishrei moladot occur in what cyclical pattern?

Although, from month to month, the halaqim ascend by exactly one unit, going from 0 to 17 in perfect cyclical order, the moladot of Tishrei do not exhibit such easy relationship.

As explained in Cycles and Moladot, all periods of Hebrew years that are exact multiples of 19 years have exactly one number of months, namely, the number of 19 year periods multiplied by 235.

All other periods of Hebrew years have two numbers of months differing from each other by exactly one month.

The lower number of months, for some period of HY Hebrew years is given by INT(235 * HY / 19). The larger number of months is just one more month than that.

Consequently, the determination of some formula which will help establish the Tishrei moladot that have whole numbers of hours will be left to the mathematically skilled who just might enjoy that challenge.

Using empirical computer methods, it is easy to determine that the next molad of Tishrei to have a whole number of hours will be the molad of Tishrei 6027H (Mon 1 Oct 2266g). That molad will occur at exactly 1d 11h.

What is particularly enticing in this phenomenon is that all of the whole hour Tishrei moladot occur in an exact cyclical pattern.

The pattern can easily be determined by examining a small number of consecutive Hebrew years for Tishrei moladot that contain only whole hours.

These Hebrew years begin with 2H, 264H, 1923H, 2185H, 4106H, 4368H, 6027H, 6289H, 8210H, ...

From this rather small sample it appears that the indicated cycle of whole hour Tishrei moladot is {... 262, 1659, 262, 1921,...} years apart.

A bit more research indicates this to be a valid pattern for the entire Hebrew calendar cycle of 689,472 years.

Question 227

The whole day Tishrei moladot occur in what cyclical pattern?

Although, from month to month, the halaqim ascend by exactly one unit, going from 0 to 17 in perfect cyclical order, the moladot of Tishrei do not exhibit such easy relationship.

As explained in Cycles and Moladot, all periods of Hebrew years that are exact multiples of 19 years have exactly one number of months, namely, the number of 19 year periods multiplied by 235.

All other periods of Hebrew years have two numbers of months differing from each other by exactly one month.

The lower number of months, for some period of HY Hebrew years is given by INT(235 * HY / 19). The larger number of months is just one more month than that.

Consequently, the determination of some formula which will help establish the Tishrei moladot that have whole numbers of days will be left to the mathematically skilled who just might enjoy that challenge.

Using empirical computer methods, it is easy to determine that the first molad of Tishrei to have a whole number of hours will be the molad of Tishrei 51,171H (Mon 15 Apr 47,411g). That molad will occur at exactly 1d 0h 0p.

What is particularly enticing in this phenomenon is that all of the whole day Tishrei moladot occur in an exact cyclical pattern.

The pattern can easily be determined by examining a small number of consecutive Hebrew years for Tishrei moladot that contain only whole days.

These Hebrew years begin with 51171H, 57458H, 63745H, 70032H, 149667H, 155954H, 162241H, 168528H, 248163H, ..., 661008

From this rather small sample it appears that the indicated cycle of whole day Tishrei moladot is {... 6287, 6287, 6287, 79635,...} years apart.

A bit more research indicates this to be a valid pattern for the entire Hebrew calendar cycle of 689,472 years.

<--->

Question 228

How are the significant days of Hanukkah, that is, the first day of Hanukkah and Rosh Hodesh Tevet, related to Pesach?

Referring to both The First Day of The Month and The Roshei Hadashim, it is possble to note that according to the fixed Hebrew calendar method

```
the first day of Rosh Hodesh Tevet can never be on Sunday
the first day of Hanukkah          can never be on Tuesday
the first day of Tevet             can never be on Thursday
the first day of Tevet             can never be on Shabbat
```
However, Sunday, Tuesday, Thursday, and Shabbat are the only days of the week on which can occur the first day of Pesach.

It is this surprising set of facts derived from the modern fixed Hebrew calendar which give rise to this startling relationship between the significant days of Hanukkah and the festival of Pesach.

Correspondent Larry Padwa opened a topic of such interest that it will lead to the next few Weekly Questions.

```Hi Remy,

The archives of the Weekly Questions are extremely interesting. On
occasion, reviewing old questions can suggest related new ones, as in
the following instance.

In questions 143 and 145, you and your correspondents discuss the
questions of the smallest span of years containing all six possible
year lengths (6), and the largest span of years not containing all six
possible year lengths (43).

A pair of related questions is to identify the smallest span of years
containing all fourteen Keviot, and the largest span not containing all
fourteen Keviot.
```

Thank you Larry Padwa for this particularly interesting suggestion.

Referring to The Qeviyyot, a qeviyyah (pl. qeviyyot) is defined to be a pair of numbers indicating the length of the Hebrew year and the day of the week on which started the Hebrew year of that length.

Weekly Question 229

What is the smallest possible span of Hebrew years to contain all 14 of the qeviyyot?

Referring to The Qeviyyot, a qeviyyah (pl. qeviyyot) is defined to be a pair of numbers indicating the length of the Hebrew year and the day of the week on which started the Hebrew year of that length.

Correspondent Larry Padwa opened a topic of such interest that it will lead to the next few Weekly Questions.

Larry Padwa correctly assumed the following.

```Hi Remy,

It couldn't be less than seventeen, since:

a) Seven of the fourteen keviot are leap years.

b) it is impossible to squeeze seven leap years into a sequence smaller
than seventeen. To see this, consider a sequence beginning at (say)
year 3 of a Machzor Katan.

Then seven leap years take us to year 19, yielding a sequence of seventeen
years. We cannot do any better than beginning at year 3 (or any other leap
year that follows it preceeding leap by three years). If we begin at any
other leap year (say year 8 or year 19), then we will require a sequence
of 18 years to get seven leap years. And beginning at a common year will
also require at least 18.

So seventeen is the shortest possible. But that in itself does not
prove that a sequence of seventeen years with all fourteen keviot is in fact
attainable.

-Larry

```
As usual, correspondent Larry Padwa made the correct assumptions! The actual size of the required span is eighteen Hebrew years. The first such span begins with Shabbat 1 Tishrei 888H (20 Aug -2873g).

Correspondent Robert E. Heyman sent this very correct calculation result.

```According to my calculations, the smallest possible year span
is 18, and the first such span began in the year 888.

Robert
```
Thank you very much correspondents Larry Padwa and Robert E. Heyman for sharing these very interesting results.

Weekly Question 230

The shortest spans of Hebrew years containing all of the 14 qeviyyot begin in which year, or years, of the mahzor qatan (19-year Hebrew calendar cycle)?

Referring to The Qeviyyot, a qeviyyah (pl. qeviyyot) is defined to be a pair of numbers indicating the length of the Hebrew year and the day of the week on which started the Hebrew year of that length.

Correspondent Robert E. Heyman sent this very correct calculation result.

```According to my calculations, the smallest possible year span
is 18, and the first such span began in the year 888.

Robert
```
The year shown by correspondent Robert E. Heyman, namely 888H (Sat 20 Aug -2873g), is the 14th year of the mahzor qatan (19-year cycle).

Very intriguingly, and by one of these delightful Hebrew calendar coincidences, all of the shortest spans of Hebrew years containing all of the 14 qeviyyot begin uniquely in the 14th year of the mahzor qatan (19-year cycle).

Perhaps one day, a mathematically inclined Hebrew calendar expert will find the reason behind this mysterious phenomenon.

Weekly Question 231

How many different sequences of qeviyyot can be found for the shortest spans of Hebrew years containing all of the 14 qeviyyot?

``` First  Begun 21 Jun 1998