Booklets and Articles Index  
                         THE HEBREW CALENDAR:
                      A Mathematical Introduction

                      Prepared by: JOHN A. KOSSEY
                        Editor: HERMAN L. HOEH

                             FIRST EDITION
                       AMBASSADOR COLLEGE PRESS
                         Pasadena, California
                          1971, 1974 Edition

                        ----------###----------

                               PROGRAM I

              USING THE TIME UNITS OF THE HEBREW CALENDAR
 

INTRODUCTION

Why should YOU study the Hebrew calendar?

     One of the major identifying signs of the Church of God is the
observance of the Sacred Festivals. As you study the twenty-third
chapter of Leviticus, you will notice that God employed a calendar to
indicate when each holy day must be properly kept during the year. The
Jews were given the responsibility of preserving that calendar for the
rest of the world.

     Since it is the responsibility of the Church to announce the time
of each festival to the congregations, detailed understanding of the
Hebrew calendar is not even necessary for a lay church member. When a
holy day is to be kept is not for the individual Christian to decide.

     On the other hand, the education you are privileged to receive as
an Ambassador College student equips you with a special depth of
biblical understanding. Shallow or sketchy knowledge of basic
background areas would lessen one's effectiveness. But working out the
calendar principles yourself is going to widen your perspective.

     You already know that the holy days portray God's master plan of
salvation for mankind. Shouldn't you also have a working knowledge of
the very calendar which houses God's Sacred Festivals?

     This is why a study of the Hebrew calendar is included in
Theological Research I-II.
 

THE PURPOSE OF THESE LEARNING PROGRAMS

     Your study of the Hebrew calendar in this course has two major
facets. One is the historical development of the calendar. This is the
primary function of the class lectures. The other is for you to achieve
needed computational facility with the calendar itself.

     Fortunately, the Hebrew calendar requires surprisingly little
mathematical sophistication. A fifth grade background in arithmetic
will suffice! Nevertheless, a certain number of skills and concepts
must be learned for you to become adept at working with the Hebrew
calendar. These programs are designed to provide you with that
understanding and practice.

     Just what will you be able to accomplish when you complete this
series of learning programs?

     For any year, such as 4 BC, 31 AD, 1520 AD, and 1979 AD, you will
correctly determine the dates on a common Roman calendar of the holy
days listed in Leviticus 23.

     How long will this operation require? With nothing but a pencil
and a blank sheet of paper, you might need anywhere from thirty to
forty-five minutes. If you use a table of reduced numbers (which is
included in one of the programs), it might take you only ten to fifteen
minutes.

     That skill is the OVERALL GOAL of this series of programs. Another
less tangible aim is to give you the confidence that you can actually
SUCCEED in working a calendar problem!

     To that end, each learning program takes a necessary part of the
main goal, and gives you the practice needed to become adept at it.
Success will breed success as you progress!

     The first page of each learning program has a clear statement of
what you must be doing by the time you complete the program. You might
think of each program as a "checkpoint" on route to your destination.
Be sure you can accomplish each program goal before going on to the
next one.

     One word of caution. Have you ever learned mathematics simply by
glancing over the textbook or watching someone else work through a
problem? No, you can't! The exercises in each program are entirely for
your benefit. In most cases, these exercises will be worked out in
detail later in the program. This is for you to have a model with which
to compare your own procedures and to check your work immediately for
errors.

     But WORK you must! Proverbs 4:13 says to "take fast hold of
instruction; let her not go." Learning the mathematical operations of
the Hebrew calendar will take ACTIVE EFFORT. With this diligence, you
will achieve the exhilaration only success can bring.

                        ----------###----------

                               Contents
                                                        Program

Introduction                                                i
Using the time units of the Hebrew calendar                 1
Calculating the day of the week of the molad Tishri         2
Calculating the day of the month of the molad Tishri        3
Using tables to find the molad Tishri                       4
Making Roman leap year and Julian/Gregorian corrections     5
Applying the postponement rules to find Tishri one          6
Counting the days of the week and the days of the month     7
Determining the dates of the annual festivals               8

                        ----------###----------

PERFORMANCE GOAL 1A:

     Without hesitation or uncertainty you will write (or recite) from
memory in any order the following time relationships:

     1 part (chalyek)      =   76 moments (regaim)
     1 hour                = 1080 parts (chalakim)
     1 day                 =   24 hours
     1 week                =    7 days
     1 lunar month         =   29 days 12 hours 793 parts
     1 common year         =   12 lunar months
     1 intercalary year    =   13 lunar months
     1 nineteen year cycle =  235 lunar months, or 12 common
                                 years & 7 intercalary years
 

PERFORMANCE GOAL 1B:

     With only paper and pencil (or pen), you will accurately add,
subtract, multiply, and divide the time relationships listed in lA as
necessary to compute specified problems. Large numbers will be "reduced
to lowest terms" when requested. For example,

     28 hours reduces to 1 day, 4 hours
     31 hours 650 parts reduces to 1 day, 7 hours, 650 parts.
 

PERFORMANCE GOAL 1A

     In order for you to become adept with calendar calculations, you
must become very familiar with a minimal number of time relationships.
Some of these you already know; others require memorization. These
numbers will occur so frequently that you simply must know them.
Otherwise, the lessons will take much longer to understand and you will
feel frustrated in the process. Learn them now -- for your own benefit.

     For a detailed yet concise description of the time elements of the
Hebrew calendar, consult either of the following references:

          "The Jewish Encyclopedia", volume 3, "Calendar" (either new
or old edition).

          Burnaby, Sherrard Beaumont, "Elements of the Jewish and
Muhammadan Calendars". London: George Bell & Sons, 1901. 554 pages.
(See chapter II, page 21.)
 

     Here is a brief explanation of some of these time elements to
assist your memorization.

     DAY: Genesis 1:5 shows that the day begins in the evening. "And
the evening and the morning were the first day. " Although each day
begins at sunset, 6 PM is the arbitrary commencement of a new day for
CALENDAR CALCULATIONS.

     Christ put his divine approval upon dividing the day into
twenty-four hours. See John 11:9, which states, "Are there not twelve
hours in the day?" Context shows that this verse refers to the daylight
portion of a twenty-four hour period. For any given day the periods of
darkness and light are usually unequal. The total length of a day,
however, is always 24 hours--except for Divine intervention!

     HOUR: Instead of being divided into minutes and seconds, the hour
is divided into parts, or chalakim. One hour consists of 1080 parts, or
3600 seconds. Using parts instead of minutes and seconds has the
advantage of eliminating fractions. (The smaller unit, the moment or
rega, is seldom needed for fundamental calculations.) Both the hour and
the part are considered fixed units anywhere on earth, just as the
minute and the second are non-varying time elements.

     MONTH: A lunar month is the time needed for the moon to revolve
around the earth. Even though this period varies from month to month,
29 days 12 hours 793 parts is the traditional average used for
calculation. Actual calendars cannot be based upon 29 1/2 days, so the
Hebrew calendar incorporates months of 29 and 30 days.

     YEAR: The Hebrew calendar has two basic types of years, common and
intercalary. (The latter term is also called "embolismic.") An
intercalary Hebrew year will have 30 additional days, so it can also be
called a leap year. By contrast, recall that the Roman leap year has
366 days instead of 365. You should also remember that the following
terms are interchangeable for the Hebrew calendar:

     LEAP (year or month) = embolismic (year or month) =
INTERCALARY(year or month)

     Common years will have 353, 354, or 355 days. Leap years may have
383, 384, or 385 days.

     19 YEAR CYCLE: The Western world is accustomed to a solar year of
365 1/4 days, since the Roman calendar in common use is solar. This
means that a given month of the year will always occur during the same
season. January, for example, is invariably a winter month.

     On the other hand, the Hebrew year by itself does not closely
match the length of a solar year. Twelve months which are each
approximately 29 1/2 days results in a year which has only 354 days --
about eleven days less than a solar year of 365 1/4 days. A common
Hebrew year is thus SHORTER than a Roman year.

     What about a Hebrew leap year? Thirteen months of 29 1/2 days is
383 1/2 days, which is LONGER than a solar year.

     God has ordained that the holy days must be kept "in their
seasons" (Lev. 23:4). He also appointed BOTH the sun and the moon "for
signs, and for seasons, and for days, and years" (Genesis 1:14). This
means that the calendar which God designed to house His sacred
festivals would be luni-solar. The months must occur at the proper
times for the holy days to fall within the proper season of the year.

     How, then, are the Hebrew lunar months related to the solar year?
Every 19 solar years (of 365 1/4 days), the moon revolves around the
earth 235 times, each "lunation" being on the average 29 days, 12
hours, 793 parts. This remarkable astronomical relationship makes it
possible to combine common years and leap years together within a
fundamental pattern that repeats itself every nineteen years:

     12 common years (12 months each)is 144 months   (each month is
      7 leap years (13 months each)  is  91 months     29d, 12 h,
     -----------------------------------------------   793p)
     19 Hebrew years is 235 months  = 19 solar years

     You should be aware, however, that 235 lunar months is about an
hour and a half less than 19 Julian years. (To be precise, 235
lunations is 1 hour, 485 parts LESS than 19 Julian years.)

     The 19-year cycle is also known as the cycle of Meton, or the
Metonic cycle.

     Be sure you have MEMORIZED the time relationships listed at the
beginning of this section! Don't assume that you know them "pretty
well."
 

PERFORMANCE GOAL 1B

     The source of many errors in calendar calculations is faulty
arithmetic. Most occur in the basic addition, subtraction, and
multiplication operations. All of these skills you learned before high
school. Consequently, some review and practice are absolutely
necessary. WORK the problems; don't be deluded into thinking that
glancing over them will suffice.

ADDITION: The key to successful computation rests in one simple rule:

                         add only likes together

Just as 3 apples and 4 oranges don't equal 7 apples, neither does
adding 11 hours to 4 days equal 15 hours. Never put a number down
without being positively certain that you know what it represents. The
best way is to label every number: 671 p is 671 parts. Another way is
to keep like quantities in clearly defined columns:

                     days      hours      parts
                    ------    -------   -------
                       1         12        103
                       3         15       1186

This is the same as:   1 d       12 h      103 p
                       3 d       15 h     1186 p

Select a format you like and use it consistently. Here are some sample
problems:
 

          5 d       15 h        175 p
        + 1 d        7 h        801 p
        ------------------------------
          6 d       22 h        976 p        So long as you add only
                                             like quantities, no
                                             outstanding difficulties
                                             will happen.

          6 d       237 h      1189 p
        +24 d       107 h      2159 p
        ------------------------------
         30 d       344 h      3348 p        The numbers can become
                                             very large, as you can
                                             see. Later on you will
                                             learn how to reduce these
                                             to lowest terms.

          8 d        13 h       198 p
         18 d        23 h       432 p
         13 d        45 h      1835 p
      +   7 d        96 h       103 p
       -------------------------------
         46 d       177 h      2568 p        Adding a whole string of
                                             numbers together will
                                             often occur. Don't be
                                             afraid to double and
                                             triple-check your
                                             arithmetic!

Solve the following problems below. If you aren't "comfortable" by the
time you complete them, try to make up a few of your own for additional
practice.

1.1)          6 d          17 h            879 p
     +        3 d          14 h            198 p
     --------------------------------------------

1.2)         67 d          95 h            777 p
     +      275 d         777 h           2589 p
     --------------------------------------------
 

1.3)         55 d         178 h           976 p
             31 d           1 h           134 p
            178 d          23 h          1937 p
            225 d           9 h         11581 p
            308 d         768 h           649 p
     +       29 d          12 h           793 p
     ----------------------------------------------------------------
 

1.4)        131 d          29 h           433 p
            227 d           8 h           191 p
              0 d           1 h           485 p
     +      130 d          12 h           883 p
     ----------------------------------------------------------------
 

The answers to these problems appears below:
1.1)        9 d           31 h          1077 p
1.2)      342 d          872 h          3366 p
1.3)      826 d          991 h         16070 p
1.4)      488 d           50 h          1992 p

REDUCTION: Before attempting to subtract or multiply these numbers, it
will be helpful for you to understand how they are reduced.

You have learned how to reduce quantities such as inches to feet and
yards, or seconds to hours and minutes, back in junior high
mathematics:

     45 inches reduces to     1   yard    0 feet   9 inches
     49 inches reduces to     1   yard    1 foot   1  inch
    436 seconds    "       "  0   hours   7 min.  16  seconds
   3600 seconds    "       "  1   hour    0 min.   0  seconds

Reduction of large numbers is not difficult. An EQUIVALENT way of
saying exactly the same thing is all that you are doing. Both
quantities are equal.

The primary units in your calculations will be days, hours, and parts.
Each of these quantities are related to each other, as you have already
learned from performance goal 1A.
 

1. 5) Complete the following:

     ________ parts = 1 hour

     ________ hours = 1 day

The answers, of course, are 1080 parts in an hour, and 24 hours in a
day. This means that 1080 parts can be written as:

          0   days       1   hour       0   parts

In like manner, 24 hours can be shifted to the days column:

          1   day        0   hours      0   parts.

Take a quantity like 3 d, 49 h, 1400 p. How do you reduce it? What you
want to do is transfer all EXCESS WHOLE HOURS contained in the parts
column to the hours column. Then you will take out all the WHOLE days
contained in the hours column and move them to the days column. Just
like reading Hebrew, you work from right to left!

To reduce numbers, apply the following operations:

     1)   Divide the parts in the parts column by 1080. The whole
number in the quotient represents hours.

     2)   Add the whole number in the quotient to the hours column.
This becomes the "revised" hours. Place the remainder in the "reduced"
parts column. If remainder is 0, put this number in that column.

     3)   Take the revised number of hours and divide this by 24. The
whole number in the quotient is the number of whole days.

     4)   Add the whole number of days to the days column, and place
the remainder in the "reduced" hours column.

-----------------------------------------------------------------------
Reduced numbers for days, hours, parts will have:
          any number of days 23 or less hours 1079 or less parts
-----------------------------------------------------------------------
     Here's how to apply these rules to reduce 3 d, 49 h, 1400 p:

1)             1 h                         1 h  320 p
           ---------
     1080 / 1400 p
            1080
           ---------
             320
 

2)            1 h  320 p
     +  3 d  49 h
     --------------------
        3 d  50 h  320 p
 

3)         2                               2 d  2 h
         --------
     24 / 50 h
          48
         --------
           2
 

4)       2 d  2 h  320 p
     +   3 d
     --------------------
         5 d  2 h  320 p                   REDUCED

Problems with large numbers are handled the same way. The answer to 1.
2 is  342 d  872 h  3366 p.
 

1)             3 h                         3 h  126 p
           ------------
     1080 / 3366 p
            3240
           ---------
             126
 

2)               3 h  126 p
     +  342 d  872 h
     -----------------------
        342 d  875 h  126 p
 

3)           36 d                          36 d  11 h
          ---------
      24 /  875 h
             72
          ---------
            155
            144
          ---------
             11
 

4)      36 d  11 h  126 p
     + 342 d
     ----------------------
       378 d  11 h  126 p
 

Reducing numbers isn't difficult. Your accuracy will be enhanced if you
consistently stick to a single format. Slopping numbers down
haphazardly on the page is inviting computational errors.

1.6) Reduce 5, 796 parts.
1.7) Reduce 579, 600 parts.
1.8) Reduce 85 d  91 h  150,000 p.
1.9) Reduce 567 d  5228 h  254,404 p.

The answers to 1.6 - 1.9 are on the next pages. Work these problems on
separate paper, then compare your calculations with the complete
arithmetical details supplied.
 

1.6            5 h                          0 d  5 h  396 p answer
           ---------
     1080 / 5796 p
            5400
           ---------
             396
 

1.7            536 h                    536 h  720 p;  536 hours
          ---------------                              must be reduced.
     1080 / 579600 p
            5400
           ---------
             3960
             3240
            --------
              7200
              6480                22 d     22 d  8 h  720 p answer
             -------          --------
               720        24 / 536 h
                               48
                              -------
                                56
                                48
                               ------
                                 8
 

1.8       85 d 91 h 150,000 p

     (1)        138 h                      138 h  960 p;  add this to
           ------------                                   85 d 91 h.
     1080 / 150,000 p
            108 0
           -----------
             42 00
             32 40
             ---------
              9600
              8640
             ---------
               960
 

     (2)        138 h  960 p
          85 d   91 h
          -------------------
          85 d  229 h  960 p
 

     (3)         9 d                       9 d  13 h;  add together the
              --------                                 reduced 960 p,
          24 / 229 h                                   9 d  13 h, and
               216                                     85 d.
              --------
                13 h
 

     (4)        13 h  960 p
          85 d
          ------------------
          94 d  13 h  960 p             answer
 

1.9       567 d  5228 h  254,404 p
 

     (1)            235 h                  235 h  604 p
                -----------
          1080 / 254404 p
                 2160
                ---------
                  3840
                  3240
                ---------
                   6004
                   5400
                ---------
                    604 p
 

     (2)          235 h  604 p
          567 d  5228 h
          ---------------------
          567 d  5463 h  604 p
 

     (3)        227 d                      227 d  15 h
              --------
          24 / 5463 h
               48
              --------
                66
                48
              --------
                183
                168
              --------
                 15 h
 

     (4)  227 d  15 h  604 p
          567 d
          --------------------
          794 d  15 h  604 p
 

MULTIPLICATION: After adding days, hours, and parts, you will find that
multiplication is quite easy. The important key to correct
multiplication is that ALL TERMS must be multiplied! Forgetting to
multiply days while you complete the operation for the hours and parts
can easily happen if you aren't wary.

Look at a sample problem:

     2 d  16 h  595 p
                x 3
     -----------------
     6 d  48 h 1785 p    Notice that the multiplier operates
                         on the days, the hours, and the parts.

Multiplication can help you solve many problems that would take much
longer by straight addition. Here is an example:

How many days, hours, parts are in an AVERAGE common year?

You should easily remember:

          The number of lunar months in a common year;
          The number of days hours and parts in a lunar month.

If you cannot recall these numbers, return immediately to the first
page of this section and drill intensely on the time relations listed
in 1A.

A common year has 12 months. Each month has 29 d, 12 h, 793 p.
Therefore an average common year has:

       29 d   12 h   793 p
            x 12 months
      --------------------
       58     24    1586
       29     12     793
      -----  -----  ------
      348 d  144 h  9516 p

If this number were to be used in a series of computations, it could
stand "as is." On the other hand, when it is the final answer, can't
you see the need to reduce such a number? You may want to verify that
an average common year has 354 d  8 h  876 p. (This is worth
remembering.)

1.10  How many days, hours, and parts are in an average intercalary
year?

1.11  What number of d, h, p are in the 19 year cycle?
 

1.12  7 x (5 d 21 h 589 p) =  _____________

1.13  (4 d  8 h  876 p)  x 12 =  _____________

1.14  (18 d 15 h  589 p)  x  6 =  _____________
 

Problems 1.10 - 1.14 are worked out for you below.
 

1.10   29 d   12 h     793 p
          x 13 months
       ----------------------
       87     36     2,379
       29     12       793
       ----------------------
      377 d  156 h  10,309 p    This reduces to 383 d  21 h  589 p.
 

1.11 You may calculate this problem by two methods:
     a) add the lengths of 12 common years and 7 leap years together,
using the example and 3.10.
     b) Find the length of 235 lunations.

     Here's how method b) is solved:

         29 d    12 h    793 p
                       x 235 months
       ---------------------
        145      60     3965
        87      36     2379
       58      24     1586
      -----    -----  ------
       6815 d  2820 h 186355 p             This reduces to
                                             6939 d  16 h  595 p.
 

1.12    5 d   21 h   589 p
                    x  7
     ----------------------
       35 d  147 h  4123 p                 This reduces to
                                              41 d  6 h  883 p.
 

1.13    4 d    8 h   876 p
                    x 12
     ----------------------
        8     16    1752
       4      8     876
     ----------------------
       48 d   96 h 10512 p                 This reduces to
                                             52 d  9 h  792 p.

1.14   18 d  15 h   589 p
                   x  6
     ----------------------------------
      108 d  90 h  3534 p                  This reduces to
                                             111 d  21 h  294 p.
 

SUBTRACTION: This is the most error-prone operation for many students.
Two facets of subtraction are particularly troublesome:

     a) borrowing
     b) negative numbers

First look at a straight-forward subtraction problem:

         5 d   21 h  589 p
     -  (4 d   19 h   98 p)
     -----------------------
         1 d    2 h  491 p

Notice that the minus sign affects all the quantities in the second
line. Obviously, adding one of the units and subtracting the others
isn't kosher! The first principle to keep in mind:

     SUBTRACT ALL TERMS called for.

Practice on two of these problems, comparing your answer with that
given.
 

1.15     6 d  6 h  883 p
     -  (1 d  5 h  785 p
     ---------------------
                                           answer: 5 d  1 h  98 p
 

1.16 -  (  74 d  14 h   45 p)
          111 d  21 h  294 p
     -------------------------             answer: 37 d  7 h  249 p
                                           (You CAN subtract "upside
                                             down.")
 

In the next example, "borrowing" will be necessary:

          8 d  23 h  403 p
     -   (6 d  22 h  528 p)
     -----------------------
 

Since 1080 parts are in an hour, 8 d  23 h  403 p can be changed to
8 d  (23-1) h  (403 + 1080) p, or 8 d  22 h  1483 p. Be sure you
understand that neither expression is different. Borrowing merely
places the number in a more convenient form for subtraction.

By borrowing quantities as needed, the problem becomes ordinary
subtraction:

          8 d  22 h  1483 p
     -   (6 d  22 h   528 p)
     ------------------------
          2 d   0 h   955 p

In order to firm up the concept of borrowing in your mind, examine
another illustration:

         3 d   3 h     0 p
     -  (1 d  18 h  6000 p)
     -----------------------
 

On this problem, you cannot simply take one hour and transfer 1080
parts. A quick estimate will tell you that at least 6 hours need to be
changed to parts. Only 3 are in the hours column. Where do you get
them?

First, make up for the lack of hours in the second column by converting
a day (or more!) into hours.  3 d  3 h  0 p transfers to

     2 d  (24 + 3) h  0 parts, or  2 d  27 h  0 p.

Next borrow 6 hours and change them to parts:

     2 d  (27-6) h  (6480 + 0) p or,
     2 d      21 h        6480 p

Finally, subtract as required by the original problem:

         2 d  21 h  6480 p
     -  (1 d  18 h  6000 p)
     -----------------------
         1 d   3 h   480 p

Once in a while, you will find that in borrowing parts, you loose too
many hours for subtracting the hours. (Instead of 18 hours, suppose you
needed to subtract at least 22 hours in the above example.) When that
happens, simply borrow again from the next column to the left.

Borrowing is a reverse of reduction. Remember that both quantities are
equivalent -- before reduction and after reduction, or before and after
borrowing. A "golden rule" for these operations:

     ----------------------------------------------------------------
        WHAT YOU ADD TO ONE COLUMN MUST BE SUBTRACTED FROM ANOTHER.
     ----------------------------------------------------------------

1.17    40 d  6 h   104 p
     - (35 d  2 h  1141 p)
     ----------------------
                                           answer:  5 d  3 h  43 p
 

1.18     136 d  123 h  5291 p
     - (  89 d   29 h  7679 p)
     --------------------------
                                           answer:  47 d  91 h  852 p
 

1.19     21 d  30 h   811 p
     -  (17 d  69 h  2050 p)
     ------------------------
                                           answer:  2 d  7 h  921 p
 

The last complication to hurdle with subtraction is negative numbers.
By borrowing, you were able to keep numbers positive. But sometimes the
arithmetic is simplified by allowing numbers to become negative-without
borrowing.

There is nothing mysterious about negative time. All that it means is
time BEFORE a prescribed reference point. Such a reference point is the
blast-off of a moon launch. Events before lift-off are considered
negative times, as you heard on many telecasts: "t MINUS eight minutes
and holding." After the rocket lifts off, time becomes positive -- with
reference to the ZERO lift-off.

An expression like days, hours, parts refers to a particular instant in
time. When you are computing a problem, some of these numbers will
occasionally become negative. Nonetheless, the expression is still
identifying a precise moment. For example look at:

     5 d  -2 h  -810 p

Does this specified time occur on the fifth day or on the fourth? The
answer is the fourth day. This is immediately clear if you change the
expression to all positive signs by "borrowing. " These are the steps:

           5  d        -2  h          -810  p
        (5-1) d  (-2 + 24) h          -810  p
           4  d        22  h          -810  p
           4  d   (22 - 1) h  (-810 + 1080) p
           4  d        21  h           270  p

So  5 days  -2 hours  -810 parts is the same as  4 days  21 hours  207
parts.

You should become familiar enough with these negative expressions that
you accurately complete computations involving them. In ordinary
subtraction, all numbers were positive. Now you will find that
subtraction and addition of numbers of either sign frequently occurring
when you eventually calculate the days, hours, parts, of the month for
a conjunction of the moon.

The next example illustrates how this type of problem is worked:

            36 d   22 h   1284 p
           -17 d +  3 h -  541 p
          ------------------------
            19 d   25 h    743 p    or    20 d  1 h  743 p

When a number is not preceded by a sign, it is taken to be positive. A
few rules apply for these problems:

     adding two negatives together  (e.g. -5 + -3)  results in a
negative number larger than either (-8).

     adding a negative number and a positive number together is the
same as subtracting. The sign of   the  larger value (called the
absolute value in mathematics) is the sign of the final number.

          (-30  +  50  =  +20)
          (-48  +  38  =  -10)

     When you subtract a negative number from another negative number,
you change the sign of the number you are subtracting, and then add as
above. (-5  -  -3  =  -5  +  +3  =  -2)

Adding two negative numbers together results in a larger negative
number. Subtracting two negative numbers results in a smaller negative
number -- closer to zero. Any time you are working with negative
numbers, be sure that you carry the signs with you. Although you need
not specifically mark positive numbers, this may be advisable for you
initially in order to keep the signs clear in your mind. Whenever you
are subtracting, remember that the subtraction sign affects every
number (term) in the expression:

          45 d  23 h   150 p
      - ( -8 d  15 h  -125 p)
      ------------------------

Notice how the numbers subtracted change signs:

          45 d   23 h     150 p
          +8 d  -15 h  +  125 p
          ----------------------
          53 d    8 h     275 p

Try the following problem:

1.20     - 18 d  - 2 h  -  780 p
         -  0 d  - 1 h  -  485 p
         - 97 d  - 2 h  -  756 p
         + 74 d + 14 h  +  196 p           The answer appears below.
         + 13 d
         + 36 d + 22 h  + 1284 p
          -----------------------
 

1.20   The problem becomes easier if you total the negative quantities
separate from the positive, and then combine.

          - 18 d  -  2 h  - 780 p              74 d    14 h   196 p
             0 d  -  1 h  - 485 p              13 d
          - 97 d  - 22 h  - 756 p              36 d    22 h 1284 p
          ------------------------      -------------------------------
          -115 d  - 25 h - 2021 p            123 d     36 h 1480 p

           123 d    36 h   1480 p
          -115 d  - 25 h - 2021 p
          ------------------------
             8 d    11 h -  541 p          This reduces (by taking one
                                             hour and changing it to
                                             parts: + 1080 - 541) to
                                             8 d  10 h  539 p.

1.21 Work the next problem without reducing until the final step:

           210 d  - 15 h  -  971 p
        - ( 37 d    19 h  - 1185 p )
          --------------------------
                                           Change signs and subtract.
                                             Your result should be
                                             173 d  -34 h  +214 p.
                                             This reduces to
                                             171 d  14 h  214 p.
 

1.22  Add 92 d  581 h  471 p  to eight times (-10 d  -21 h  -204 p).
Then subtract  -152 d  - 589 h  188 p  from the sum. Reduce at the
final answer. How many days over a full number of weeks is this?

     First multiply: - 10 d  -  21 h  -  204 p
                                       x 8
                    --------------------------
                     - 80 d  - 168 h  - 1632 p
     Add:              92 d    581 h     471 p
                    ---------------------------
                     + 12 d  + 413 h  - 1161 p
 

     Subtracting  -152 d  -589 h  188 p, you change the signs and add:

                       12 d  +  413 h  - 1161 p
                    + 152 d  +  589 h  -  188 p
                    ----------------------------
                      164 d  + 1002 h  - 1349 p

     Borrow 2 hours, and then divide 1000 hours by 24 to convert the
hours column to a number less than 24. This gives:

                    205 d  16 h  811 p

To find how many days this is over a full number of weeks, you divide
by 7 the REDUCED number of days. 205 / 7 leaves a remainder of 2. Two
days plus 16 hours and 811 p is the time over a full number of weeks.

                              *   *   *

     After working the problems in this section, you should feel
confident about your ability to successfully add, subtract, multiply,
and reduce days, hours, and parts. Some may still require additional
drill on these operations. Design your own problems for more practice.
Have another student check your work, or see your instructor for
assistance.

                        ----------###----------

                              PROGRAM II

             CALCULATING THE DAY OF WEEK FOR MOLAD TISHRI
 

PERFORMANCE GOAL 2:

     Given a required Roman year, you will correctly calculate the day
of the week, hours, and parts for the molad Tishri of that Roman year.
This will be accomplished WITHOUT CHARTS AND TABLES.

     Why do you need to be concerned with the molad of Tishri? The
answer is that you must know when it occurs before you can determine
the date of the Festival of Trumpets. And all other holy days within a
Roman year (January-December) are ultimately referenced to that holy
day. The molad of Tishri is prerequisite to most calculations involving
the Hebrew calendar. Correct calculation of the molad of Tishri is thus
essential for determining what dates on the Roman calendar we commonly
use for God's Sacred Festivals to occur.

     Certain definitions and concepts need to become crystallized in
your mind:

     What is a molad?
     When is Tishri?
     What is a bench mark?
     How is time reckoned in the calculations?
     What is meant by the "advancement" of the molad?

Let's find the answers to these questions.

     What is the molad Tishri?

     TISHRI is the seventh month of the sacred calendar. The computed
time for the conjunction of the sun, moon, and the earth is called a
MOLAD, from the Hebrew MOLED (plural, MOLEDOTH). This word means
renewal, or rejuvenescence.

     Molad of Tishri is the computed time of the new moon of the month
of Tishri, which corresponds to September/October. As Tishri is also
the first month of the civil Hebrew year, the molad Tishri is also the
calculated astronomical commencement of the year.

     Another term which you will be using is bench mark. All this means
is a point of reference from which measurements can be made. Any known
molad (expressed as day of the month, day of the week, hours, and
parts, e.g., October 6, Sunday, 23 h, 204 p in 3761 BC) can serve as a
bench mark. The most practical choice for a bench mark, however, will
be the molad Tishri of year one in a 19 year cycle. 3761 BC is such a
year.

     In order to avoid a mental mix-up later, let's clarify how we
reckon time. Just what does an expression like 4 d  7 h  503 p mean?
Such an expression can be taken either of two ways: a) as an interval
of time b) a time in the week.

     Consider case a). Here, when you are speaking of a length of time,
you assume that you are starting from a reference point of 0 days, 0
hours, 0 parts. This is just the same as clocking a track runner on a
stop watch that starts ticking at the sound of the gun. A certain time
span is involved, figuring from a bench mark of 0 days, 0 hours, 0
parts.

     With case b), you are still reckoning time, but from a DIFFERENT
reference point. In the calendar calculations you are working in
Theological Research, the week begins at Sunday midnight. Sunday is
regarded as the first day of the week; midnight the zeroth hour, zero
parts:

          The week begins: Sunday: 1 d  0 h  0 p.

     An expression like 1 day 13 hours 0 parts MUST BY DEFINITION OF
THE STARTING POINT refer to Sunday, the 13th hour after midnight, or
Sunday 1 PM.

     Likewise, 1 day 13 hours 179 parts refers to a time slightly later
than 1 PM on Sunday.

     The reason why we arbitrarily begin Sunday as 1 day 0 hours 0
parts is so that there will be an exact coincidence with the days of
the week: 1 d is Sunday; 2 d is Monday; 4 d is Wednesday; 7 d is
Sabbath. So long as all the numbers are reduced, you merely look at the
number of days, and these correspond to the day of the week. (For the
same reason of SIMPLICITY, we begin the day with midnight instead of 6
PM so far as the CALCULATIONS are concerned. Since the bench mark is
expressed by this same reckoning, the final results will be exactly the
same.)

     By contrast, if we decided to define the beginning of the week,
Sunday, as the zero day, and Saturday sunset (say 6 PM) as the start of
Sunday, we would have the expression  0  d  0 h  0 p = 6 PM Saturday
evening. Then  1 d  13 h  179 p would have an entirely different
meaning, if it referred to a time in the week. 1 d would then be
Monday; 13 h  179 p would be slightly after 7 AM. Do you see how much
more complicated your work would then become?

     Here is another distinction that can help you understand the
difference in meaning between an interval of time, and a time in the
week:

   a) an interval of time:

     parts are added to the hours       are added to the days

   b) a time in the week:

     parts are added to the hours       WITHIN the day of the week.

     If 4 days 7 hours 503 parts refers to an interval of time, then
503 parts plus 7 hours is added to 4 days. Therefore the interval of
time goes as far as the 7th hour(and 503 parts) of the fifth day.

     If 4 days 7 hours 503 parts is to be taken as a point in the week,
then 503 parts plus 7 hours is WITHIN the 4th day of the week.

     Remember this difference in what is meant comes about by the way
the starting point is DEFINED, and for no other reason.

     During your calculation of the advancement of the molad over a
full number of weeks, you may come up with a number like 0 days 4 hours
and 71 parts. Be comforted by the fact that the expression is simply an
interval of time by the way Sunday midnight is defined.

     How can you relate a time interval to an expression of time during
a week?

     YOU MUST ADD AN INTERVAL OF TIME to the BENCH MARK before you can
determine a real day of the week. The bench mark will implicitly tell
you where the week begins. The bench mark for the year 3761 BC is
Sunday, the 23rd hour 204 parts. (We could have called that bench mark
Monday if we started the day at 6PM instead of midnight. But then we
would be confronted with a more complicated interpretation of what the
time expressions mean.)

     Now transform each of the following expressions to a) a time
interval b) a time during the week. Use this format:

a: _______ days, plus _______ hours _______ parts of the _______ day

b: _______ (day of the week), between _______ & _______ AM or PM

2.1  4 d   3 h  191 p
2.2  1 d  16 h  304 p
2.3  6 d   7 h    8 p
2.4  7 d  22 h    5 p
2.5  2 d  13 h  871 p

The answers are given below.

2.1  a)   4 days plus 3 hours 191 parts of the 5th day
     b)   Wednesday, between 3 & 4 AM

2.2  a)   1 day plus 16 hours 304 parts of the 2nd day
     b)   Sunday, between 4 & 5 PM

2.3  a)   6 days plus 7 hours 8 parts of the 7th day
     b)   Friday, between 7 & 8 AM

2.4  a)   7 days plus 22 hours 5 parts of the 8th day
     b)   Sabbath, between 10 & 11 PM

2.5  a)   2 days plus 13 hours 871 parts of the 3rd day
     b)   Monday, between 1 & 2 PM

     One particularly important time interval will occur throughout
your experience with the Hebrew calendar. In order to calculate the
molad Tishri, you will be working with two molads:

          * a known molad, such as 3761 BC -- a bench mark
          * the molad of the Roman year which you are determining

     The time interval between these two molads is the ELAPSED TIME.
Since you are dealing with the molad of Tishri in both the required
year and the bench mark, the elapsed time will be a whole number of
years, such as 4520 years, 1503 years, 38 years.
     Let's investigate another feature of the Hebrew calendar which we
can call MOLAD "ADVANCEMENT." Now you know that the MOLAD itself
doesn't really move, since the term is defined as the calculated
conjunction of the sun, moon, and the earth. We are using a figure of
speech very similar to sun "rise."

     Here's an illustration of what is meant by the advancement of the
molad. In 3761 BC, the molad Tishri was on Sunday  1 d  23 h  204 p. In
1980 AD the molad Tishri will be Tuesday  3 d  23 h  206 p. Although
thousands of years have elapsed over the time span, the APPARENT
"advancement" in the week of the second molad is only 2 days 0 hours
and 2 parts.

     Why does the molad occur on different days of the week? The length
of an average lunar month is  29 d  12 h  793 p. How much greater than
four full weeks is this?
 

        29 d  12 h  793 p
      -(28 d   0 h    0 p)
     -----------------------------
         1 d  12 h  793 p;  about a day and a half.

     The molads of two successive months cannot occur on the same day
of the week because of this EXCESS over 28 days. In one month, if the
molad were Monday 8 AM (2 d  8 h  0 p), the molad of the next month
would be:

         2 d   8 h     0p
       +(1 d  12 h  793 p)
     -----------------------------
         3 d  20 h  793 p,  or Tuesday, between 8 and 9 PM.

     Although the second molad occurred  29 d  12 h  793 p  after the
first one, the second molad was "displaced" WITH REFERENCE TO THE WEEK
by  1 d  12 h  793 p. Merely as a convenient label, we will refer to
that apparent shift as MOLAD ADVANCEMENT. But the molad doesn't move;
only the time of its occurrence IN THE WEEK apparently advances.

     The TOTAL MOLAD ADVANCEMENT is simply the EXCESS over the number
of full weeks in the elapsed time from the bench mark to the molad
Tishri of the desired Roman year.

     Just as monthly molads will occur on different days, the molad of
Tishri will advance in the week over the previous molad of Tishri. If
in three years the total advance were 13 days, the Molad would be 13
days later. But in terms of the day of the week, this would be 13-7, or
6 days later in the week.

     Now you will learn how to calculate the day of the week for the
molad Tishri.

     You will find it easier to understand how to determine the day of
the week for a given molad if the steps are explained first without the
mathematical details:

     In order to determine the day of the week for the molad of Tishri,
you must find the TOTAL ADVANCEMENT of the molad that occurs within the
time span involved from a known molad, or bench mark.

     What will make up that time span? From that bench mark, a certain
number of years will elapse to the particular year in question:

     molad of Tishri                                molad of Tishri
       (bench mark)                                 of required year
                              (elapsed time)
           *________________________________________________*
 

     Later on in this program, you will learn how to express elapsed
time as:

            a whole multiple of 19 year cycles,
     plus   the excess number of common years,
     plus   the excess number of leap years.

     Logically, the total advancement of the molad, or excess over full
weeks in the elapsed time, for the year in question can be found by
adding:

          the advancement due to whole multiples of 19 year cycles,
      +   the advancement due to the number of common years,
      +   the advancement due to the number of leap years.

     An example will clarify this. If the elapsed time is 156 years,
there are 8, 19 year cycles, 4 common years, and 2 leap years (you'll
see how this is done later). The total advancement of the molad will
be:

          8 times the advancement of one 19 year cycle
       +  4 times the advancement of one common year,
       +  2 times the advancement of one leap year.

     The final step is simply adding the total excess over full weeks
to whatever bench mark you started with. This figure will give you the
day of the week of the molad Tishri in question.

     Be sure that you understand the qualitative elements connected
with the advancement of the molad. A molad "advances" with respect to a
known molad because of the excess time in one average lunar month over
a full number of weeks. All you are doing is adding the total
advancement that occurs within the time interval from the bench mark to
the molad Tishri of the year in question. You are actually finding the
excess over the full weeks from the bench mark to the molad of the
required year.

     Your success at determining the day of the week of a required
molad impinges upon:

          a)   Correctly finding the ELAPSED TIME from a bench mark to
the required year.

          b)   Expressing the elapsed time in terms of multiples of 19
year cycles
                     +   common years in the remainder
                     +   leap years in the remainder

          c)   Calculating the molad "advancement" attributable to each
element of b), then adding these together, reducing as necessary.

          d)   Adding the reduced advance of the molad c) to the bench
mark.
 

ELAPSED TIME, 2A

     Three possibilities exist for the elapsed time from the bench mark
to the required year:

          a1)  Both years are AD dates.
          a2)  Both years are BC dates.
          a3)  One year is BC and the other is AD

(Although you might work backwards in time, this program will only
consider problems in which the bench mark is the EARLIER of the two
years.)

     a1: Bench mark and required year both AD dates. The elapsed time
is simply the difference between the two years. If the bench mark is
1845 AD and the required year is 1931, the elapsed time is:

                    1931    -    1845    =    86 years

If the bench mark is 895 AD and the required year is 1751, the elapsed
time is:

                    1751    -    895    =    856 years

     a2: Both years are BC dates. Many calculations use 3761 BC as a
bench mark. To be mathematically consistent, it is helpful to place a
negative sign ( - ) before all BC years. Since you will still be
subtracting in order to find the difference, the second number will
become positive -  ( - )  =  +. As you are primarily concerned with
years after 3761 BC, or -3761, the year in question will always be more
positive (closer to zero). As a check when both years are BC, expect
the elapsed time to be SMALLER than 3761 years. (Use only the "absolute
value" for the elapsed time, which means you can disregard the final
negative sign.)

Using 3761 BC as a bench mark, what is the elapsed time to 585 BC?

               -3761  -  (-585)  =  -3761  +  585  =  -3176

The elapsed time is 3176 years.

What is the elapsed time to 1486 BC?

               -3761  -  (-1486)  =  -3761  +  1486  =  -2275

The elapsed time is 2275 years.

     a3: Bench mark is BC and the required year is AD. 3761 BC is
frequently used as bench mark for AD years. Only one thing is really
different here. There is no year allotted for 0 AD or 0 BC.
Mathematically, the number system does have a 0. What do you do?

Graphically, the two systems look like:

Calendar:      5 BC   4 BC   3 BC   2 BC   1 BC   1 AD   2 AD   3 AD
Number Scale: -5     -4     -3     -2     -1      0      1      2

The number scale has one more place, a zero, than the calendar.
Therefore in time intervals that cross AD/BC, you must SUBTRACT ONE
from the answer you compute arithmetically. (The time span from 4 BC to
2 AD is only five years.) BE SURE YOU REMEMBER TO SUBTRACT ONE FROM THE
MATHEMATICAL COMPUTATION! But only when you cross over from BC to AD.

What is the time elapsed from 3761 BC to 1000 AD?

          -3761  -  +1000  =  -3761  -1000  =  -4761

The elapsed time, however, is 4761    -    1 year, or 4760 years.

What is the time span (elapsed time) to 1974 AD from 3761 BC?

        -3761 BC  -  (+1974)  =  -3761  -  1974  =  -5735

The elapsed time is 5735  -  1, or 5734 years. As a check, you should
be aware that in going from BC to AD, the elapsed time will be a larger
number (taken as an absolute value) than either the bench mark or the
required year.

Drill yourself on elapsed time calculations with the following
problems. Remember that if you are wrong here, all the rest of your
calculations w ill be off!

               Bench mark     Required year  Elapsed time

     2a.1        1845 AD         1971 AD
     2a.2        1883 AD         1945 AD
     2a.3        3761 BC         1442 BC
     2a.4        3761 BC            4 BC
     2a.5        3761 BC          721 BC
     2a.6        1861 BC          604 BC
     2a.7        3761 BC           31 AD
     2a.8        3761 BC         1859 AD
     2a.9        3761 BC         1982 AD

See problems 2b.1 to 2b.9 for the elapsed time.

EXPRESSING ELAPSED TIME, 2B

     Once you've determined the number of years between the bench mark
and the required year, you need to express that time in terms of the
Hebrew calendar.

     On a practical basis, 19 year cycles are convenient. Find the
number of 19 year cycles in the elapsed time, divide the elapsed time
by 19. The QUOTIENT is the number of 19 year cycles.

     Usually, however, this division will result in a remainder, a
number from 1 to 18. This remainder will tell you the number of years
elapsed in the next cycle.

     As you already know each 19 year cycle consists of 12 common years
and 7 leap years.

=====================================================================
     SINCE 142 AD (see footnote 1), the years in a cycle that are leap
years are:     3    6    8    11    14    17    19
=====================================================================

Be sure to memorize these numbers! (Of course, the years 1, 2, 4, 5, 7,
9, 10, 12, 13, 15, 16, and 18 are common.)

======================================================================
     Before 142 AD, the leap years were:  2  5  7  10  13  16  18
======================================================================

For any remainder you acquire after dividing the elapsed time by 19,
you:

     1)   Decide whether the required year is before 142 AD, or 142 AD
and after.
     2)   Count the number of leap years that fit in the remainder (or
you could count the common years).
     3)   The number of common years will be the remainder minus the
number of leap years.

For example, the elapsed time from 1845 AD to 1975 AD is 130 years.
This is 130/19 time cycles, or six 19 year time cycles plus 16 years
remainder.

The leap years for 1975 (after 142 AD) are 3, 6, 8, 11, 14. Five leap
years altogether. Since 1975 has 16 elapsed years and there are five
leap years thus far, there must be 16 - 5 or 11 common years.

--------------------------------------------------------------------
     (footnote 1) There is some evidence that an adjustment to the
Hebrew calendar may have taken place during the patriarchate of Simon
III (140-163). See Cyrus Adler, "Calendar, History of," in "The Jewish
Encyclopedia" (New York: Funk and Wagnalls, 1907), Vol. 3, p. 500.
--------------------------------------------------------------------

Here is another example. If the bench mark is 3761 BC and the required
year is 27 AD, the elapsed time is:

     -3761  -  (+27)  =  -3788 years; the elapsed time is 3787 years,
since you go from BC to AD.

      3787 / 19  =  199  19 year cycles, plus 6 elapsed years.

For 27 AD (before 142 AD), the leap years of the cycle are 2 and 5.
With two leap years, there must be  6  -  2, or 4 common years.

Practice expressing elapsed time in terms of 19 year cycles, the number
of leap years, and the number of common years. See problems 2a.1 to
2a.9.

          Elapsed   19 year   # of leap    # common
           time      cycles    years        years

2b.1        126 yrs
2b.2         62
2b.3       2319
2b.4       3757
2b.5       3040
2b.6       1257
2b.7       3791
2b.8       5619
2b.9       5742
 

Check answers against those below.

2b.1        6 cycles     4 leap    8 common
2b.2        3   "        1   "     4   "
2b.3      122            0         1
2b.4      197            5         9
2b.5      160            0         0
2b.6       66            1         2
2b.7      199            4         6
2b.8      295            5         9
2b.9      302            1         3

"ADVANCEMENT" OF THE MOLAD. 2c

     From your previous program (1A & 1B), you are equipped to find out
the required information regarding the advancement of the molad. Since
a lunar month has 29 days, 12 hours 793 parts, it is 1 day, 12 hours,
and 793 parts in excess of a full number of weeks. As you saw before, a
monthly molad (two successive months) advances 1 d  12 h  793 p.

     Knowing this, you can easily determine how much a molad "advances"
in a common year of 12 months, in a leap year of 13 months, or in a 19
year cycle of 235 months.

How much does the time of a molad "advance" in the week during a common
year?

     1 d   12 h  793 p   (the monthly "advancement",
                           or excess over 4 weeks)
                x 12          (number of lunar months in a common year)
     -----------------
     2     24   1586
     1     12    793
     -----------------                    6 days             8 hours
    12 d  144h  9516 p                 -------           ---------
                                   24 / 152 h      1080 / 9516 p
                                        144               8640
                                       -------            -------
    18 d    8 h  876 p   reduced          8 h              876 parts

     After 12 months, the molad "advancement" is 18 d  8 h  876 p.
Since full weeks will not affect the days of the week, you can divide
the reduced number of days by 7. So with respect to the week the molad
"advancement" for a common year is:

          4 d  8 h  876 p.

     You should notice an alternative way of arriving at the same
number. How much does a common year exceed the number of full weeks in
the year? Multiply the length of an average lunar month by 12 months:

     12  x  (29 d  12 h  793 p)  =  348 d  144 h  9516 p

Reduce this number: 354 d  8 h  876 p

Divide the reduced number of days by 7 to eliminate full weeks:

          4 d  8 h  876 p.

2c.1  Verify that an average leap year exceeds a full number of weeks
by 5 d  21 h  589 p.

2c.2  Verify that a 19 year cycle exceeds a full number of weeks by  2
d  16 h  595 p.

>From here on, you already are competent to handle the details of
multiplying the advancement of the molad, and then adding. For 4 BC (as
in 2a.4 and 2b.4) you will do the following:

     197  x (2 d  16 h  595 p)  molad advancement of 19 year cycles
       5  x (5 d  21 h  589 p)  molad advancement of leap years
       9  x (4 d   8 h  876 p)  molad advancement of common years

Multiplying and adding you will discover that this is:

          567 d  5228 h  254,404 p.

When reduced, and divided by seven (only the full reduced days are
divided by 7), the advancement of the molad over a week is:

          3 d  15 h  604 p.
 

ADDING REDUCED ADVANCEMENT OF MOLAD TO THE BENCH MARK, 2d

     The bench mark for 3761 BC is Sunday  23 h  204 p. This is
reckoning from midnight. As Sunday is the first day of the week, the
bench mark can be expressed as:

          1 d  23 h  204 p.

>From here on, you merely add the reduced advancement of the molad to
the bench mark. For 4 BC, this is:
 

              3 d  15 h  604 p
           +  1 d  23 h  204 p
          ----------------------------
              4 d  38 h  808 p
     or       5 d  14 h  808 p
 

     The fifth day of the week is Thursday, so the molad occurred on
Thursday, the 14th hour (2 PM), 808 parts.

So long as the final d, h, p, are reduced, you need only be concerned
(for now) with the first column.
 

2d.1  Verify that the molad for 721 BC is 5 d  7 h  364 p.

2d.2  Verify that the molad for 31 AD is 5 d  23 h  941 p.

2d.3  At this point, you should test yourself to be sure you can
fulfill the goal of program 2. Calculate the DAY OF THE WEEK of the
molad Tishri for 1996 AD without consulting this program. You'll need
to have memorized (or else work out again) the molad "advancement" for
a 19 year cycle, a common year, and a leap year, as well as remember a
bench mark.

Compare your calculations to the answer below, which is worked out in
detail.

Calculate the DAY OF THE WEEK of the molad Tishri for 1996 AD.

     -3761
     -1996
     -----------
      5757 yrs
        -1                302 cycles
     -----------         ----------
      5756 yrs      19 / 5756
                         57
                         --------
                           56
                           38
                          ------
                           18 elapsed yrs; 6 leap yrs; 12 common yrs
 
 

     2 d  16 h      595 p              5 d  21h    589 p
           x  &nbs