I was searching the internet and came across a Christian Chat conversation from 2010 at

Proving that the Hebrew Masoretic Text is right and the Greek Septuagint is wrong

I decided to join Christian Chat in search of an theoretical mathematician who will review the problem discussed in the following article. Please email at [email protected] if you find someone.

The following describes the problem and why I need a theoretical mathematician who will go public.

Genesis Accuracy: University Finds Mathematical Proof

By Gioacchino Michael Cascione [email protected] author of “In Search of the Biblical Order”

One of the last places one would look for a complex mathematical formula is in the Book of Genesis. While completing the last chapter of my new book, “Repetition in the Bible: The Phenomenon of Divine Communication,” a mathematical formula connecting Genesis 5 and 11 was observed.

Without taking the time to give a lengthy explanation, there is mathematical equation in the ages of the 10 patriarchs from Adam to Noah and in the 10 patriarchs from Shem to Abraham. The following is a chart listing their ages, their ages at the birth of their notable sons, and the years they lived after the birth of their notable son as found in the Bible.

Patriarchtotal ageage at birth of sonyears after son’s birthPatriarchtotal ageage at birth of sonyears after son’s birthAdam 930 130 800 Shem 600 100 500 Seth 912 105 807 Arpachshad 438 35 403 Enosh 905 90 815 Shelah 433 30 403 Cainan 910 70 840 Eber 464 34 430 Mahalalel 895 65 830 Peleg 239 30 209 Jared 962 162 800 Reu 239 32 207 Enoch 365 65 300 Serug 230 30 200 Methuselah 969 187 782 Nahor 148 29 119 Lamech 777 182 595 Terah 205 (70) 135 Noah 950 500 450 Abraham 175 100 75 85753171

The initial pattern was found in the 7 headed beasts with 10 horns in Revelation (Rev. 12:3; 13:1; 17:1, 3). These Hebraic symbols can either be added for a total of 17 or multiplied for a total of 70. The same pattern is found in the literal dates for the beginning of the flood on the 17^{th}day of the 2^{nd}month (Gen. 7:11), and the Ark’s landing on Ararat on the 17^{th}day of the 7^{th}month (Gen. 8:4). This is a total of 150 day as explained by the great rabbinic scholar Umberto Cassuto, a key source for this research.

The total ages of the 10 patriarchs from Adam to Noah are 8575 years. This total is divisible by 7. The ages of the 7 oldest patriarchs, when added to 8575 creates a list of 17 numbers for a total of 15,113 years. This number is divisible by 17. The odds of such a coincidence are 119 to 1.

The same approach was applied to the second list of patriarchs from Shem to Abraham whose total ages add up to 3,171 years. This number is also divisible by 7. The 7 largest numbers were added to 3,171 for a total of 5,814. This number is also divisible by 17. The odds of both lists exhibiting these same properties are 14,161 to 1, if the individual experimenting with simple math knew what he was looking for. However, because the second list is smaller than the first, the odds are higher. In regular math common denominators increase, not decrease.

The calculations in these two lists share identical properties. Hence there is no question that they are governed by the same mathematical formula. (More unusual properties of these numbers are listed at the end of this article.)

The question is, “If these lists were humanly devised, what would the formula look like to arrive at these numbers?” How was it done? With this question in mind, assistance was requested from a Doctoral candidate in engineering at Iowa State University. After a number of attempts he recommended a 4^{th}year math wiz, Matt Swanson [email protected] for a solution to the problem. After examining the problem Swanson consulted a doctoral student in mathematics. She and Swanson were able to get close but they could not find the exact solution which they determined was based in modular mathematics.

They both recommended that I locate a theoretical mathematician at another university because Iowa State excels in statistical and scientific mathematics. Swanson also provided 3 pages of mathematical equations to help me present the problem to a doctor of theoretical mathematics.

After searching for an expert at another university in a different state, the secretary directed me to 2 of 8 theoretical mathematicians on their faculty, who were still available in June of this year. I did not explain what the problem was, other than I needed to consult a theoretical mathematician. The doctor I met graciously gave his time but he will not allow me to give out his name. He solved the problem. He also consulted an expert in the history of mathematics who wrote the following reply from his cell phone:

“It is my understanding that the Chinese Remainder Theorem was unknown in the Mediterranean basin prior to when they learned it from the Chinese. Quotient and remainders were important to the Babylonians, and the Euclid algorithm, as a method for finding common measures, is probably even older. But there are Chinese instruments and tables for using the CRT for counting after 520 AD, but I am pretty sure no middle or European counterparts identified as such. If they did use it, there should at least be tables that make it practice. Calculating using an algorithm would not have [been] practical for everyday problems.”

In other words Moses was the greatest Mathematician of the ancient world by 2000 years, or someone else rewrote these numbers 500 years after Christ, or Moses had help from another source.

The calculations I received from the anonymous professor were sent to Swanson for verification. Swanson states that the calculations are correct and the anonymous professor found the solution. By coincidence the anonymous professor is an Orthodox Jew who reads Hebrew and views his findings as further affirmation of the authenticity of the Masoretic Text and that Moses is an historic figure.

I need a PhD in Mathematics to examine these calculations, and go on record that they are correct.

Calculations from Anonymous University Professor on Ages of the Antediluvian and Postdiluvian Patriarchs

“I worked a bit on the problem, and realized that my initial thought was not quite right. I think that the student you talked with was on the right track. So far this is what I can do:

If you specify 9 numbers, and look for a 10th which satisfies your rules about divisibility, and where the 10th number is not among the 3 smallest numbers in the set of ten, then you can find the 10th number using this formula:

76*(the sum of the 9 numbers)+ 42*(the sum of the 6 largest numbers of the 9 you specify) - 119*n

where n is a whole number. In your case for the antediluvian ages, the sum of the 9 smallest ages is 7606, and the sum of the larger 6 of those is 5569, so the formula is 76*7606+42*5569-n*119=811954-119*n. The remainder of 811954 modulo 119 is 17. You get that with n=6823. However, if we need our number to be among the 7 largest ones, it must be at least 895. With n=6815, we get 969. with n=6816, we get 850 which is too small. 969 is the smallest solution that satisfies both the divisibility constraints and also is at least 895. Of course we can keep adding 119 to 969 and we will get an infinite number of additional solutions 1088, 1207, 1326,...

When I try the same rule with the post-diluvian numbers, I get 76*2571+42*2043-119*n=281202-119*n. Possible values that show up are 243, 362, 481, 600, 719, 838 etc. The desired value of 600 shows up, but, it is not the first solution that is sufficiently large. Even if we had specified that our missing number had to be the largest age, the constraint would have been satisfied by the value of 481.

I have also looked at the case where the missing number is one of the 3 smaller ones. In that case, you get a different formula, but the solution is not so special.

The formula is:

118*(sum of the 9 specified numbers)+84*(sum of the 7 largest specified numbers)-119*m where m is a whole number. You can check that this works with the antediluvian numbers if you exclude 365. In that case you get the sum of the 9 numbers as 8210 and of the 7 larger numbers as 6538. The formula becomes 1517972-119*m. Using m=12753 we recover the value of 365, but, other possible values are 8, 127, 246, 484, 603, and 722.

If you want to allow the missing number to be either in the set of the 3 smallest or in the 7 largest, I do not have a single formula that I can use. I suspect that the unknown n's and m's in my formulas could be defined a bit better if I can use some more complicated functions in the formula (the floor function or some other sort of rounding function).

76*(the sum of the 9 numbers) + 42*(the sum of the 6 largest numbers of the 9 you specify) - 119*n is arrived at.

We are looking for a number (call it X) so that when it is added to the sum of the 9 numbers (call that sum S) is divisible by 7, and when twice that number (2X) is added to S and also to the sum of the 6 larger numbers of the 9 specified (call that sum L) is divisible by 17. We can write that as X+S = 0 modulo 7 and 2X+S+L=0 modulo 17.

We can rewrite those modular equalities as X=-S mod 7 and 2X=-S-L mod 17.

Now we need to use a very neat trick that allows us to effectively divide through by 2 when we work modulo 17. Since 2*9=18=1 modulo 17, multiplying by 9 modulo 17 accomplishes the same thing as dividing by 2. That allows us to write X=9*2*X=-9(S+L) mod 17.

That means we are looking for a number X which is -S modulo 7 and which is -9(S+L) modulo 17.

There is a famous theorem called the Chinese remainder theorem first published in the 3rd to 5th centuries by Chinese mathematician Sun Tzu (according to Wikipedia) . It says that if you have two simultaneous congruences, and the moduli do not share any common factors (in this case that is that 7 and 17 have no factors in common), then you can find a unique solution modulo the product of the moduli (in this case modulo 119). Say we have congruences X=A mod 7 and X=B mod 17. If we find find numbers c and d so that 7c=1 modulo 17 and 17d=1 modulo 7, then setting X=A*(17*d)+B*(7*c), we see that modulo 7, X=A*(17*d)=A*1=A, and modulo 17, X=B*(7*c)=B*1=B. It is also true that adding or subtracting any multiple of 7*17=119 will not change this result.

In our case, it is easy to check that 7*5=35=1+2*17=1 modulo 17, and 17*5=85=1+12*7=1 modulo 7 so we can use c=d=5. (There is a quick algorithm based on the Euclidean algorithm that can be used to find the right values for c and d). To get the congruences we need for our X, we use X=-S(17*5)-9(S+L)(7*5) modulo 119.

I want to simplify this a bit, so I group together all the terms involving S to get X=(-17*5-9*7*5)S-(9*7*5)L modulo 119.

Finally I reduce the coefficients modulo 119 using that -17*5-9*7*5=-400=76-4*119=76 modulo 119, and -9*7*5=-315=42-3*119=42 modulo 119

This gives the formula X=76*S+42*L mod 119 or equivalently X=76*S+42*L-119*n for some whole number n.

It does all seem pretty complicated, but, each step just relies on the properties of modular arithmetic. Most of the ordinary things that you can do with algebraic equations also work for arithmetic modulo some number (called the modulus). Division is the operation that is a bit tricky, but as you saw, we replaced dividing by 2 by multiplying by 9 modulo 17. That kind of trick can always be worked providing the number you want to invert shares no factors with the modulus.

If you prefer, you can just produce the formula and verify directly that it produces the desired results:

We can check that the calculation works directly by adding X to S X+S=76*S+42*L-119*n+S=77*S+42*L-119*n=7(11*S+6*L-17*n) and of course that is divisible by 7. Adding 2X to S+L we get 2X+S+L=154*S+84*L-2*119*n+S+L=153*S+85*L-2*119*n=17(9*S+5*L-14*n) and that is divisible by 17.

Your other constraint for the 10th number is that it should be larger than the smallest 3 numbers from your original 9 in the sum S. (If it were not, then the new number would not need to be added twice, and the calculation would change.) This can always be arranged by adding multiples of 119 to X (i.e. making the value of n smaller in the formula for X).

Regarding the math that the Hebrews or other ancients knew, you may want to consult a math historian to get more depth. Sometimes special cases or more cumbersome variants of algorithms or theorems may have been known, and those may not make it into Wikipedia etc. I would like to ask one of my colleagues about this issue if that is OK with you. I do not need to describe your observations, but, I would need to explain the question of whether the ancients might have been able to solve this sort of divisibility problem.”

The reply was published earlier in the article.

More properties of these lists:

The following observations about these lists were not included as constraints in the formula. The first list is divisible by 7 cubed. As a Hebraic symbol, 7 cubed can also be viewed as 7 + 3 =10. Also 7 cubed = 343, which as a Hebraic symbol also equals 10. The total of the 7 largest numbers in the first list is divisible by 7 as is the total of the 3 smallest numbers. The age of Lamech, Noah’s father, is 777.

The total of the 7 numbers from the bottom up in the second list is divisible by 7.

By coincidence the key number that must be present in the first list, in order for the formula work is 969, which happens to be the age of Methuselah. A key number that must be in the algorithm of both lists is 119, the number of years Nahor lived after the birth of Terah.

We suspect there are more numerical relationships in these numbers. However, the Septuagint changes the numbers, adds 1466 years to the age of the earth, and does not exhibit any numerical properties in the list.