9 1 1

[Scott]

1 3 + 1 8 + 2 6 + 5 7

1 3 1 8 + 2 6 + [ 2 6 x 2 6 ] = 2 0 2 0


3 5 x 5 7 + 2 5 = 2 0 2 0

Nov.7th 1 9 9 5 + 2 5 yrs, is, Nov.7th 2 0 2 0,...the 2 6 th Nov.7th

1 3 1 8 + [ 2 6 x 2 6 ] = 1 9 9 4{ + 2 6 = 2 0 2 0


1 + 2 + 3 + 4 . . . . . . + 3 6 = 6 6 6

1 3 1 8 + 3 6 + 6 6 6 = 2 0 2 0

Post #91 top segment



3 .(1 4 . . . . 9 9 4 digits . . . . 2 1 6 4)(2 0 1 9 . . . 6 + 6 + 6 x 5 7 digits . . . .}digit # 2 0 2 0

The first Rapture was Enoch,... 9 8 7 years after Adam's creation.

[ 2 6 x 2 6 ] + 3 1 1 = 9 8 7

Nov.7th is day 3 1 1 of the year, except in a leap year, Nov.6th is the first 3 1 1 days of the
year completed, and Nov.7th begins.
__________________________________________________ ____________________________

3 . 1 4 . . . . . . (8 6 . . . . . . . 2 3 6 8 digits . . . . . 6 6)6

When you see this string of numbers below, occurring for the first time in Pi,
they finish at digit # 1, 4 6 9, 6 6 6

2 3 6 8 8 6

2 3 6 8 - [ 8 6 + 6 6 6 ] = [ 1 5 9 0 + 2 6 ]


[ 6 x 2 6] + [ 6 x 2 6 ] + [ 6 x 2 6 ] + 1 5 9 0 + 6 6 6

= 2 7 2 4

November Seventh Two Thousand Twenty

= 2 7 2 4

[Scott]

Count the number of the beast + 6 6 6 = 2 3 4 6

1 5 9 0 + 7 5 6 = 2 3 4 6

1 5 9 0 + 6 6 6 = 2 2 5 6



3 .(1 4 . . . . . 2 6 1 digits. . . . . . 2)(3 4 6 . . . . 1 9 9 5 digits . . . . . . }digit # 2 2 5 6


There are 1 9 9 5 digits{the match to Enoch's 9 8 7), from the first occurrence
of 2 3 4 6, to reach a number of digits after the decimal point that will accomodate my name,
1 5 9 0, and the number 6 6 6!


Enoch was Raptured 9 8 7 yrs after Adam and Eve's creation
This number divides evenly by 7, and has digits that add to, [ 8 + 8 + 8 ] ,
and multiply to an answer consisting of a 4, 5, and 0, which are
the numbers in the set{1234567890) that aren't utilized until 6 6 6 is counted
and its answer subtracted from it. { Post # 62, and 71, pages 7, 8 )

There are, i discovered, only 2 other numbers that have those 3 features,


7 9 8
9 8 7
1 9 9 5


7 x 9 x 8 = 5 0 4
7 9 8 - 5 0 4 = 2 9 4


9 x 8 x 7 = 5 0 4
9 8 7 - 5 0 4 = 4 8 3{69x7)


2 9 4 + 4 8 3 = 7 7 7


1 x 9 x 9 x 5 = 4 0 5
1 9 9 5 - 4 0 5 = 1 5 9 0


1 5 9 0 + 7 7 7 = 2 3 6 7
1 5 9 0 + 7 7 8 = 2 3 6 8

There are 7 7 7 numbers standing between the English sum of my name,
and the Greek sum of Jesus Christ
or,

7 7 7 < 1 5 9 0 > 2 3 6 8

Scripture refer's to Jesus Christ as being,

"The Bread of God" = 1 9 9 5 in Greek{John 6:32)

The FLOOD ended 1 6 5 7 years from Adam and Eve's creation,

1 6 5 7 + 2 3 6 8 yrs + 1 9 9 5 yrs = NOW

That's the new mankind's timeline,....within the old mankind's timeline,
that began with the first Adam bringing death upon us!

Post #'s 2, 46, 61, and 75

[Scott]

The differences between the 3 numbers, Enoch's being the middle one, and their counted results,
have a combined total of 2 3 6 7.
At the time when i discovered only 1 number larger than 9 8 7, had the same 3 features as 9 8 7,
and its 4 0 5 result subtracted from itself like i did with 6 6 6{6 6 6 - 2 1 6 = 4 5 0),
produced 1 5 9 0, i knew nothing about English gematria, and had no idea
that 1 5 9 0 was the sum of my name, but when i did find out 1 8 yrs later, i did this,

1 9 6 3 + 1 9 9 5 = [ 1 5 9 0 + 2 3 6 8 ]

==================================================

The "Revelation 1 2 Sign" that John wrote about in the 1 1 7 9 th chapter

of The Bible, is believed by many to have taken place on 9 / 2 3 / 2 0 1 7

"September Twenty Third Two Thousand Seventeen" = 9 + 2 3 + 2 0 1 7 + 1 1 7 9

A=6, B=12, C=18, D=24, . . . . . . Z=1 5 6


The Book that is all about who Jesus Christ is, and why he came, contains

1 1 8 9 chapters

2 3 6 8 - 1 1 8 9 = 1 1 7 9

(Chapter}1 1 7 9 + 1 1 8 9{Chapter) = 2 3 6 8

Thats a special zone of The Bible, those last eleven chapters,....we've entered that zone!

==================================================

6 6 6 first occurs starting at digit # 2 4 4 0

2 1 6 first occurs starting after digit # 9 9 0

9 9 0 + 2 4 4 0 or 2 4 4 0 + 9 9 0 = 3 4 3 0

7 x 7 x 7 x 1 0 = 3 4 3 0

So if we count backwards 9 9 0 digits (2 1 6 occurs at 9 9 1 going forward) from digit # 3 4 3 0,
we are in a sense, meshing 2 1 6 with 6 6 6,

6 2 1 6 6 6

Those last two 6's of 6 6 6 would be occupying the same position as the 1 and 6 of 2 1 6



Now heres the number that occurs at the 6 6 6 th digit,

3 .(1 4 . . . . . . . . . . . . 6 6 6 digits . . . . . . . . . . 3)4 3 0{ 1st occurrence of 3 4 3 0

Both 3 4 3 and 3 4 3 0 first occur at digit # 6 6 6




6 6 6 + 2 1 6 + 1 6 6 6 + 6 6 6 + 2 1 6 = 3 4 3 0

1 6 6 6 + 3 6 + 6 6 6 = 2 3 6 8

1 5 6 + [ 9 0 6 + 2 3 6 8 ] = 3 4 3 0

[ 1 5 6 + 8 6 ] + 9 0 6 + [ 1 5 9 0 + 6 6 6 + 2 6 ] = 3 4 3 0

With digit # 6 6 6 being digit 1, to the end of this string, 7 5 1 5 9 0 6),

is 4 2 0 1 digits

This is what you see at 2 1 6's first occurrence, that completes at digit # 9 9 3

3 . 1 4 . . . . . 2 1 6 4 2 0 1 9 8 9



3 . 1 4 . . . . . 2 1 6( . . . . 2 0 2 0 digits . . . . 0 9 5 1){my name in reverse


The first occurrence of 2 1 6, to digit # 2 3 6 8,
covers this many digits,

[ 2 6 x 2 6 ] + 2 6 + [ 2 6 x 2 6 ]



3 .(1 4 . . . . . . . . . 4 8 7 0 digits. . . . . . . . . 7 5 1 5 9 0 6)

2 5 0 2 + 2 3 6 8 = 4 8 7 0

"You have discovered the date of The Rapture" = 2 5 0 2

[Scott]

6 + 6 + 6 yrs, and nineteen 9 1 1's

have passed since, and including, THE 9 1 1

This Sept.11th 2 0 2 0, will be 9 + 1 1,....9 1 1's, including THE 9 1 1


Because its a leap year this year, 1 1 / 0 7 takes place on the day of the year that is normally 1 1 / 0 8
for 3 out of every 4 years, and this is what separates the number 1 1 0 8 from 2 0 2 0

1 1 0 8 < 9 1 1 > 2 0 2 0

7 8 9 - 1 2 3 = 6 6 6

7 8 9 + 1 2 3 = 9 1 2

[Scott]

In Hebrew these words,

"The Beginning" = 9 1 1

There are 5 7 days from the end of 9 / 1 1 to the end of 1 1 / 0 7


My name, 1 5 9 0, begins to occur for the first time at digit # 4 8 6 6

and completes at digit # 4 8 6 9 after the decimal point,

3 .(1 4 . . . . . . . 2 3 6 8 + 1 5 9 0 + 9 1 1 . . . . . . 1 5 9 0)

3 .(1 4 . . . . . . . . 3 1 4 + 5 x 9 1 1 . . . . . . . . 1 5 9 0)

1 6 x 5 7 = 9 1 2


3 . 1 4 1 5 9 2 6
Three point One Four One Five Nine Two Six = [ 1 5 9 0 + November Seventh{1122) ]

3 .(1 4 . . . . . . . . . 1 5 x 3 1 4 + 1 5 9 . . . . . . . . . 1 5 9 0)


The Twenty Sixth November Seventh = 2 4 4 2

6 6 6 completes its first occurrence at digit # 2 4 4 2


Heres the first occurrence of 3 1 4,

3 . 1 4 . . . . . . . . . . . . . . . . . . (3 1 4 . . . . . . . . . . . 2 6 + 2 7 2 4 . . . . . . . 1 5 9 0)

The 2 6 th Nov.7 th is "November Seventh Two Thousand Twenty" = 2 7 2 4


1 7 6 + 3 1 4 + 1 7 6 = 6 6 6

Gen.7 :1 7 and Rev. 1 3 : 1 8,
1 7 6 verses Forward, and 1 7 6 verses Backwards

[Scott]

1 5 x 9 0 + 1 7 6 + 6 6 6 + 1 7 6 = 2 3 6 8

2 3 x 6 8 + 2 6 = 1 5 9 0


3 6 x 2 6 + [ 2 6 + 1 3 8 0 + 2 6 ] = 2 3 6 8

My son's name is 1 3 8 0

1 3 8 0< 9 8 7 >2 3 6 8

2 3 6 8

2 8 x 3 6 = 9 8 8


9 8 8 + 1 3 8 0 = 2 3 6 8

========================================

1 5 9 0 + 6 6 6 + [ 6 + 6 + 6 x 2 6 ] = 2 7 2 4

I'll be 5 8 years from conception this Spring,

5 8 + 1 5 9 0 + 1 0 1 8 + 5 8 = 2 7 2 4

1 5 x 9 0 + 1 0 1 8 = 2 3 6 8

2 6 + 2 + 6 + Twenty Six + Two + Six = [ 1 5 9 0 + 5 8 ]

========================================

The morning of Sept.1 1 + 5 7 days, is the morning of Nov.7 ,

which is 3 1 1 days of the year completed, because of it being a leap year

this year,

Adding 5 7 to 2 3 1 1 = 2 3 6 8

1 3 + 1 8 + 2 6 = 5 7

"The Last Fifty Seven days" = 1 5 9 0

2 6 x 2 6 + 3 1 1 = 9 8 7, Enochs Rapture year

[Scott]

It was day 8 6 of year 1 9 9 1 when God had me

discover the 6 6 6, 2 1 6, and 4 5 0 design in the number set,

and discover the only number larger than 9 8 7(Enoch), that matches 9 8 7, which is, 1 9 9 5

Here is the measurement of digits from the first occurrence of 6 6 6 to

the first occurrence of 1 3 1 8,


3 . 1 4 . . . . . . . . . . 6 6 6( . . . . . . 6 6 6 + 2 1 6 + 1 9 9 1 digits . . . . . . 1 3 1 8)1 4 8 0

3 . 1 4 . . . . . . . . . .(6 6 6 . . . . . . 6 6 6 + [ 6 6 6 + 2 1 6 ] + 6 6 6 . . . . . 1 3 1 8 1 4 8 0)
Page 8, Post # 75{1480)


1 2 3 - 4 5 - 6 - 7 8 9 - 0

3 - 2 - 1 - 3 ^

1 2 3 + 6 6 6 = 7 8 9

6 x 6 x 6 = 2 1 6

6 6 6 - 2 1 6 = 4 5 0

123 456 789 0

"The 6 6 6 and 4 5 0 code" = 1 5 9 0 {Post # 94)


The first 8 6 digits of Pi, + 1 1 0 7 + 2 0 2 0 = digit # 3 2 1 3


The first occurrence of 1 3 1 8{Rev.13:18) is followed by the second occurrence

of the Greek sum of "CHRIST"{Savior)= 1 4 8 0

which completes 6 6 6, 2 1 6, and 1 9 9 5 digits

from the completion of the first occurrence of 6 6 6


4 5 0 + 2 1 6 = 6 6 6

3 . 1 4 . . . . . . . . . . (1 5 9 0 . . . . . . 4 5 0 digits . . . . . . . 1 3 1 8)

6 6 6 digits beginning with the four digits that are the first occurrence of my name,

are dividing at 4 5 0 and 2 1 6 right at the completion of the first

occurrence of 1 3 1 8

[Scott]

The English sum of "Jesus Christ" = 9 0 6, it first occurs at digit # 1 2 9 3

The Greek sum of "Jesus Christ" = 2 3 6 8, it first occurs at digit # 7 4 8 1


The following string of numbers that i've explained in several posts, completes

its first occurrence at digit # 4 8 7 0,


3 .(1 4 . . . . . . . . . . . . . . . 7 5 1 5 9 0 6)

From the first digit of 9 0 6, to the first digit of 2 3 6 8,

3 . 1 4 . . . . . . . . . . . . . . . 9(0 6 . . . . . . . . 1 3 1 8 + 4 8 7 0 . . . . . . . . . 2)3 6 8


9( . . . . . . . )3 6 8

1 5 9 0 + 1 3 1 8 + 4 8 7 0 + 1 5 9 0 = 9 3 6 8

I was born Jan.1st 1 9 6 3, 1 9 6 2 completed.

1 9 6 2 + 1 5 9 0 + 1 3 1 8 = 4 8 7 0




The English sum of "God" = 1 5 6 completes its first occurrence at digit # 2 3 2 9

The Hebrew sum of "Elohim" = 8 6 begins to first occur at digit # 7 4


3 . 1 4 . . . . . . . . (8 6 . . . . . . . . 1 5 9 0 + 6 6 6 . . . . . . . 1 5 6)

==================================================

January First Nineteen Sixty Three = 2 4 0 6 (+ 1 5 9 0 = 3 9 9 6

3 9 9 6 begins its first occurrence at digit 2 1 1 3 to digit 2 1 1 6

3 . 1 4 . . . . . . . . . . . . . . . 3 9 9 6(. . . . . . . . 2 6 + 2 7 2 4 . . . . . . . . . . 1)5 9 0

I will be 2 1 1 3 0 days old at 3:06 am Nov. 7th 2020

==================================================

3 . 1 4 1 5 9 2 6 . . . . . . . (8 6 . . . . . . . 2 3 6 8 . . . . . . . 6 6)6

That's the first occurrence of both 8 6 and 6 6 6

[Scott]

The Beast = 4 8 0

4 8 0 + 1 5 9 + 2 6 = 6 6 5

4 8 0 < 1 5 9 + 2 6 > 6 6 6

I think....that if we are approximately in the middle of the final 7 yrs
in man's timeline, then we must be approximately
at the end of the final 7 yrs in God's accurate timeline,
which means the 2nd half of The Tribulation is about to
begin with The Rapture of The Body and Bride of Jesus Christ!

Even though God wanted us to keep a timeline, he knew it wouldn't be accurate,
but he's working with it, along with the one he's keeping!

===========================================

In Greek, "Jesus" = 8 8 8
and "Christ" = 1 4 8 0

In Pi, 1 4 8 0 occurs first, starting at digit # 1 0 3
and 8 8 8 finishes occuring at digit # 4 7 5 3

3 . 1 4 . . . . . . . . . . 1(4 8 0 . . . . . . . 1 5 9 0 + 3 0 6 0 . . . . . . . 8 8 8)

"Saturday November Seventh Twenty Twenty" = 3 0 6 0


===========================================


Here are the 3 special numbers again, from post # 1 0 2

7 9 8 + 9 8 7 + 1 9 9 5 = 3 7 8 0

Heres the order they occur in Pi,


3 . 1 4 . . . . . . . 7 9 8 . . . . . . . . . 2 0 1 9 9 5 . . . . . . . . . . 9 8 7

All 3 have completed their first occurrences by digit # 1 5 1 2

Apocalypse = 1 5 1 2, is the word in the Greek N.T. with the largest gematria sum.

3 . 1 4 . . . . . . . 7(9 8 . . . . . . . . . 1 4 1 2 digits . . . . . . . 9 8)7
or,
3 . 1 4 . . . . . . . (7 9 8 . . . . . . . . . 1 4 1 2 digits . . . . . . . 9)8 7


3 .(1 4 ...........................................1 4 1 2 digits + 2 3 6 8 digits } digit # 3 7 8 0

( 2 3 6 8, post # 1 0 2 )
============================================


3 . 1 4 . . . . . . . 6 6 6( . . . . 2 4 2 4 . . . . . 1)(5 9 0 . . . . . 2 4 2 4 . . . . . 2 0 2 0 2 0)

Thats the first occurrence of 6 6 6 and 1 5 9 0,

and the first and second occurrence of 2 0 2 0 meshed together,...."Emphasis" on the year!

( 2 0 2 0, post # 73 )( 2 4 2 4. post # 70 )


============================================


November Seventh Two Thousand Twenty = 2 7 2 4

The second occurrence of 2 0 1 9 is meshed with the first occurrence of 1 9 9 5

2 0 1 9 9 5

3 .(1 4 . . . . . . . 7 0 5 digits . . . . . . 2 0 1 9)(9 5 . . . . . . . . 2 0 1 9 digits . . . . . . . }digit # 2 7 2 4

God clearly had this equation in mind, when deciding the location that 2 0 1 9 9 5 in the digits of Pi would occupy,

The number 2 7 2 4 - The number 2 0 1 9 = 7 0 5
and digit 7 0 5 in Pi, (when backing up 2 0 1 9 from digit 2 7 2 4) is the 9
in the second occurrence of 2 0 1 9


3 .(1 4 . . . . . . . . 7 0 4 . . . . . . . . . . 2 0 1)(9 9 5 . . . . . . . . . 2 0 2 0 . . . . . . . . . }digit # 2 7 2 4

[Scott]

3 .(1 4 . . . . . . . 3 6 + 6 6 6 digits . . . . . . . 2)(0 1 9 9 5 . . . . . . . 1 3 1 8 digits . . . .}digit # 2 0 2 0

That's the 2nd occurrence of 2 0 1 9

==============================================

9 1 1 first occurs starting at digit # 1 5 3 3


3 . 1 4 . . . . . . . . . . . . . . . . . . (9 1 1 . . . . . . . . . 9 1 0 digits . . . . . . . . . . 6 6 6)

That's the first occurrence of 9 1 1 and 6 6 6

A 9 precedes 9 1 1,

3 . 1 4 . . . . . . . . . . . . . . . . . (9 9 1 1 . . . . . . . . . 9 1 1 digits . . . . . . . . . . 6 6 6)


6 6 6 completes its first occurrence at digit # 2 4 4 2, which is Position 2 4 4 4 when the 3 and

Decimal point are included, and position 2 4 4 4 minus 9 1 1 positions, is digit # 1 5 3 3,

which is the 9, in the first occurrence of 9 1 1