### Bible Wheel Book

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1 0 5 0{R.A.M.) occurs first starting at digit # 6 8 1

6 8 1 + 2 0 2 0 = 2 7 0 1{Hebrew sum of the words in Gen.1:1)

Count the number of the beast(=1680)) + 6 6 6{that complete's at digit # 2 4 4 2) = 2 3 4 6

2 3 4 6 occurs first at starting at digit # 2 6 0

Digit # 2 5 9 + 2 4 4 2 = digit # 2 7 0 1

6 6 6 + 2 1 6 = 8 8 2

The first occurrence of 8 8 2 starts at digit # 3 7 2

Digit # 2 7 0 1 is 3 7 2 digits after digit # 2 3 2 9,

and digit # 2 3 2 9 is the completion of the first occurrence of 1 5 6.

3 .(1 4 . . . . 7 3 . . . )(8 6 . . . . 1 5 9 0 + 6 6 6 . . . . 1 5 6} digit # 2 3 2 9

digit # 2 3 2 9 + 3 7 2 digits = digit # 2 7 0 1

7 3 + [ 3 6 x 7 3 ] = 2 7 0 1
Last edited by Scott; 07-15-2020 at 10:57 PM.

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2 0 2 0 - 1 3 1 8 = 7 0 2 - 6 6 6 = 3 6 { 1 + 2 + 3 . . . . + 3 6 = 6 6 6 )

1 3 1 8 + 6 6 6 = [ 1 9 4 8 + 3 6 ]

Israel was reborn as a nation at the time God decided, in the year 1 9 4 8.

1 9 4 8 doesn't occur till almost the 25000 digit mark,

but 1 9 4 7, the nearest completed year, completes it's first

occurrence at digit # 2 0 5 7{3 7 + 2 0 2 0),

which is, 7 3 + 1,3 1 8 + 6 6 6 digits.

3 .(1 4 . . . . 7 3 . . . )(8 6 . . . . 1 3 1 8 + 6 6 6 . . . .}digit # 2 0 5 7

4 9 0{ 7 0 x 7 ) occurs first right after digit # 9 0 6 {p.17, #164),

and completes at digit # 9 0 9{post # 209)

In 1 9 4 8 the record number of digits known was 2 0 3 7{ 1 1 + 7 + 2 0 1 9 )

They didn't know that 4 0 5 digits later 6 6 6 completes its first occurrence at digit # 2 4 4 2
Last edited by Scott; 07-18-2020 at 11:02 AM.

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November Eighth Two Thousand Sixteen = 2 4 4 2 = [ 6 6 6 + 1 1 1 0 + 6 6 6 ]

Donald John Trump = 1 1 1 0

The first occurrence of this, 2 3 6 8 1 1 1 0, is followed by,

2 3 6 8 1 1 1 0 2 0 1 7

His Inauguration was 2 0 1 7

I consider these obvious "designs",....to be warning's to us about him!

He was born on 6 / 1 4 of 1 9 4 6

6 6 6 + 6 1 4 + 6 6 6 = 1 9 4 6

If we place a span of 2 4 4 2 days finishing on 1 1 / 0 7 / 2 0 2 0

then 5 / 1 4 / 2 0 1 8 would be day 1 5 3 5.

In Pi, digit # 1 5 3 5 is the completion of the first occurrence of 9 1 1
Last edited by Scott; 07-18-2020 at 05:47 PM.

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The first occurrences of 2 1 6 and 6 6 6,

cover a span of,

5 4 1{Israel) + 9 1 1 digits.

2 1 6 starts right after digit # 9 9 0

6 6 6 ends at digit # 2 4 4 2

5 4 1 + 7 7 7 = 1 3 1 8

2 1 6 completes at digit # [ 7 7 7 + 2 1 6 ]

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

9 1 1 + 6 6 6 = 1 5 7 7

The first occurrence of 1 5 7 7 starts right after digit # 3 2 7 9

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

digit # 4 8 6 9 is where number 1 5 9 0 completes its first occurrence.

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

1 5 7 7 completes its first occurrence at digit # [ 1 5 6 + 1 1 0 7 + 2 0 2 0 ]

So it's 2 3 6 8 + 9 1 1 digits to reach the start of the first occurrence of

1 5 7 7, which is 9 1 1 + 6 6 6, and then 1 5 9 0

digits from there to reach the completion of the first occurrence of

the number 1 5 9 0 at digit # 4 8 6 9

2 3 6 8 + 9 1 1{6 6 6 + 9 1 1 begins) + 1 5 9 0 = 4 8 6 9

I was completely unaware of most of what i've posted regarding 9 1 1

when i started this thread and titled it 9 1 1,....God knew what he was about to reveal to me!!
Last edited by Scott; 07-21-2020 at 10:39 AM.

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Revelation chapter Thirteen verse Eighteen = [ 1 5 6 + 2 4 4 2 ]

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

<><><><><><><><><><><><><><><><><>

Richard Amiel McGough = 1 0 5 0

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

If i take Jesus Christ from the center of the span of digits our 2 names cover,

it leaves 2 spans of 9 1 1 at each end, one that begins with the gematria sum of his name,

and one that ends with mine,

3 . 1 4 . . . . . . . .(1 0 5 0 . . . . 9 1 1 + 2 3 6 8 + 9 1 1 . . . . 7 5 1 5 9 0 6)
7 5 6, post # 204

3 . 1 4 . . . . . . . .(1 0 5 0 . . . . 9 1 1 + 9 1 1 . . . .}digit # 1 5 9 0 + 9 1 1 + 1
Last edited by Scott; 07-23-2020 at 10:29 AM.

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(1 0 5 0 . . . . . 9 1 0 . . . . . } digit # 1 5 9 0

Revelation chapter thirteen verse eighteen = 2 5 9 8

1(0 5 0 . . . . . . . . . . . 2 5 9 8 + 1 5 9 0 . . . . . . . . . 7 5 1 5 9 0)6

<><><><><><><><><><><><><><>

As i was turning 1 0 3 1 2 days old, at about 3:00 am on March 27th 1991,

i discovered something about the 3 1 1th day of the year, Nov.7th, which is day 3 1 2

this year because its a leap year, but 11/7 begins after 311 days, and i also discovered

the 6 6 6, 2 1 6, and 4 5 0

design in the number set, and, that only 1 number greater than Enoch's Rapture year, 9 8 7,

has the same features as 9 8 7, and that number is 1 9 9 5

I went to bed when the sun was coming up, thinking the Rapture could possibly be Nov. 7th 1 9 9 5,

now more than 2 9 years after that night, i looked for 1 0 3 1 2 and this is what followed,

1 0 3 1 2 0 3 1 1 9 9 5

The 5 of 1 9 9 5 is digit # 2 3 8 9 6 8

My son's name adds to 1 3 8 0, this is its first occurrence, and the first occurrence of 8 6,

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

<><><><><><><><><><><><><><><>

January First Nineteen Sixty Three = 2 4 0 6

November Seventh Twenty Twenty = 2 4 0 6

I counted my birth date in digits, from the completion of the first occurrence of my name 1 5 9 0,

and it brought me to the first occurrence of 1 5 5 9 0

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

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

That's the first and second occurrence of 2 0 2 0
Last edited by Scott; 07-24-2020 at 10:00 AM.

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The number of digits leading up to that first occurrence 1 5 9 0 thats nestled into

a 7 5 6, is, 4 8 6 3 digits.

1 4 x 4 8 6 3 + 3 0 7 2 digits,

brings us to the last digit of the first occurrence of 1 5 9 0 0 ,

1 5 9 0 0 is followed by 6 2 4 2 0 2 0

6 / 2 4 / 2 0 2 0

June Twenty Fourth Two Thousand Twenty = 3 0 7 2

3 .(1 4 . . . . 4 8 6 3 . . . )(7 5 1 5 9 0 . . . 1 4 x 4 8 6 3 + 3 0 7 2 . . . . 1 5 9 0 0)6 2 4 2 0 2 0

8. Senior Member
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Jesus Christ = 7 5 4 / Hebrew

Jesus Christ = 2 3 6 8 / Greek

Jesus Christ = 9 0 6 / English

Digit # 7 5 4 + 2 3 6 8 + 9 0 7 digits = [ 2 4 3 9 + 1 5 9 0 digits ]

7 5 4, 2 3 6 8, p.20 post 192
9 0 7 p.21 post # 202

The first occurrence of 6 6 6 begins right after digit # 2 4 3 9

and the first occurrence of 6 6 6 is what the 1 5 9 0 digits begin with.

1 5 9 0 completes its first occurrence at digit # 4 8 6 9

Digit # 1 5 9 0 is the second 7, in what is both the first occurrence

of 7 7 7, and 7 7 7 7

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

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

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

7 + 7 + 7 x 3 6 + 7 + 7 + 7 = 7 7 7

4 x 8 x 6 x 9 = [ 1 5 6 + 9 0 6 + 6 6 6 ]

3 . 1 4 1 5 9

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

3 1 4 1 - [ 1 5 6 + 9 0 6 + 6 6 6 ] = 1 4 1 3
Last edited by Scott; 08-01-2020 at 04:47 PM.

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The number 2 1 6 occurs first in Pi from digit # 9 9 1 to # 9 9 3

Digit # 7 7 7 + 2 1 6 digits = digit # 9 9 3

The 2 1 6 digits end with the first occurrence of the number 2 1 6

Digit # 7 7 7 + 1 5 9 0 digits{that begin with 216) = 2 3 6 7 digits.

The 4 digit number occurring for the first time starting at digit # 2 3 6 8 is, 3 5 3 4

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

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