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Pattern 65<: Eight Word Subsets Have Both End-Letter Sums Divisible by 7/h1>
Pattern Statement:

Genesis 1:1 contains seven words, producing 127 possible nonempty word subsets. For every subset, add the Hebrew alphabet positions of the first letters of its selected words. Separately, add the alphabet positions of their last letters.

Eight subsets have both a first-letter sum and a last-letter sum divisible by 7.

Why this pattern matters

The calculation examines every possible nonempty subset under one fixed rule. Eight subsets satisfy two simultaneous conditions: their first-letter sum and their last-letter sum are both divisible by 7.

How this pattern is measured

Number the seven words from 1 through 7. Generate all 2 7 −1=127 nonempty subsets, with each word used no more than once. For each subset:

1. Add the Hebrew alphabet positions of the selected words’ first letters.
2. Add the alphabet positions of their last letters.
3. Count the subset only when both totals are divisible by 7.
Word order is preserved, but it does not affect the sums.

| Word | First-letter position | Last-letter position | | ---: | --------------------: | -------------------: | | 1 | 2 | 22 | | 2 | 2 | 1 | | 3 | 1 | 13 | | 4 | 1 | 22 | | 5 | 5 | 13 | | 6 | 6 | 22 | | 7 | 5 | 18 |

| Selected words | First-letter sum | Last-letter sum | | ---------------- | ---------------: | --------------: | | 1, 5 | 7 | 35 | | 2, 5 | 7 | 14 | | 3, 6 | 7 | 35 | | 1, 3, 5, 6 | 14 | 70 | | 1, 4, 6, 7 | 14 | 84 | | 2, 3, 5, 6 | 14 | 49 | | 2, 4, 6, 7 | 14 | 63 | | 1, 2, 4, 5, 6, 7 | 21 | 98 |

Result for Genesis 1:1

8 of the 127 nonempty word subsets satisfy both conditions

First-letter sum ≡ 0 (mod 7)
Last-letter sum ≡ 0 (mod 7)

21 subsets have a first-letter sum divisible by 7.
20 subsets have a last-letter sum divisible by 7.
8 subsets satisfy both conditions.

Read detail page for Pattern 65

Probability note

(1 in 1e2[141]).
1 in 10^141 probability that all patterns so far occurred by chance.

If the first- and last-letter sums behaved like independent uniformly distributed residues modulo 7, a particular subset would satisfy both conditions with probability 1/49. Across 127 subsets, the expected count would therefore be approximately:

127/49=2.59.

The observed count is 8.

Probability note: Under a simple uniform model, approximately 2.59 of the 127 subsets would be expected to satisfy both divisibility conditions; Genesis 1:1 has eight. Because the subsets reuse the same seven words and are not statistically independent, the binomial model is not appropriate. A Hebrew control-corpus test is needed for a reliable probability estimate.

This page uses the probability notation provided in the original pattern sequence.

Supporting tables

First and last letter places are highlighted.

Genesis 1:1
Hebrew WordNumber of Letters
בראשית6
ברא3
אלהימ5
את2
השםימ5
ואת3
הארצ4
Total Words: 7Total Letters: 28
Traditional Hebrew Letter ValuesWord Total
2 200 1 300 10 400 913
2 200 1 203
1 30 5 10 40 86
1 400 401
5 300 40 10 40 395
6 1 400 407
5 1 200 90 296
Verse Value Total2701
Hebrew Alphabet PositionsWord Total
2 20 1 21 10 22 76
2 20 1 23
1 12 5 10 13 41
1 22 23
5 21 13 10 13 62
6 1 22 29
5 1 20 18 44
Verse Position Total298
This pattern is one part of the larger theory and should be evaluated together with the full set of patterns, rules, and control tests.
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