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Repeating hexadecimal digits

Here we will display factor tables of hexadecimal numbers with repeating digits.

We will look at numbers having patterns of the forms   0AAA...AABh   and   0ABB...BBBh   where A and B are hexadecimal digits greater than zero and having no factor in common.

For example, we exclude numbers like 0222...224h and 0222...226h etc. which are obviously just multiples of the sequences 0111...112h and 0111...113h.

Some examples

0111...113h(16n + 29) / 15
0111...115h(16n + 59) / 15
0111...117h(16n + 89) / 15
....
0133...333h(18 * 16n-1 - 3) / 15 ⇒ (6 • 16n-1 - 1) / 5
0155...555h(20 * 16n-1 - 5) / 15 ⇒ (4 • 16n-1 - 1) / 3 ⇒ (24n-2 - 1 ) / 3 ⇒ (22n-1 - 1) • (22n-1 + 1) / 3
0177...777h(22 * 16n-1 - 7) / 15
....

Repeating decimal digits

On the net, there are interesting tables of factors of decimal numbers consisting of various patterns of repeated digits.

On Makoto Kamada's site for example, there are tables of factors of numbers like 21, 221, 2221, 22221, etc. (up to several hundred digits long).

In fact, all combinations of decimal digit patterns of the form AAA...AAB and ABB...BBB are listed (plus some others as well).

Repeating hexadecimal digits

Here we will instead factor hexadecimal numbers with similar repeating digits. See links to the tables, below.

A little algebra

We can convert numbers of the form 0AAA...AABh into decimal easily, so that modern integer factoring algorithms like the Number Field Sieve can be used.

Assume that A and B are hexadecimal digits and that N = 0AAA...AABh where there are n-1 'A's.

  1. N = 16 • (0AAA...AAAh) + B
  2. N = 16 • A • (0111...111h) + B
  3. 15 • N = 16 • A • (0FFF...FFFh) + 15 • B
  4. 15 • N = 16 • A • (16n - 1 - 1) + 15 • B
  5. N = (A • 16n + (15 • B - 16 • A)) / 15

Similarly for numbers of the form 0ABB...BBBh, we have:

  1. N = ((15 • A + B) • 16n-1 - B) / 15

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© 2014 Bob Backstrom — Updated: Mon Sep 11, 2017.