Composites of two digits.

[Ed Pegg]

Here's the start of a sequence.  
111199 = 11*11*919  
119911 = 11*11*991  

Composite numbers using only two digits with a prime factorization using the same two digits.  I was only able to find these two values.  Are there any more?

For composites of three digits, solutions are much more common.  Here's a range.  

11111255 = 5 2222251
11111337 = 3^4 137177
11133551 = 11 13^2 53 113
11199661 = 11 61 16691
11255555 = 5 2251111
11339919 = 3^4 139999
12113312 = 2^5 31 12211
12215522 = 2 11 555251
13317171 = 3 7^2 17 73^2
13731373 = 73 137 1373
13771377 = 3^4 17 73 137
13773771 = 3^2 11 373^2
14441471 = 11^2 41^2 71
15157175 = 5^2 11 55117
15995195 = 5 59^2 919
17331733 = 73 137 1733
17737137 = 3^4 11 17 1171
17772722 = 2 11^2 271^2
19199719 = 7^2 11 179 199
19933199 = 11 13 139393
21229219 = 11 1929929
21911219 = 11 1991929
22272372 = 2^2 3^2 23 37 727
22717712 = 2^4 17^5
23313312 = 2^5 3^3 11^2 223
28188182 = 2 11 1281281
31111119 = 3^2 13 139 1913
31111717 = 7^2 13^3 17^2
31115513 = 11^2 13 131 151
31135131 = 3^3 1153153
31313711 = 11^2 13 17 1171
31373137 = 73 137 3137
32233327 = 7^2 23 37 773
32999999 = 29^2 39239
33227222 = 2 7 2373373
33353515 = 5 13 513131
33362236 = 2^2 23 362633
33713371 = 73 137 3371
33777173 = 73 337 1373
37137737 = 7^2 13 173 337
37711377 = 3^2 11 113 3371
44144313 = 3 431 34141
44343477 = 3^3 347 4733
44444413 = 13 43^4
51333315 = 3 5 11 311111
55151195 = 5 11^2 91159
69766669 = 7 9966667
69976669 = 7 9996667
69997669 = 7 9999667
72332332 = 2^2 23 337 2333
73111113 = 3^3 31 113 773
73317331 = 73 137 7331
73717371 = 3^4 7 13 73 137
77137137 = 3^3 7 11^2 3373
77173333 = 137 317 1777
77173337 = 73 337 3137
77879899 = 7877 9887
79333779 = 3 7 3777799
79337937 = 3 7 3777997
91889891 = 919 99989
91944941 = 11 419 19949
96967699 = 97 999667
98118889 = 11 8919899
98881189 = 11 8989199
99392339 = 293 339223
99779729 = 7^3 97 2999

[Gareth McCaughan]

On 12/01/2026 00:00, Ed Pegg wrote:
> Here's the start of a sequence.
> 111199 = 11*11*919
> 119911 = 11*11*991
>
> Composite numbers using only two digits with a prime factorization
> using the same two digits.  I was only able to find these two values. 
> Are there any more?

110111111 = 11*10010101.

No others with up to 9 digits. My code for checking these is
super-stupid and inefficient so I haven't gone further.

Handwaving: there are exponentially many n-digit numbers all of whose
base-100 digits are 0 or 1, and of order 1/n of numbers that big are
prime, and all of these will pair up with 11 to make the above happen.
So I expect Ed's sequence to be infinite.

(Perhaps these numbers are less likely to be prime than random n-digit
numbers. But I wouldn't expect them to be _enough_ less likely to be
prime, for large n, to invalidate the handwaving. Butbut I am not an
expert on this stuff.)

--
g

[David desJardins]

On Sun, Jan 11, 2026 at 4:00 PM Ed Pegg <edp...@gmail.com> wrote:

Here's the start of a sequence.  
111199 = 11*11*919  
119911 = 11*11*991  

Composite numbers using only two digits with a prime factorization using the same two digits.  I was only able to find these two values.  Are there any more?

Depending on the exact definition, some or all of these might qualify:

4 = 2 * 2

8 = 2 * 2 * 2

9 = 3 * 3

22 = 2 * 11

25 = 5 * 5

32 = 2 * 2 * 2 * 2 * 2

33 = 3 * 11

55 = 5 * 11

77 = 7 * 11

121 = 11 * 11

333 = 3 * 3 * 37

777 = 3 * 7 * 37

1111 = 11 * 101

1331 = 11 * 11 * 11

Then I also found these:

110111111 = 11 * 10010101

1111011011 = 11 * 101001001

11000111111 = 11 * 1000010101

110011011011 = 11 * 10001001001

110111101111 = 11 * 10010100101

111100011011 = 11 * 10100001001

111101010101 = 101 * 1100010001

111101111011 = 11 * 10100101001

(12 13-digit solutions using {0,1})

(18 14-digit solutions using {0,1})

(17 15-digit solutions using {0,1})

(33 16-digit solutions using {0,1})

As Gareth says, there are probably infinitely many like this, mostly with 11 as a factor. I don't know about digit pairs other than {0,1}.

[Ed Pegg]

That's some excellent hand-waving, Gareth.

[Jon Wild]

On 2026-01-11 8:55 PM, David desJardins wrote:

> On Sun, Jan 11, 2026 at 4:00 PM Ed Pegg <edp...@gmail.com> wrote:
>
> Here's the start of a sequence.
> 111199 = 11*11*919
> 119911 = 11*11*991
>
> Composite numbers using only two digits with a prime factorization
> using the same two digits.  I was only able to find these two
> values.  Are there any more?
>

>> As Gareth says, there are probably infinitely many like this, mostly
>> with 11 as a factor. I don't know about digit pairs other than {0,1}.

If pairs that avoid the digits {0,1} are found to be more interesting,
and if we avoid examples like David's where one side of the equals sign
only features one of the digits, then you could turn to other bases. I
didn't program a search but doodled by hand in base 5 to find that:

233 = 2 x 2 x 32

(In base 10 this is 68 = 2 x 2 x 17)

Looked quickly in base 8 and couldn't see any small examples. It's not
surprising that the smaller the base, the greater the chance that you
get those digit collisions. In the limit, binary, every factorization
has Ed's property.

Jon

[Robert Israel]

For factorizations with only 0 and 1, consider most (but not all) the terms of  A203897.

Thus if p is a prime consisting of digits 0 and 1, with no adjacent 1's and at least one run of at least two 0's, then 11 * p is an example.

Heuristically we expect there to be infinitely many of these.

Cheers,

Robert

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[Daniel Mondot]

We can continue:

(48 17-digit solutions using {0,1})

(80 18-digit solutions using {0,1})

As Gareth says, there are probably infinitely many like this, mostly with 11 as a factor. I don't know about digit pairs other than {0,1}.

No, while about half of them factor with 11, most of the rest factor with 101, but there are exceptions, 

for instance, from the 18-digit group:

111100111111111111 = 11*101*100000100010001 which is the first number made of zeroes and ones, that factors into 3 numbers made of zeroes and ones.

Furthermore, some higher numbers with have neither 11, nor 101 as factors:

for instance:

11011111110010111010101 = 10010101*1100000001000001
111101101101010111011001 = 101001001*1100000000010001
1101110011001111101010101 = 1100010001*1001000000000101
1111010101011101110011001 = 1100010001*1010000000001001
11011100110010111101010101 = 1100010001*10010000000000101
100110010110111010101001111 = 10001000011*10010000000000101
100111011010011010111100101 = 10001100001*10010000000000101
11000011111111110100010101 = 100000010101*110000100000001
1000000111010101110110111101 = 100000010101*10000000100001001
11111110101110101110000001 = 101010000001*110000100000001
1010100010111101111110001001 = 101010000001*10000000100001001
1100011111111111100111100101 = 1000010100101*1100000001000001
111100101110111110100001001 = 1010000001001*110000100000001
10010100111011111011010111100101 = 1000010001100001*10010000000000101
11000000110101111101000110011001 = 10000000100001001*1100000000010001
11001100010111011101100001100001 = 10001000000100001*1100000001000001

PS: besides the (infinite) group of numbers using 0s and 1s, I haven't found any new numbers in the other groups, with 18 digits or less.

Daniel. 

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