From Newsgroup: rec.puzzles

From '536 Puzzles and Curious Problems' by Henry Ernest Dudeney.

What are the factors (the numbers that will divide it without any remainder) of this number --1000000000001? This is easily done if you
happen to know something about numbers of this peculiar form. In fact, it
is just as easy for me to give two factors if you insert, say, one hundred
and one noughts, instead of eleven, between the two ones.
There is a curious, easy, and beautiful rule for these cases. Can
you find it?
--
David Entwistle
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From Newsgroup: rec.puzzles

On Wed, 3 Dec 2025 19:02:35 -0000 (UTC), David Entwistle wrote:

There is a curious, easy, and beautiful rule for these cases. Can you
find it?

Having looked at this, think 'Dudeney easy', not 'easy easy'...
--
David Entwistle
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From Newsgroup: rec.puzzles

In article <10grnkm$67am$1@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

There is a curious, easy, and beautiful rule for these cases. Can you
find it?

Having looked at this, think 'Dudeney easy', not 'easy easy'...

Any easier question is why all numbers of the form abcabc are
divisible by 7, 11, and 13.

-- Richard
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From Newsgroup: rec.puzzles

On Thu, 4 Dec 2025 12:10:48 -0000 (UTC), Richard Tobin wrote:

Any easier question is why all numbers of the form abcabc are divisible
by 7, 11, and 13.

Yes, thanks. That is easier and does shed some light.

I'm sure there will be some mistakes in the following, but hopefully my intended meaning is clear.

SPOILER.
POILER.
OILER.
ILER.
LER.
ER.
R.
.

In his answer Dudeney says:

If the number of noughts enclosed by the two ones is 2 added to any
multiple of 3, two factors can always be written down at once in this
curious way. 1001 = 11 x 91; 1000001 = 101 x 9901; 1000000001 = 101 x
999001; 1000000000001 = 10001 x 99990001. The last is our required answer,
and 10001 = 73 x 137. The multiple of 3 in 11 is 3: therefore we insert 3 noughts in each factor and one more 9.

I'm not sure I understand that last bit (from 'The multiple ...'), but get
the general idea.

I also see that...

If the number of noughts is 0, 2, 4, 6 ... then 11 is a factor.
If the number of noughts is 1, 5, 9, 13 ... then 101 is a factor.

In the cases where none of the above apply (3, 7, 15 ...), then I'm not
sure what the rule is, but the ones I have looked at do have factors.

I'm unclear whether we can say that, other than 11 and 101, all have
factors and none of the numbers of this form are prime.
--
David Entwistle
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From Newsgroup: rec.puzzles

On 05/12/2025 14:00, David Entwistle wrote:

On Thu, 4 Dec 2025 12:10:48 -0000 (UTC), Richard Tobin wrote:

Any easier question is why all numbers of the form abcabc are divisible
by 7, 11, and 13.

Yes, thanks. That is easier and does shed some light.

I'm sure there will be some mistakes in the following, but hopefully my intended meaning is clear.

SPOILER.
POILER.
OILER.
ILER.
LER.
ER.
R.
.

In his answer Dudeney says:

If the number of noughts enclosed by the two ones is 2 added to any
multiple of 3, two factors can always be written down at once in this
curious way. 1001 = 11 x 91; 1000001 = 101 x 9901; 1000000001 = 101 x
999001; 1000000000001 = 10001 x 99990001. The last is our required answer, and 10001 = 73 x 137. The multiple of 3 in 11 is 3: therefore we insert 3 noughts in each factor and one more 9.

Well, I have to say that's not very enlightening - just some instructions with no underlying
explanation or intuition. :(

My explanation would be to think of all these examples as "well known" polynomial factorisations.
(By well known, I mean brighter high-school maths students would probably have come across these.)
The polynomials factorisations are clearer and easier to understand than the corresponding number
factorisations which can all be obtained by substituting x with various powers of 10.

So our "well known" polynomial factorisation is:

x^n + 1 = (x + 1) (x^(n-1) - x^(n-2) + x^(n-3) - ... - x + 1)

Note the alternating signs or the right, so n must be odd for this to work.

Concrete example: x^3 + 1 = (x + 1)(x^2 - x + 1). If we multiply out the product on the right,
all the terms cancel except the first and last, giving the left hand side.

Our plan is to use one of these factorisations, applying x = 10^m. Example: using the x^3 + 1
factorisation just given, with x = 10^2 = 100 gives:

100^3 + 1 = (100 + 1) (100^2 - 100 + 1) i.e.
1000001 = 101 * (10000 - 100 + 1)
= 101 * 9901

So what n and x = 10^m do we want to use to factor 1000000000001 = 10^12 + 1?

Note, Dudney talks about counting the number of zeros in the number, but what is more useful is the
power of 10 involved: 12 in our case (rather than 11). If we choose n and m we will end up with a
factorisation for (10^m)^n = 10^(m*n), so we want m*n = 12. Also remembering n must be odd, we can
use n=3, m=4. Hey, that's our favourite example, but substituting x=10000 rather than x=100.

10000^3 + 1 = (10000 + 1) (10000^2 - 10000 + 1) i.e.
1000000000001 = 10001 * (100000000 - 10000 + 1)
= 10001 * 99990001

Clearly this approach works using n=3 when factoring 10^k + 1 whenever k is a multiple of 3. Then
we choose m = n/3 as we did for 10^12 + 1. [This is much clearer to me than Dudney saying it
applies "the number of noughts enclosed by the two ones is 2 added to any multiple of 3"!]

And also, Dudney has focused on n=3, but we could also use n=5,7,9,11,... so we can factor 10^k+1
whenever k is a multiple of one of these odd numbers. Example: n=5, m=2 (so x=100)

100^5 + 1 = (100 + 1) (100^4 - 100^3 + 100^2 - 100 + 1) i.e.
10000000001 = 101 * (100000000 - 1000000 + 10000 - 100 + 1)
= 101 * 99009901

I'm not sure I understand that last bit (from 'The multiple ...'), but get the general idea.

I also see that...

If the number of noughts is 0, 2, 4, 6 ... then 11 is a factor.

those are k = 1,3,5,7... cases above where we can use one of our polynomials with x=10.

If the number of noughts is 1, 5, 9, 13 ... then 101 is a factor.

those are basically k = 6,10,14... cases above where k has an odd factor

In the cases where none of the above apply (3, 7, 15 ...),

k = 2, 4, 8, 16, (i.e. powers of 2)

These would need a bit more thought. They must be well understood by number theory maths people.

Mike.

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From Newsgroup: rec.puzzles

On 06/12/2025 01:39, Mike Terry wrote:

On 05/12/2025 14:00, David Entwistle wrote:

On Thu, 4 Dec 2025 12:10:48 -0000 (UTC), Richard Tobin wrote:

Any easier question is why all numbers of the form abcabc are divisible
by 7, 11, and 13.

Yes, thanks. That is easier and does shed some light.

I'm sure there will be some mistakes in the following, but hopefully my
intended meaning is clear.

SPOILER.
POILER.
OILER.
ILER.
LER.
ER.
R.
.

In his answer Dudeney says:

If the number of noughts enclosed by the two ones is 2 added to any
multiple of 3, two factors can always be written down at once in this
curious way. 1001 = 11 x 91; 1000001 = 101 x 9901; 1000000001 = 101 x
999001; 1000000000001 = 10001 x 99990001. The last is our required answer, >> and 10001 = 73 x 137. The multiple of 3 in 11 is 3: therefore we insert 3
noughts in each factor and one more 9.

Well, I have to say that's not very enlightening - just some instructions with no underlying
explanation or intuition.a :(

My explanation would be to think of all these examples as "well known" polynomial factorisations.
(By well known, I mean brighter high-school maths students would probably have come across these.)
The polynomials factorisations are clearer and easier to understand than the corresponding number
factorisations which can all be obtained by substituting x with various powers of 10.

So our "well known" polynomial factorisation is:

a x^n + 1a =a (x + 1) (x^(n-1) - x^(n-2) + x^(n-3) - ...a - x + 1)

Note the alternating signs or the right, so n must be odd for this to work.

Concrete example:aa x^3 + 1a = (x + 1)(x^2 - x + 1).a If we multiply out the product on the right,
all the terms cancel except the first and last, giving the left hand side.

Our plan is to use one of these factorisations, applying x = 10^m.a Example: using the x^3 + 1
factorisation just given, with x = 10^2 = 100 gives:

a 100^3 + 1a =a (100 + 1) (100^2 - 100 + 1)aaaa i.e.
a 1000001aaa =a 101 * (10000 - 100 + 1)
aaaaaaaaaaaa =a 101 *aa 9901

So what n and x = 10^m do we want to use to factor 1000000000001 = 10^12 + 1?

Note, Dudney talks about counting the number of zeros in the number, but what is more useful is the

Sorry - should be Dudeney [+same below]

power of 10 involved:a 12 in our case (rather than 11).a If we choose n and m we will end up with a
factorisation for (10^m)^n = 10^(m*n), so we want m*n = 12.a Also remembering n must be odd, we can
use n=3, m=4.a Hey, that's our favourite example, but substituting x=10000 rather than x=100.

a 10000^3 + 1aaa = (10000 + 1) (10000^2 - 10000 + 1)aaaa i.e.
a 1000000000001a =a 10001 * (100000000 - 10000 + 1)
aaaaaaaaaaaaaaaa =a 10001 *aa 99990001

Clearly this approach works using n=3 when factoring 10^k + 1 whenever k is a multiple of 3.a Then
we choose m = n/3 as we did for 10^12 + 1.a [This is much clearer to me than Dudney saying it
applies "the number of noughts enclosed by the two ones is 2 added to any multiple of 3"!]

And also, Dudney has focused on n=3, but we could also use n=5,7,9,11,... so we can factor 10^k+1
whenever k is a multiple of one of these odd numbers.a Example:a n=5, m=2a (so x=100)

a 100^5 + 1aa =a (100 + 1) (100^4 - 100^3 + 100^2 - 100 + 1)aaaa i.e.
a 10000000001 =aaaaa 101 * (100000000 - 1000000 + 10000 - 100 + 1)
aaaaaaaaaaaaa =aaaaa 101 *aa 99009901

I'm not sure I understand that last bit (from 'The multiple ...'), but get >> the general idea.

I also see that...

If the number of noughts is 0, 2, 4, 6 ... then 11 is a factor.

those are k = 1,3,5,7... cases above where we can use one of our polynomials with x=10.

If the number of noughts is 1, 5, 9, 13 ... then 101 is a factor.

those are basically k =a 6,10,14... cases above where k has an odd factor

In the cases where none of the above apply (3, 7, 15 ...),

k = 2, 4, 8, 16,aa (i.e. powers of 2)

These would need a bit more thought.a They must be well understood by number theory maths people.

Mike.

A quick search shows I'm wrong on this - it's not really understood!

11 and 101 are prime, and it's known 10^2^n + 1 is not prime for the first few n (11 and 101
excepted), but that's about it...

Mike.








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From Newsgroup: rec.puzzles

On Wed, 3 Dec 2025 19:02:35 -0000 (UTC), David Entwistle wrote:

From '536 Puzzles and Curious Problems' by Henry Ernest Dudeney.

For no other reason than curiosity, I've printed out a table indicating
prime values (X) and composite values (0). The rows represent the number
base: 2 (top row) through 10 (bottom row). The columns are, left to right,
1 + base^n for n = 10 (left), through 2 (right). So the top row represents
the numbers (base 10) 1025, 513, 257, 129, 65, 33, 17, 9, 5. Of which 257,
17 and 5 are prime. Written in binary these are: ... 10001, 1001, 101.

The next row represents the numbers (base 10) 59050, 19684, 6562, 2188,
730, 244, 82, 28, 10. Written base 3, these are: ... 10001, 1001, 101.

The bottom row represents the numbers (base 10) 10000000001, 1000000001, 100000001, 10000001, 1000001, 100001, 10001, 1001, 101. Of which only 101
is prime.

Obviously all rows representing an odd bases don't include any primes, as
1 + base^n is even when the base is odd.

I'm not sure what going on with the row representing 1 + 8^n, this series continues without primes up to 1 + 8^20, which is as far as I've looked.

0 0 X 0 0 0 X 0 X
0 0 0 0 0 0 0 0 0
0 0 X 0 0 0 X 0 X
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 X 0 X
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 X
--
David Entwistle
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From Newsgroup: rec.puzzles

On 07/12/2025 08:21, David Entwistle wrote:

On Wed, 3 Dec 2025 19:02:35 -0000 (UTC), David Entwistle wrote:

From '536 Puzzles and Curious Problems' by Henry Ernest Dudeney.

For no other reason than curiosity, I've printed out a table indicating
prime values (X) and composite values (0). The rows represent the number base: 2 (top row) through 10 (bottom row). The columns are, left to right,
1 + base^n for n = 10 (left), through 2 (right). So the top row represents the numbers (base 10) 1025, 513, 257, 129, 65, 33, 17, 9, 5. Of which 257,
17 and 5 are prime. Written in binary these are: ... 10001, 1001, 101.

The next row represents the numbers (base 10) 59050, 19684, 6562, 2188,
730, 244, 82, 28, 10. Written base 3, these are: ... 10001, 1001, 101.

The bottom row represents the numbers (base 10) 10000000001, 1000000001, 100000001, 10000001, 1000001, 100001, 10001, 1001, 101. Of which only 101
is prime.

Obviously all rows representing an odd bases don't include any primes, as
1 + base^n is even when the base is odd.

I'm not sure what going on with the row representing 1 + 8^n, this series continues without primes up to 1 + 8^20, which is as far as I've looked.

In another post I outlined how simple polynomial factorisations could be applied to numbers in base
10 like 1000000000001 = 10^12 + 1, but these factorisations apply just as well for any other base,

So just as 10^n + 1 has factors when n is a multiple of any odd number, this applies for numbers
other than 10, like 2.

In your case, 1 + 8^n = 1 + 2^3n, and of course 3n is a multiple of 3 which is odd. So all these
numbers have "polynomial" factorisations, and we can use our

1 + x^3 = (1 + x)(1 - x + x^2)

factorisation. So:

1 + 8^n = 1 + (2^n)^3
= (1 + 2^n)(1 - 2^n + (2^n)^2)
= (1 + 2^n)(1 - 2^n + 2^2n)

Mike.


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From Newsgroup: rec.puzzles

David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> writes:

On Wed, 3 Dec 2025 19:02:35 -0000 (UTC), David Entwistle wrote:

From '536 Puzzles and Curious Problems' by Henry Ernest Dudeney.

For no other reason than curiosity, I've printed out a table indicating prime values (X) and composite values (0). The rows represent the number base: 2 (top row) through 10 (bottom row). The columns are, left to right,
1 + base^n for n = 10 (left), through 2 (right). So the top row represents the numbers (base 10) 1025, 513, 257, 129, 65, 33, 17, 9, 5. Of which 257, 17 and 5 are prime. Written in binary these are: ... 10001, 1001, 101.

This is an extremely well-studied, and well factored, form. Check out
the Cunningham factorisation project, which dates back half a century.

Phil
--
We are no longer hunters and nomads. No longer awed and frightened, as we have gained some understanding of the world in which we live. As such, we can cast aside childish remnants from the dawn of our civilization.
-- NotSanguine on SoylentNews, after Eugen Weber in /The Western Tradition/
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