On Sunday, November 7, 2021 at 3:12:51 AM UTC-3, Richard Hertz wrote:

<snip>
> y = (2b - 1) = (2ᶜ - 1)/(2a - 1)
>
> could be explored as a matrix in the memory with:
>
> - a range of values for c between 1 and N (you choose the limit, depending on the available memory). They are in a ROW.
>
> - under each row, you expand colums for value of a between 1 and M (memory again) containing (2ᶜ - 1)/(2a - 1) values
> on each cell of the N x M matrix. You keep a column for values of a for separate.
>
> - now, if you launch a program to find cells with integers ONLY (up to the 5th.decimal, for sure), the program will deliver
> matches that render couples of (a, c) values exactly (which is a double column).
>
> - now, it only rest to solve y = (2b - 1) = (2ᶜ - 1)/(2a - 1) to obtain exact values (a, b, c). You then derive x and y from them.
>
> Glad to contribute with this interesting puzzle.

I didn't realize immediately that the equation

y = (2b - 1) = (2ᶜ - 1)/(2a - 1)

which generate N x M values in a matrix with N rows for c (1, 2, 3, 4, 5, 6, 7, ............)
and M columns for a (1, 2, 3, 4, 5, 6, 7, ............) generates values for y which are integers BUT MIGHT BE EVEN TOO.

So, every cell contains the EXACT value of y on the full spectrum of numbers: rational, irrational and INTEGER.

Due to that fact, each cell of the matrix has to be checked using TWO consecutive rules:

1) That y, the value at the cell, is an integer: Check that the decimal part is 0.00000........... (I thought of checking up to the 5th. decimal)

2) Having the results (y, a, c), you STILL have to verify that y is an ODD number, by verifying that in y = (2b - 1), b IS AN INTEGER.

These two filters are required because you could get y = 214 (even integer), but it doesn't satisfy that b = 107.5 (non integer).

Only after steps 1) and 2) are executed and verified, and you have the set (a, b, c), you can generate x and y that verify

x.y = 2ᶜ - 1
x = (2a - 1)
y = (2b - 1)

which, I think, was the original question from Sylvia.

Solving c in one single step would require a 3D matrix N x M x Q cells (N for c, M for a, Q for b). Then, any integer found is odd.

c = log₂ (4ab - 2a - 2b + 2)

One advantage is to select ranges for N, M and Q so you can explore BIG integers. You don't need to start from 1.
Rules for delimiting ranges can be developed with certainty.

And this point is where I stop. I understand (by googling it) that this problem is managed within Number Theory and Abstract Algebra,
as I could watch a good number of problems posting similar challenges.

Sylvia Else wrote:

> On 07-Nov-21 11:03 am, Gregor Bicha wrote:
>> Sylvia Else wrote:
>>
>>> On 07-Nov-21 10:33 am, rotchm wrote:
>>>> For instance, for 7 the formula works and gives 9. But 1 also
>>>> produces the result: 7*1 = 111.
>>>> Better yet, for 97, the formula works too but doesn't give the
>>>> smallest; the smallest is 2901803883615. Indeed,
>>>> 97 * 2901803883615 = FFFFFFFFFFFF
>>>
>>> What I'm hoping[*] for is that David will find a method of determining
>>> y
>>> for arbitrarily large x. Then I'll steal the method and use it to get
>>> famous in an apparently unrelated field.
>>
>> lol, but this rotchm is the most stupid uneducated sack of shit
>> disturbing with nonsense this group.
>>
>> but the problem is simple, since in binary you only have 1s and 0s, the
>> same is applicable to 0s, as is it to 1s.
>>
> rotchm appears to have figured this out already. It's far from clear to
> me that you even understand the proposition.

rotchm and you, stupid bitches, understood nothing, but evt the need of a
formula. You are the one implying the outcome of 1s, idiot. Your f are 1s
all along. Go back and read your shit. And therefore, the same procedure
may be applied to get 0s instead of 1s.

you stupid inbreed idiot.

rotchm wrote:

> On Saturday, November 6, 2021 at 8:39:36 PM UTC-4, Sylvia Else wrote:
>
>> rotchm appears to have figured this out already. It's far from clear to
>> me that you even understand the proposition.
>
> Of course the troll does not understand the question. Trolls are idiots
> by

you inbreed cretin, you are the one understanding jackshit. Enough
looking around your crackpot "papers". The problem is about getting 1s
related to a start integers. Go back and read what your bitch wrote.

go report this, you stinking unemployable sack of shit.

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Michael Moroney wrote:

> On 11/6/2021 8:40 PM, Sylvia Else wrote:
>> On 07-Nov-21 11:03 am, Gregor Bicha wrote:
>
>>> lol, but this rotchm is the most stupid uneducated sack of shit
>>> disturbing with nonsense this group.
>>>
>>> but the problem is simple, since in binary you only have 1s and 0s,
>>> the same is applicable to 0s, as is it to 1s.
>>>
>>>
>> rotchm appears to have figured this out already. It's far from clear to
>> me that you even understand the proposition.
>
> Again, that's the nymshifting troll, whose only purpose is to generate
> noise in here. It deliberately "misunderstood" the problem to get
> people squabbling about what an idiot it is.

not me, go back and read it. The bitch said about and integer and F-s in
binary. The f-s are 1, hence the same procedure, whatever is used is
seemingly to be applied for zeros. You are an idiot.

Actually I am talking with 3 trolls idiots here.

On 11/5/2021 11:00 PM, Sylvia Else wrote:
> For any odd number x, there exists another odd number y, such that the
> product of x and y expressed in binary is a contiguous sequence of 1s,
> though I'll express them in hex.
>
> For example
>     3 * 1 =       3
>     5 * 3 =       f
>     7 * 9 =      3f
>     9 * 7 =      3f
>     11 * 93 =   3ff
>     13 * 315 =  fff
>
>     97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
>

There are some solutions for 2^c-1 where c is even.

If c is even, let c = 2d.

So, 2^2d-1 can be factored into (2^d+1)*(2^d-1).

In addition, the triple (2^d-1), 2^d, (2^d+1) are 3 adjacent integers.
One of them must be a multiple of 3. 2^d cannot be a multiple of 3
since it is strictly a power of 2, so either (2^d-1) or (2^d+1) is a
multiple of 3. So for c even, there are (at least) two pairs of odd
numbers whose products are all 1s. One is {(2^d+1),(2^d-1)} and the
other is {(2^d±1)*3,(2^d∓1)/3}

On Sunday, November 7, 2021 at 10:43:19 AM UTC-3, rotchm wrote:

<snip>

> Eeech... You havent tried hard havent you!
> With c = 8, what do you get? [do it by hand, its more precise than Excel].
> Now, you see whats going on? See the pattern? Can you explain it now?
> What are thus the other solutions?
>
> Now, can you generalize to any "a" ? Yes/no/why not/counterexample ?
>
> And the question of interest is:
>
> For any [odd] M, is there an [odd] N such that M*N = 11...1_₂?
> Yes or no?

I don't understand what you pretend to state.

Anyways, try with this:

1) Use Excel 2010 or better.

2) Start with cell A2, B2 and write there this formula (you have to translate to English the functions):

=SI((LOG((4*$A2*B$1-2*$A2-2*B$1+2);2)-TRUNCAR(LOG((4*$A2*B$1-2*$A2-2*B$1+2);2)))=0;LOG((4*$A2*B$1-2*$A2-2*B$1+2);2);"")

3) Use the row A = 1 to write values of a, since B column, 1, 2, 3, 4, 5, 6, .......,59, 60

4) Use the column A, from A = 2, to write values of b: 1, 2, 3, 4, 5, 6, .......,59, 60

5) Freeze pannels, so you can always watch column A and row for A = 1.

Watch what happens.

Only values of c that verify the equation will appear. The rest of cells will be BLANK.

Then you will check how the values of c appear, from 1 to 12, for specific sets (a, b) that satisfy the equation.

Once you have this template, you can play with ranges for a and b at will. For instance, starting at a = 243, etc.

If you develop any further this template (with aditional restrictions) you even can extract "c" in a separate column (along with a and b).

And this is using Excel, which is VERY limited in memory usage.

If you write a little program, but using a LARGE matrix, every solution (a, b, c) will appear. You only have to eliminate the swap values
(a, b) & (b, a), which are the same numbers. Only take one set and discard the other.

Dono. wrote:

> On Sunday, November 7, 2021 at 7:46:08 AM UTC-8, rotchm wrote:
>> On Sunday, November 7, 2021 at 10:17:09 AM UTC-5, Dono. wrote:
>>
>> > b=(2^(c-1)+a-1)/(2a-1)
>> >
>> > The above is a diophantic equation that admits integer solution only
>> > if a=EVEN. This observation halves the number of searches.
>> Let c = 6 and a = 5 [which is an odd integer]. b=(2^(c-1)+a-1)/(2a-1) =
>> 4 [an integer].
>>
>> Thus, there are solutions when "a" is odd.
> you are right

and wrong, a 4 will never give you an 1 in that nible, idiot. The sylvia
said something about f-s in hex, those are 1s in binary.

let me conclude, you are both idiots, and should be reported.

rotchm wrote:

> On Sunday, November 7, 2021 at 10:17:09 AM UTC-5, Dono. wrote:
>
>> b=(2^(c-1)+a-1)/(2a-1)
>>
>> The above is a diophantic equation that admits integer solution only if
>> a=EVEN. This observation halves the number of searches.
>
>
> Let c = 6 and a = 5 [which is an odd integer]. b=(2^(c-1)+a-1)/(2a-1) =
> 4 [an integer]. Thus, there are solutions when "a" is odd.

imbecile, a 4 gives you zeros in binary. You stinking stupid sob. Go
report yourself.

Richard Hertz wrote:

> On Sunday, November 7, 2021 at 10:43:19 AM UTC-3, rotchm wrote:
>
> <snip>
>
>> Eeech... You havent tried hard havent you!
>> With c = 8, what do you get? [do it by hand, its more precise than
>> Excel].
>> Now, you see whats going on? See the pattern? Can you explain it now?
>> What are thus the other solutions?
>> Now, can you generalize to any "a" ? Yes/no/why not/counterexample ?
>> And the question of interest is:
>> For any [odd] M, is there an [odd] N such that M*N = 11...1_₂?
>> Yes or no?
>
> I don't understand what you pretend to state.

the crackpot is a cretin, what would you expect.

Michael Moroney wrote:

>>     97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
>>
>>
> There are some solutions for 2^c-1 where c is even.
>
> If c is even, let c = 2d.
>
> So, 2^2d-1 can be factored into (2^d+1)*(2^d-1).

choose d=4, and what was the latest you've been proved an idiot.

On 2021-11-07 18:04:51 +0000, Gregor Bicha said:

> Michael Moroney wrote:
>
>> On 11/6/2021 8:40 PM, Sylvia Else wrote:
>>> On 07-Nov-21 11:03 am, Gregor Bicha wrote:
>>
>>>> lol, but this rotchm is the most stupid uneducated sack of shit
>>>> disturbing with nonsense this group.
>>>>
>>>> but the problem is simple, since in binary you only have 1s and 0s,
>>>> the same is applicable to 0s, as is it to 1s.
>>>>
>>>>
>>> rotchm appears to have figured this out already. It's far from clear to
>>> me that you even understand the proposition.
>>
>> Again, that's the nymshifting troll, whose only purpose is to generate
>> noise in here. It deliberately "misunderstood" the problem to get
>> people squabbling about what an idiot it is.
>
> not me, go back and read it. The bitch said about and integer and F-s in
> binary. The f-s are 1, hence the same procedure, whatever is used is
> seemingly to be applied for zeros. You are an idiot.
>
> Actually I am talking with 3 trolls idiots here.

Very carefully and cogently argued. Who could possibly disagree?

--
Athel -- French and British, living mainly in England until 1987.

Athel Cornish-Bowden wrote:

>>> Again, that's the nymshifting troll, whose only purpose is to generate
>>> noise in here. It deliberately "misunderstood" the problem to get
>>> people squabbling about what an idiot it is.
>>
>> not me, go back and read it. The bitch said about and integer and F-s
>> in binary. The f-s are 1, hence the same procedure, whatever is used is
>> seemingly to be applied for zeros. You are an idiot.
>>
>> Actually I am talking with 3 trolls idiots here.
>
> Very carefully and cogently argued. Who could possibly disagree?

this is what "sylvia" wrote, all 1s in binary. You too stupid, or lack something intelligible to say? A 1,3,7 or f are 1s in binary.

> For example
>     3 * 1 =       3
>     5 * 3 =       f
>     7 * 9 =      3f
>     9 * 7 =      3f
>     11 * 93 =   3ff
>     13 * 315 =  fff
>     97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF

On 08-Nov-21 5:10 am, Michael Moroney wrote:
> On 11/5/2021 11:00 PM, Sylvia Else wrote:
>> For any odd number x, there exists another odd number y, such that the
>> product of x and y expressed in binary is a contiguous sequence of 1s,
>> though I'll express them in hex.
>>
>> For example
>>      3 * 1 =       3
>>      5 * 3 =       f
>>      7 * 9 =      3f
>>      9 * 7 =      3f
>>      11 * 93 =   3ff
>>      13 * 315 =  fff
>>
>>      97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
>>
>
> There are some solutions for 2^c-1 where c is even.
>
> If c is even, let c = 2d.
>
> So, 2^2d-1 can be factored into (2^d+1)*(2^d-1).
>
> In addition, the triple (2^d-1), 2^d, (2^d+1) are 3 adjacent integers.
> One of them must be a multiple of 3.  2^d cannot be a multiple of 3
> since it is strictly a power of 2, so either (2^d-1) or (2^d+1) is a
> multiple of 3. So for c even, there are (at least) two pairs of odd
> numbers whose products are all 1s. One is {(2^d+1),(2^d-1)} and the
> other is {(2^d±1)*3,(2^d∓1)/3}

OK, it doesn't look as if David is going to make an appearance.

I came upon this not very remarkable, and probably widely known,
property while pondering the factorisation problem. Not that I expect to
find a solution to that either, but it helps pass the time when I can't
sleep.

Anyway, there a theorem in number theory that says that if a and x have
no common factors, then a^n = 1 mod x, where n is the number of numbers
less than x that have no common factors with x. Where a is two, it will
never have a common factor with an odd number x.

It follows that 2^n = 1 mod x for the appropriate n. If x happens to be
prime, then n = x - 1.

If 2^n = 1 mod x, then 2^n - 1 = 0 mod x. That is, 2^n - 1 is divisible
by x. So we get y by dividing 2^n - 1 by x.

As rotchm has observed, it's not necessarily the smallest value for y.
Also y found by the above goes up horribly fast with increasing x.
Leaving that aside, if one could find y by some other means, one would
then have the value of n, which would break RSA encryption. However, as
usual, one quickly realises that finding y is just as intractable is the
original factorisation problem.

Sylvia.

On Sunday, November 7, 2021 at 1:22:45 PM UTC-5, Richard Hertz wrote:
> On Sunday, November 7, 2021 at 10:43:19 AM UTC-3, rotchm wrote:

> > For any [odd] M, is there an [odd] N such that M*N = 11...1_₂?
> > Yes or no?
> I don't understand what you pretend to state.

Its what the OP asked. did you not understand his question??
For instance, given say, M=117, can you find an integer N such that

117 * N = 11...1_₂ ?

> Anyways, try with this:
>
> 1) Use Excel 2010 or better.

No, Excel does not always give the (correct) answers. And Excel will not prove it for all M's.
Excel will not find the counterexample (you will need a much bigger "matrix").

> If you write a little program, but using a LARGE matrix, every solution (a, b, c) will appear.

Or, just some math to find a general formula.
Or a little program that doesn't use any matrix and hardly any memory; there is a simple procedure.
Or by hand, via a different and simple highschool procedure; for instance, I applied this procedure to M=117 and
by hand I got the answer in a few seconds.

On Sunday, November 7, 2021 at 6:14:42 PM UTC-5, Sylvia Else wrote:

> As rotchm has observed, it's not necessarily the smallest value for y.
> Also y found by the above goes up horribly fast with increasing x.
> Leaving that aside, if one could find y by some other means,

Its a trivial exercise (even by hand) to find N if MN = 11...1_₂
There is a simple procedure to find N.

On Sunday, November 7, 2021 at 8:15:27 PM UTC-3, rotchm wrote:

<snip>

> > If you write a little program, but using a LARGE matrix, every solution (a, b, c) will appear.

> Or, just some math to find a general formula.

> Or a little program that doesn't use any matrix and hardly any memory; there is a simple procedure.

> Or by hand, via a different and simple highschool procedure; for instance, I applied this procedure to M=117 and
> by hand I got the answer in a few seconds.

Why are you so upset?.
Do you have any problem with me?

Be precise, because you are behaving as an asshole and it is starting to irritate me: high school, by hand, few seconds?

On Sunday, November 7, 2021 at 3:40:40 PM UTC-8, Richard Hertz wrote:
> ... high school, by hand, few seconds?

Yes, for example, suppose you want a number N such that 97*N has the form 1111...111 in binary. The smallest such N is 2901803883615. On the other hand, if you want 97*N to have the form 1111...1111 in decimal, the smallest such N is

1145475372279495990836197021764032073310423825887743413516609392898052691867124856815578465063

Do you see how to compute this?

This sheds some light on a recent discussion in another thread about the difference between scientists and mathematicians on one hand, and engineers on the other. An engineer might try to use a "brute force" method of trial and error on a computer with an Excel spreadsheet, searching for an odd integer y for a given odd x such that xy is a binary repunit. This approach has some obvious disadvantages: First, it's ridiculously laborious and inefficient; second, it can never prove that there is such a y for every odd x; and third, it gives no real insight into the solutions.

Now, an amateur mathematician (nearly 400 years ago) found the simple theorem that gives the explicit answer, and another mathematician (about 300 years ago) refined that theorem to give the minimum y value explicitly for any x, and they did this without any electronic computers at all. Moreover, the theorem they proved is used very extensively in modern technology for things like digital data error detection and corrections codes, and so on. Of course, many of the engineers who use CRCs are not experts in how or why they work. They rely on the mathematicians.

On Sunday, November 7, 2021 at 3:40:40 PM UTC-8, Richard Hertz wrote:
> ... high school, by hand, few seconds?

Yes, for example, suppose you want a number N such that 97*N has the form 1111...111 in binary. The smallest such N is 2901803883615. On the other hand, if you want 97*N to have the form 1111...1111 in decimal, then the smallest such N is

1145475372279495990836197021764032073310423825887743413516609392898052691867124856815578465063

Do you see how to compute this easily?

This sheds some light on a recent discussion here about the difference between scientists and mathematicians on one hand, and engineers on the other. An engineer might try to use a "brute force" method of trial and error on a computer with an Excel spreadsheet, searching for an odd integer y for a given odd x such that xy is a binary repunit. This approach has some obvious disadvantages: First, it's ridiculously laborious and inefficient; second, it can never prove that there is such a y for every odd x; and third, it gives no real insight into the solutions.

Now, an amateur mathematician (nearly 400 years ago) found the simple theorem that gives the explicit answer, and another mathematician (about 300 years ago) refined that theorem to give the minimum solution, and they did this without any electronic computers at all. Moreover, the theorem they proved is used very extensively in modern technology for things like digital data error detection and corrections codes, and so on. Of course, many of the engineers who use CRCs are not experts in how or why they work. They rely on the mathematicians.

On Sunday, November 7, 2021 at 6:40:40 PM UTC-5, Richard Hertz wrote:
> On Sunday, November 7, 2021 at 8:15:27 PM UTC-3, rotchm wrote:

> > Or by hand, via a different and simple highschool procedure; for instance, I applied this procedure to M=117 and
> > by hand I got the answer in a few seconds.
> Why are you so upset?.

I am not. You must be confusing me with someone else.

> Do you have any problem with me?

Nope.

> Be precise, because you are behaving as an asshole and it is starting to irritate me: high school, by hand, few seconds?

I am trying to hint you that the OP's problem is much simpler than it looks. I am trying to convince you to
*ponder* it, to work at it via a different approach. Inspiring one to work "harder" or "differently" is not being an "asshole". But I do get it that simple minds may think so.

On Sunday, November 7, 2021 at 9:03:09 PM UTC-3, Al Coe wrote:

<snip>

Obviously, I touched a sensitive nerve on mathematicians, WHICH I never pretended to do.

I have respect for the realm of PURE MATHEMATICS, which IS NOT the realm of APPLIED MATHEMATICS, got it?

The ridiculous number that you posted probably has a perfect meaning in the fantasy world of pure math, NOT IN THE REAL WORLD.

1145475372279495990836197021764032073310423825887743413516609392898052691867124856815578465063, if fed to a PC,
is truncated to 114547537227949 . 10E+79 (15 decimal digits plus powers of 10).

This is because a standard i3 FPU dedicates 50 bits to the mantissa and 14 bits to the exponent (signs included).

And N decimal digits uses about 3.32.N bits.

To manage a 10E+93 number like you posted would require a FPU managing a mantissa of 309 bits plus K bits for exponents > 330 bits
per WORD. And such STUPID CPU/FPU will not exist in the foreseeable future.

As far as I know, the most advanced commercial/scientific computer operating today uses 128 bits CPU/FPU. And this is
UNALTERED for the last 30 YEARS, when IBM delivered his RISC RS/6000 in 1991.

I don't deny that a 1024 bits CPU can be built FOR SPECIAL PURPOSES. I don't have the slightest doubt about it.
Using the monstrous power of PGA today, you COULD device 8 x 128 word sliced CPU in a single chip, managing mantissas of
900 bits (aprox. 300 decimal digits), if you want, have the money, the time and the purpose.

So, deal with this fact: with a 128 bits CPU, you can manage no more than 30 decimal digits for a mantissa. 40 decimal digits
if you use it in integer mode. And that's all.

For the rest of the problems with precision above 40 decimal digits, use an abacus and tell me how did it work for you.

OR, you can expend your time using decomposition by factoring large numbers in products of smaller numbers. There are
a huge amount of solutions developed in the XX century, including the use of logarithms in-between.

Now, don't be so defensive, as if I called names to your mother. Live and let live OR deal with a confrontation.

I don't like confrontations unless the other part is calling a fight. Is that what happens with you and the other one?

On Sunday, November 7, 2021 at 9:13:48 PM UTC-3, rotchm wrote:

<snip>

> > Be precise, because you are behaving as an asshole and it is starting to irritate me: high school, by hand, few seconds?
> I am trying to hint you that the OP's problem is much simpler than it looks. I am trying to convince you to
> *ponder* it, to work at it via a different approach. Inspiring one to work "harder" or "differently" is not being an "asshole". But I do get it that simple minds may think so.

No, imbecile, you are using the down-play game.

Don't be a coward, and admit it.

And I don't need that you tell me to work harder, because that is trolling. As you see, you pulled the "simple minds" shit.

I'll remember your attitude in the future. You passed below my attention radar till now. I'll see how this issue develops.

And, as a hobby, I'm OK with what I did. Did you try the Excel formula? It works nice, but only up to c < 500 (the key is using
logarithms. It does the trick for a.b < 15 decimal digits in my I3 CPU.

To see the truncation, just copy&paste in your computer this 18 digits number: 816785180559421928 and zoom out decimals.

On Sunday, November 7, 2021 at 8:10:13 PM UTC-5, Richard Hertz wrote:
> On Sunday, November 7, 2021 at 9:13:48 PM UTC-3, rotchm wrote:

> > I am trying to hint you that the OP's problem is much simpler than it looks. I am trying to convince you to
> > *ponder* it, to work at it via a different approach. Inspiring one to work "harder" or "differently" is not being an
>> "asshole". But I do get it that simple minds may think so.
>
> No, imbecile, you are using the down-play game.

No I am not. But I admit that it *looks* like it.

> And I don't need that you tell me to work harder, because that is trolling.

Which is why I did not tell you to work harder. I am telling you that it is much simpler than it looks. I've been doing problems like that by hand in high school. So this is a hint to you to refresh your memory on what you learned in high school and see what techniques you can apply to solve the OP's question.

> As you see, you pulled the "simple minds" shit.

Nowhere did I imply that you are a "simple mind".

If A implies B, that does not mean if B then A, agree?

> I'll remember your attitude in the future.

Yes, do remember, when I discuss, I mean what I write....the reader does need to understand English and (Boolean) logic.

On 08-Nov-21 10:25 am, rotchm wrote:
> On Sunday, November 7, 2021 at 6:14:42 PM UTC-5, Sylvia Else wrote:
>
>> As rotchm has observed, it's not necessarily the smallest value for y.
>> Also y found by the above goes up horribly fast with increasing x.
>> Leaving that aside, if one could find y by some other means,
>
> Its a trivial exercise (even by hand) to find N if MN = 11...1_₂
> There is a simple procedure to find N.
>

The interesting part of the problem space is where the number of bits is
in the order of 2^(M * N), rather than that the bits equal M * N. One is
trying to discover M * N where the number of bits in M * N is in the
hundreds or thousands.

In such a case, if one could discover y for some x, one would have a
possible answer. Fortunately for RSA, y would generally be so large that
it could not be represented in any plausible computer, let alone be
calculated.

Sylvia.

On 08-Nov-21 11:56 am, Richard Hertz wrote:
> On Sunday, November 7, 2021 at 9:03:09 PM UTC-3, Al Coe wrote:
>
> <snip>
>
> Obviously, I touched a sensitive nerve on mathematicians, WHICH I never pretended to do.
>
> I have respect for the realm of PURE MATHEMATICS, which IS NOT the realm of APPLIED MATHEMATICS, got it?
>
> The ridiculous number that you posted probably has a perfect meaning in the fantasy world of pure math, NOT IN THE REAL WORLD.
>
> 1145475372279495990836197021764032073310423825887743413516609392898052691867124856815578465063, if fed to a PC,
> is truncated to 114547537227949 . 10E+79 (15 decimal digits plus powers of 10).
>
> This is because a standard i3 FPU dedicates 50 bits to the mantissa and 14 bits to the exponent (signs included).
>
> And N decimal digits uses about 3.32.N bits.
>
> To manage a 10E+93 number like you posted would require a FPU managing a mantissa of 309 bits plus K bits for exponents > 330 bits
> per WORD. And such STUPID CPU/FPU will not exist in the foreseeable future.
>
> As far as I know, the most advanced commercial/scientific computer operating today uses 128 bits CPU/FPU. And this is
> UNALTERED for the last 30 YEARS, when IBM delivered his RISC RS/6000 in 1991.

The fact that the CPU only has hardware instructions for such numbers
doesn't mean that the software running on it is constrained by that limit.

As for lack of meaning in the real world, the widely used public key
encryption system RSA routinely uses calculations of numbers with a
thousand or more bits, and every PC, Mac, and smartphone does this with
no fuss at all, and without the users, including you apparently, being
aware of it.

Sylvia.

Richard Hertz wrote:
> On Sunday, November 7, 2021 at 9:03:09 PM UTC-3, Al Coe wrote:
>
> <snip>
>
> Obviously, I touched a sensitive nerve on mathematicians, WHICH I never pretended to do.
>
> I have respect for the realm of PURE MATHEMATICS, which IS NOT the realm of APPLIED MATHEMATICS, got it?
>
> The ridiculous number that you posted probably has a perfect meaning in the fantasy world of pure math, NOT IN THE REAL WORLD.
>
> 1145475372279495990836197021764032073310423825887743413516609392898052691867124856815578465063, if fed to a PC,
> is truncated to 114547537227949 . 10E+79 (15 decimal digits plus powers of 10).
>
> This is because a standard i3 FPU dedicates 50 bits to the mantissa and 14 bits to the exponent (signs included).
>
> And N decimal digits uses about 3.32.N bits.
>
> To manage a 10E+93 number like you posted would require a FPU managing a mantissa of 309 bits plus K bits for exponents > 330 bits
> per WORD. And such STUPID CPU/FPU will not exist in the foreseeable future.

I do not have such a kind of CPU/FPU on my computer, nevertheless I have
no problem to handle it:

Python 3.9.1 (default, Feb 28 2021, 08:57:14)
[Clang 10.0.0 (clang-1000.10.44.4)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> n =
1145475372279495990836197021764032073310423825887743413516609392898052691867124856815578465063
>>> n
1145475372279495990836197021764032073310423825887743413516609392898052691867124856815578465063
>>> n**2
1312113828498849932229320758034401000355582847523188331634145925427322671096362951678784037872850921563147446167249901263434405461542599743612957911324723242390723880078044530519111593969

You seem quite illiterate in computing science, Richard.