Special Factors of 3-Digit Numbers

Originally posted by BigDoggProblem
The prime factors of 999 are:

3*3*3*37 = 999

Suddenly the 37 doesn't look so out of place...

What you have is not really a proof per se.

A modulo proof is required. And will prove conclusively numbers that are n.

Originally posted by phgao
What you have is not really a proof per se.

A modulo proof is required. And will prove conclusively numbers that are n.

I didn't say it was a proof. The only 'proof' I've posted is the proof given to me by the program that I wrote - proof by empirical means.

I did work on a mathematical proof, off-forum, and established that the factors of 999 must all be solutions. However, I have not yet proven that the 'other' cases (222, 444, 555, 666, 777, 888) are the only 'other' cases possible. Perhaps I will leave that to one of the math geniuses - I enjoy the programming side of things more anyway.

Originally posted by BigDoggProblem
I didn't say it was a proof. The only 'proof' I've posted is the proof given to me by the program that I wrote - proof by empirical means.

I did work on a mathematical proof, off-forum, and established that the factors of 999 must all be solutions. However, I have not yet proven that the 'other' cases (222, 444, 555, 666, 777, 888) are the only 'ot ...[text shortened]... e that to one of the math geniuses - I enjoy the programming side of things more anyway.

yeah-a bit longer than 10 mins...

n divides 999*a, 999*b and 999*c, so if n divides 999 or if n divides all of a, b and c. thus 222 etc work. and i'm not a maths genius. although it's nice that you think of me that way 🙂

and the problem with programming occrus if you ever wish to work out out for abcde....xyz. although you may never wish to do that...

Originally posted by genius
yeah-a bit longer than 10 mins...

n divides 999*a, 999*b and 999*c, so if n divides 999 or if n divides all of a, b and c. thus 222 etc work. and i'm not a maths genius. although it's nice that you think of me that way 🙂

and the problem with programming occrus if you ever wish to work out out for abcde....xyz. although you may never wish to do that...

Sorry, you aren't the genius I was looking for, despite your handle.

For clarity, in the case of n=222, n never divides a or b or c, but only the three digit number abc.

We're still lacking adequate proof that the only exceptional cases (where n is not a factor of 999) are 222, 444, etc. Those solutions are being established by brute force, although admittedly not much force, because there are so few possibilities at the higher numbers.

I don't see why it would be difficult to have a program find x, y, and z once it has found all the solutions for n. Would you mind clarifying your last statement?

charming.

"For clarity, in the case of n=222, n never divides a or b or c, but only the three digit number abc."
is it not the other way round, that a, b and c divide n?

"I don't see why it would be difficult to have a program find x, y, and z once it has found all the solutions for n. Would you mind clarifying your last statement?"
suppose you find all cases for abc being a 3 digit number. then you are asked to find all cases for abc...xyz, a 26 digit number. you'll be at it all week! whereas if you prove it for less digits it shant be too hard to change it to hold for 26 digits. or, indeed, any number. well, it'll take less than a week solidly anyway 😛[/b]

Originally posted by genius
charming.

"For clarity, in the case of n=222, n never divides a or b or c, but only the three digit number abc."
is it not the other way round, that a, b and c divide n?

"I don't see why it would be difficult to have a program find x, y, and z once it has found all the solutions for n. Would you mind clarifying your last statement?"
suppose you ...[text shortened]... 6 digits. or, indeed, any number. well, it'll take less than a week solidly anyway 😛[/b]

is it not the other way round, that a, b and c divide n?

Your earlier statement, "n divides 999*a, 999*b and 999*c, so if n divides 999 or if n divides all of a, b and c. thus 222 etc work." (emphasis mine) made it sound like you were claiming the opposite.

You had used a, b, and c to represent digits, and x, y, and z to represent n's multipliers, so it was confusing when you later used abc...xyz to represent a 26-digit number.

I would like to know how you discount n=74 as a solution. Let's see if you can prove it without using brute force.