You have misstated the condition. The remainder is always 2. What you mean is that if you remove the 1s at the ends of the row, the remainder is zero.

Right?

OK. Now, that is a consequence of Fermat's ("little") theorem, namely that if a is an integer (for our purposes greater than 1), and p is a prime, then

a^p mod p = a.

OK so far?

You will already know that  the nth row of the triangle adds up to 2^n. OK?

Then the pth row sum mod p (p prime, remember?) will be 2 because

2^p mod p = 2.

But you have removed 2 from 2^p, leaving a mod of zero.

Which was the fact to explain. Right?

You will find Fermat's theorem discussed and proved and heavily used in any book of number theory.

It occurs to me that I was a bit abrupt in my response, but maths is largely abrupt isn't it? In some ways that is part of its charm. However, you might wonder why a non-mathematician should know that theorem at all.

The fact is that I have a sentimental attachment to that theorem because I discovered the theorem (a few centuries late perhaps, but I did!)

I was sitting on a commuter train, working on  problem concerning a game we had bought, and one of my results was... THAT!

The rest I have forgotten years ago, but I never have forgotten my reaction: I was irritated! a^p mod p = a if p is prime?

Rubbish! What a stupid idea! I had made some idiotic error!

So I went over it again a couple of times in different ways and... no error!

It took me some time to generate a mechanical model that visibly proved it to me.

(Incidentally, you might find it interesting to do some looking up on pseudo-primes...)

I was thrilled and went around the office bragging about this discovery, till a mathematician in the OR department gently told me about the theorem being known (albeit important in number theory.)

That hardly dashed me; my demonstration was my own and I knew no number theory to speak of anyway. And I was (and still am) intrigued at my original contemptuous reaction to my own discovery -- that it was nonsense, and that I only believed my own result when I had forced myself to.

Great fun,  really!

>You have misstated the condition.<

That's not quite right Jon. Ed's description is precisely what I see. If you number the top of the triangle Row Zero (which is conventional) then all the numbers in row P, where P is prime, are exactly divisible by P except for the 1s at the end.

Everything else you say is right, of course.

I've had it taught differently that the top row (1) is row 1. Row 2 being (1, 1). and the rule being 2^n-1. What I've found is that all the numbers in the row are multiples of the row number

Sorry, I misunderstood the question until Pete rattled me into reading it more carefully! There may be a connection with Fermat (the resemblance is suspicious!)

Bit of a hurry now. Back later.

Maybe!  ;-)