1994 m / (1993*1994) < k / n < (1993 m + 1993)/(1993*1994)
and so there are gaps; those gaps are 0 to 1993, 1994 to 3986, ... 1993^2 - 1, 1993^2
so for one, n must be at least 1993 for there to be one k/n in each of 1992 gaps between 0 and 1.

Also, n can't be 1993 or 1994, because then k/n would fall on m/1993 or (m+1)/1994

I feel like shit, though, and while you don't give half a damn hairy rat's ass, I am going to stop working on this problem on account of having the flu.

>>4325322
I've thought about the problem, I see some properties, but as to how to actually find such n is beyond me.

>>4325482
For the record, we can ignore any answer larger than 3987. It's obvious that if n = 3987, then it works for k = 2m.

Bounding from the other side:
n1993+1994
If n1993+1994, then we take k = 2m+1 and

m1993=2m1993+19932m+11993+19942m+21994+1994=1994m+1

>>4325487
Ah, you beat me to it.

>>4325496
Except that I said k=2m, which indeed should be k=2m+1. Cheers

3987 is the solution.

>>4325505
It's only the solution, if there is some kind of proof that it's the smallest.

>>4325505
Yeah I would also think that. But now we have to prove it. I agree that if it's not that, then the gaps will most likely not all be "aligned" properly and some values will fall between a m+11994 and then next m+11993.

>>4325322
>>4325512

for(int n = 1992; n < 4000; n++) {
boolean b = true;
for(int m = 1; m < 1993 && b; m++) {
int k = (m*n)/1993 + 1;
if (k*1994 >= ((m+1)*n)) {
b = false;
}
}
if (b) {
System.out.println("Answer:" + n);
break;
}
}
It outputs 3987.
It's an ugly proof, but a proof

>>4325556
Works for me. Having the non-computationally-challenging proof would be nice, though.

Assume 1995 =< n < 1993 + 1994

For every legit k,

k/n < (n-1)/n < 2*1993/(1993+1994)
k/n >= 3/n > 3/(1993+1994) = 1/(3*443) = 1/1329

So: 1/1329 < k/n < 2*1993/(1993+1994) for every k.

Not sure if this helps.

>>4325565
Derp, plugged in the wrong values

>>4325556
Let n be any number between 1994 and 3986.
Look at the case where m = 1992.
Since n <= 2*1993, we can see that in those cases, k must be equal to n-1. However, n-1/n >= 1993/1994, if n >= 1994.
Hence, we reach contradiction.

>>4325585
Sounds right to me. We haven't forgotten anything, have we?

Why must k=n-1?

>>4325640
Because otherwise the left does not hold.

>>4325645
Oh, fair enough

>>4325608
Nope, that's just what I was thinking of. We know it can't be smaller than 1993, and we know it can't be at least 1993 but smaller than 3987, and we know it can't be larger than 3987.