Positive integer ordered pairs in binomial coefficients (1 Viewer)

[juantheron]

 

[Paradoxica]

m9 it's been 3 years. let it go already.

http://math.stackexchange.com/questions/283977/ordered-pairs-in-binomial-coefficients

 

[InteGrand]

For anyone who might want to try this using 2016's factors, here are its factors:

1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 18, 21, 24, 28, 32, 36, 42, 48, 56, 63, 72, 84, 96, 112, 126, 144, 168, 224, 252, 288, 336, 504, 672, 1008, 2016 .

 

[Paradoxica]

For anyone who might want to try this using 2016's factors, here are its factors:

1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 18, 21, 24, 28, 32, 36, 42, 48, 56, 63, 72, 84, 96, 112, 126, 144, 168, 224, 252, 288, 336, 504, 672, 1008, 2016 .

The problem with this exact type of problem is that you have to consider every single divisor of whatever number you are solving for, thereby making it a brute force problem.

 

[InteGrand]

The problem with this exact type of problem is that you have to consider every single divisor of whatever number you are solving for, thereby making it a brute force problem.

And 2016 has too many factors to do by hand without it getting very tedious.

 

[InteGrand]

The problem with this exact type of problem is that you have to consider every single divisor of whatever number you are solving for, thereby making it a brute force problem.

And 2016 has too many factors to do by hand without it getting very tedious. (There could be a more elegant way though.)

 

[Paradoxica]

And 2016 has too many factors to do by hand without it getting very tedious. (There could be a more elegant way though.)

If there is, it is beyond all of stack exchange...

 

[seanieg89]

A rather dry problem but I was a little bored. Here goes:

64C2=2016
24C3=2024
17C4=2380
15C5=3003
14C6=3003


The above binomial coefficients xCy are each those with the least x for a given y such that the binomial coefficient is at least equal to 2016 (v easy to bruteforce with a calc by increasing x till you hit 2016).

Apart from giving us the nontrivial solution at x=64, this also tells us from monotonicity that:
-for x>64, all solutions must have y=1 or x-1.
-for x>=24, all solutions must have y=1,2,x-2,x-1
...
-for x>=14, all solutions must have min(y,x-y) =< 5.

On the y-side of things, these inequalities also tell us by monotonicity that there are no solutions for y=j or x-j where j=3,4,5 and that y=2 and y=x-2 each have a unique solution at x=64.

This tells us that apart from 2016C1,2016C2015,64C2,64C62, all remaining solutions must have y =< x =< 13.

This is very good! These binomial coefficients are small! In fact the largest of them is 13C6=1716<2016 so there is nothing left to check.

I.e. there are precisely four solutions.

 

[InteGrand]

-for x>64, all solutions must have y=1 or x.

 

[seanieg89]

Yep x-1. Will edit, thanks.

 

[juantheron]

Thanks seanieg89 for Nice explanation.