[Guest]

10 boxes each contain n balls numbered 1 to n

x balls are removed from each box (x < n), leaving n-x balls in each box, such that:

a: any combination of 5 boxes contains balls 1 to n (at least once each); and

b: any combination of 4 boxes does not contain balls 1 to n (at least one missing)

What is minimum n, and what is resulting x?

[bonanova]

Nice puzzle.

1. When the balls are removed, each box will have a different set of numbered balls.
   
   Assume the opposite: that boxes p and q have the same set of numbers.
   Take any four boxes that include p but exclude q.
   At least one numbered ball is missing [four boxes.]
   
   Add box q to the set.
   This set of [five] boxes must include all the numbers.
   But this is impossible since no new numbers were added: q duplicates p.
   .
2. Each number must remain in 6 or more boxes.
   If one number remains in only 5 boxes, the other 5 boxes do not contain every number.
   Thus of the initial 10n balls, at least 6n remain; no more than 4n balls are taken.
   
   So x <= .4n, which must be integral, so n is a multiple of 5.
   .
3. n=5 [x=2] does not work:
   
   These balls would be removed: 12 13 14 15 23 24 25 34 35 45.
   That seems nice, in that each possible combination of two balls is represented.
   However, that leaves 345 245 235 234 145 135 134 125 124 123 in the boxes.
   
   The last four boxes contain 12345.
   .
4. n=5 [x=3] does not work:
   
   Reverse the above case. The last five boxes do not contain the number 1.
   .
5. Checking out higher multiples of 5.
   

[preflop]

10 boxes each contain n balls numbered 1 to n

x balls are removed from each box (x < n), leaving n-x balls in each box, such that:

a: any combination of 5 boxes contains balls 1 to n (at least once each); and

b: any combination of 4 boxes does not contain balls 1 to n (at least one missing)

What is minimum n, and what is resulting x?

n = 5 and x = 4

seems easy unless I am missing something here.

10 boxes with 5 each = 50 total balls. n = 5.

remove 4 from each. x = 4.

we now have 10 boxes with 1 ball each. take any 5 boxes and you 1 to 5 (1 to n).

it can't be less than 5 because you need at least one ball in each box and 5 boxes to choose.

[preflop]

I am totally off - above, i misread the question.

[Guest]

Nice puzzle.

So x <= .4n, which must be integral, so n is a multiple of 5.

.

From what you wrote, I don't see why n must be a multiple of 5, since it is an inequality, not an equality. For example: 2 <= (.4)(7) is still true regardless of the fact that the right hand side is not integral. Other than that, your conclusions about the problem seem quite sound and I wish I had more time to think this problem fully through!

[Guest]

Well, we know that for any 4 boxes there must be at least 1 number missing. The remaining 6 baoxes must all contain that number. Therefore, no 2 combinations of 4 will have the same number missing. There are 10C4 combinations so therefore n = 210 and x = 5.

[Guest]

I made an error for the x value because I wasn't even thinking - I've no idea why I put 5 down. My revised answer is n = 210 (10C4) and x = 84 (9C3). Each box will appear in 9C3 combinations of 4 each with its own number missing. (See my above post) I'm tired right now so I've probably got it wrong again.

[bonanova]

I made an error for the x value because I wasn't even thinking - I've no idea why I put 5 down. My revised answer is n = 210 (10C4) and x = 84 (9C3). Each box will appear in 9C3 combinations of 4 each with its own number missing. (See my above post) I'm tired right now so I've probably got it wrong again.

Sounds good. Nice!

210 x .4 = 84 as advertised.