Compare the rule with concrete cases
Fictional seven indivisible review markers are shared at A:B=2:1. Compute ideal shares, floor each, then give remaining markers to the largest fractional remainders; ties use A before B. Ideals are 14/3 and 7/3, floors four and two, leaving one for A.
| Group | Ideal share | Floor / remainder | Final whole units |
|---|---|---|---|
| A | 14/3 | 4 / 2/3 | 5 |
| B | 7/3 | 2 / 1/3 | 2 |
| Total | 7 | 6 floors + 1 remainder unit | 7 |
Work through the distinction
The floor stage alone loses one token. Independent ceiling would produce five and three, exceeding seven. The remainder rule preserves the total and deterministic allocation.
If the total is six with the same ratio, what are the shares?
Show the worked answer
A receives four and B two. Both ideal shares are whole, so no remainder allocation is needed.
Acceptance wording to adapt
Allocate the declared whole-unit total using ideal proportional shares, integer floors and largest remainders; use the stated tie order. Final shares must be nonnegative integers summing exactly to the input total.