The winning Connecticut Play4 numbers for September 17, 2003 were: day 1-8-2-4; night 6-7-5-5.
Sum is the digits added together (5+5+0 = 10). Root sum keeps adding until one digit is left (10 → 1+0 = 1) — many regular players track it in their notebooks. Odd/Even marks each digit O or E (5-5-0 = OOE). High/Low marks 0–4 as L and 5–9 as H (5-5-0 = HHL).
Sum is the digits added together (5+5+0 = 10). Root sum keeps adding until one digit is left (10 → 1+0 = 1) — many regular players track it in their notebooks. Odd/Even marks each digit O or E (5-5-0 = OOE). High/Low marks 0–4 as L and 5–9 as H (5-5-0 = HHL).
| Date | Draw | Numbers | Sum | Root |
|---|---|---|---|---|
| Sep 17, 2003 | Night | 6-7-5-5 | 23 | 5 |
| Sep 17, 2003 | Day | 1-8-2-4 | 15 | 6 |
| Sep 16, 2003 | Night | 2-9-8-3 | 22 | 4 |
| Sep 16, 2003 | Day | 3-3-0-3 | 9 | 9 |
| Sep 15, 2003 | Night | 6-5-6-7 | 24 | 6 |
| Sep 15, 2003 | Day | 7-6-1-6 | 20 | 2 |
| Sep 14, 2003 | Night | 8-9-8-6 | 31 | 4 |
| Sep 13, 2003 | Night | 5-4-6-3 | 18 | 9 |
| Sep 12, 2003 | Night | 7-5-5-6 | 23 | 5 |
| Sep 12, 2003 | Day | 7-3-8-8 | 26 | 8 |
| Sep 11, 2003 | Night | 5-2-3-6 | 16 | 7 |
| Sep 11, 2003 | Day | 0-4-1-9 | 14 | 5 |