Pips answers for Saturday, March 28, 2026

Start with the 10 region at the top: only 4 + 6, or 5 + 5 can make it with these dominoes.

• the single-cell “>1” region at the top-left: 2
• the 10 region at the top: 4, 6
• the “=” region at the bottom-left: 6, 6
• the single-cell 3 region at the center: 3
• the single-cell “>1” region at the right: 3
• the 10 region at the bottom-right: 5, 5

1. The “=” region at the bottom-left needs every pip to match — nothing else fits. Place the 6–6 domino vertically on row 2, column 1 and row 3, column 1.
2. The 10 region at the top needs a total of exactly 10 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–4 domino vertically on row 1, column 3 and row 2, column 3.
3. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2; the single-cell “>1” region at the top-left must total more than 1 — nothing else fits. Place the 2–6 domino horizontally on row 1, column 1 and row 1, column 2.
4. Row 2, column 5 has only one open neighbor, so its domino must run to row 3, column 5; the single-cell “>1” region at the right must total more than 1 — nothing else fits. Place the 3–5 domino vertically on row 2, column 5 and row 3, column 5.
5. Row 3, column 4 has only one open neighbor, so its domino must run to row 3, column 3; the 10 region at the bottom-right needs a total of exactly 10 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–5 domino horizontally on row 3, column 3 and row 3, column 4.

Start with the “=” region at the top-right: all 3 cells must show the same number, and only 1 or 3 appear on enough dominoes to fill it.

• the “=” region at the top-left: 0, 0
• the “=” region at the top-right: 3, 3, 3
• the “=” region at the center: 1, 1, 1
• the “=” region at the bottom: 6, 6
• the single-cell 5 region at the right: 5
• the single-cell “<4” region at the bottom-left: 2
• the “=” region at the bottom-right: 3, 3

1. The “=” region at the top-right needs every pip to match — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–3 domino vertically on row 1, column 4 and row 2, column 4.
2. The “=” region at the bottom-right needs every pip to match — nothing else fits. Place the 3–5 domino vertically on row 3, column 4 and row 4, column 4.
3. The “=” region at the center needs every pip to match — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 1–3 domino vertically on row 1, column 3 and row 2, column 3.
4. The “=” region at the bottom-right needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 3–6 domino horizontally on row 4, column 2 and row 4, column 3.
5. The “=” region at the bottom needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 1–6 domino horizontally on row 3, column 2 and row 3, column 3.
6. The “=” region at the top-left needs every pip to match — nothing else fits. Place the 0–1 domino vertically on row 1, column 2 and row 2, column 2.
7. Row 1, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the “=” region at the top-left needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 0–4 domino vertically on row 1, column 1 and row 2, column 1.
8. Row 4, column 1 has only one open neighbor, so its domino must run to row 3, column 1; the single-cell “<4” region at the bottom-left must total less than 4 — nothing else fits. Place the 2–4 domino vertically on row 3, column 1 and row 4, column 1.

Start with the 11 region at the top-left: the only way to make it with these dominoes is 5 + 6.

• the 11 region at the top-left: 5, 6
• the 5 region at the top-left: 1, 4
• the 4 region at the top-right: 1, 3
• the “=” region at the top-right: 2, 2
• the 11 region at the left: 5, 6
• the single-cell 6 region at the center: 6
• the “=” region at the right: 0, 0, 0
• the “=” region at the center: 3, 3, 3
• the 5 region at the bottom-left: 2, 3
• the “=” region at the center: 5, 5, 5
• the 8 region at the bottom-right: 4, 4
• the 10 region at the bottom-left: 4, 6
• the 2 region at the bottom: 1, 1
• the 5 region at the bottom-right: 2, 3

1. The 11 region at the top-left needs a total of exactly 11 — 5 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 5–6 domino vertically on row 2, column 1 and row 3, column 1.
2. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2; the 11 region at the top-left needs a total of exactly 11 (1 of its 2 cells is already filled) — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 1–5 domino horizontally on row 1, column 1 and row 1, column 2.
3. Row 1, column 3 has only one open neighbor, so its domino must run to row 1, column 4; the 5 region at the top-left needs a total of exactly 5 (1 of its 2 cells is already filled) — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 1–4 domino horizontally on row 1, column 3 and row 1, column 4.
4. Row 4, column 1 has only one open neighbor, so its domino must run to row 5, column 1; the 11 region at the left needs a total of exactly 11 (1 of its 2 cells is already filled) — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–6 domino vertically on row 4, column 1 and row 5, column 1.
5. Row 6, column 1 has only one open neighbor, so its domino must run to row 7, column 1; the 5 region at the bottom-left needs a total of exactly 5 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–4 domino vertically on row 6, column 1 and row 7, column 1.
6. Row 7, column 2 has only one open neighbor, so its domino must run to row 7, column 3; the 10 region at the bottom-left needs a total of exactly 10 (1 of its 2 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 1–6 domino horizontally on row 7, column 2 and row 7, column 3.
7. Row 7, column 4 has only one open neighbor, so its domino must run to row 7, column 5; the 2 region at the bottom needs a total of exactly 2 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–2 domino horizontally on row 7, column 4 and row 7, column 5.
8. The single-cell 6 region at the center needs a total of exactly 6 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–6 domino horizontally on row 3, column 3 and row 3, column 4.
9. The “=” region at the right needs every pip to match (1 of its 3 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–3 domino vertically on row 3, column 5 and row 4, column 5.
10. The “=” region at the top-right needs every pip to match — nothing else fits. Place the 0–2 domino vertically on row 2, column 6 and row 3, column 6.
11. Row 1, column 5 has only one open neighbor, so its domino must run to row 1, column 6; the 4 region at the top-right needs a total of exactly 4 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–3 domino horizontally on row 1, column 5 and row 1, column 6.
12. The 8 region at the bottom-right needs a total of exactly 8 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–4 domino vertically on row 6, column 6 and row 7, column 6.
13. The “=” region at the center needs every pip to match — nothing else fits. Place the 4–5 domino horizontally on row 5, column 5 and row 5, column 6.
14. The “=” region at the center needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 3–3 domino horizontally on row 4, column 3 and row 4, column 4.
15. Row 5, column 3 has only one open neighbor, so its domino must run to row 5, column 4; the “=” region at the center needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 5–5 domino horizontally on row 5, column 3 and row 5, column 4.