Pips answers for Wednesday, August 27, 2025

Hints first, answers last — open only what you need. Puzzles edited by Ian Livengood.

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Easy Pips answer — August 27, 2025

4 dominoes · 2×4 board · by Ian Livengood

Hint 1: where to start

Start with the “=” region at the bottom: all 4 cells must show the same number, and only 0s appear on enough dominoes to fill it.

Hint 2: the pips in each region
  • the single-cell 1 region at the top-left: 1
  • the “=” region at the bottom: 0, 0, 0, 0
  • the 18 region at the top-right: 6, 6, 6
Hint 3: step-by-step solution
  1. The 18 region at the top-right needs a total of exactly 18 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 6–6 domino horizontally on row 1, column 3 and row 1, column 4.
  2. The “=” region at the bottom needs every pip to match — nothing else fits. Place the 0–6 domino horizontally on row 2, column 3 and row 2, column 4.
  3. The “=” region at the bottom needs every pip to match (1 of its 4 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–1 domino horizontally on row 1, column 1 and row 1, column 2.
  4. Row 2, column 1 has only one open neighbor, so its domino must run to row 2, column 2; the “=” region at the bottom needs every pip to match (2 of its 4 cells are already filled) — nothing else fits. Place the 0–0 domino horizontally on row 2, column 1 and row 2, column 2.
Full answer board
1=18

Medium Pips answer — August 27, 2025

8 dominoes · 5×5 board · by Rodolfo Kurchan

Hint 1: where to start

Start with the “=” region at the bottom-left: all 5 cells must show the same number, and only 1s appear on enough dominoes to fill it.

Hint 2: the pips in each region
  • the single-cell “>2” region at the top-right: 3
  • the single-cell “<5” region at the top-left: 4
  • the 12 region at the center: 6, 6
  • the “=” region at the bottom-left: 1, 1, 1, 1, 1
  • the single-cell “>4” region at the bottom-right: 5
Hint 3: step-by-step solution
  1. The 12 region at the center needs a total of exactly 12 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–6 domino horizontally on row 2, column 2 and row 2, column 3.
  2. The 12 region at the center needs a total of exactly 12 (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 vertically on row 3, column 3 and row 4, column 3.
  3. The “=” region at the bottom-left needs every pip to match (1 of its 5 cells is already filled) — nothing else fits. Place the 1–1 domino vertically on row 4, column 1 and row 5, column 1.
  4. Row 2, column 1 has only one open neighbor, so its domino must run to row 3, column 1; the single-cell “<5” region at the top-left must total less than 5 — nothing else fits. Place the 1–4 domino vertically on row 2, column 1 and row 3, column 1.
  5. Row 4, column 2 has only one open neighbor, so its domino must run to row 3, column 2; the “=” region at the bottom-left needs every pip to match (4 of its 5 cells are already filled) — nothing else fits. Place the 1–5 domino vertically on row 3, column 2 and row 4, column 2.
  6. Row 5, column 4 has only one open neighbor, so its domino must run to row 4, column 4; the single-cell “>4” region at the bottom-right must total more than 4 — nothing else fits. Place the 2–5 domino vertically on row 4, column 4 and row 5, column 4.
  7. Row 1, column 4 has only one open neighbor, so its domino must run to row 2, column 4; the single-cell “>2” region at the top-right must total more than 2 — nothing else fits. Place the 3–3 domino vertically on row 1, column 4 and row 2, column 4.
  8. Row 3, column 4 has only one open neighbor, so its domino must run to row 3, column 5 — nothing else fits. Place the 0–0 domino horizontally on row 3, column 4 and row 3, column 5.
Full answer board
>2<512=>4

Hard Pips answer — August 27, 2025

10 dominoes · 5×7 board · by Rodolfo Kurchan

Hint 1: where to start

Start with the “=” region at the center: all 5 cells must show the same number, and only 1s appear on enough dominoes to fill it.

Hint 2: the pips in each region
  • the single-cell 6 region at the top: 6
  • the single-cell 5 region at the top-right: 5
  • the “=” region at the center: 1, 1, 1, 1, 1
  • the “=” region at the top: 4, 4
  • the “=” region at the top-right: 2, 2
  • the 0 region at the bottom-left: 0, 0
  • the single-cell “>3” region at the bottom: 4
  • the single-cell 6 region at the bottom-right: 6
  • the single-cell 2 region at the bottom: 2
Hint 3: step-by-step solution
  1. The “=” region at the top 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 4–6 domino vertically on row 1, column 3 and row 2, column 3.
  2. Row 4, column 7 has only one open neighbor, so its domino must run to row 4, column 6; the single-cell 6 region at the bottom-right needs a total of exactly 6 — nothing else fits. Place the 6–6 domino horizontally on row 4, column 6 and row 4, column 7.
  3. Row 2, column 2 has only one open neighbor, so its domino must run to row 3, column 2; 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 1–1 domino vertically on row 2, column 2 and row 3, column 2.
  4. The “=” region at the center needs every pip to match (2 of its 5 cells are already filled) — nothing else fits. Place the 0–1 domino vertically on row 3, column 3 and row 4, column 3.
  5. Row 4, column 2 has only one open neighbor, so its domino must run to row 4, column 1; the 0 region at the bottom-left needs a total of exactly 0 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–5 domino horizontally on row 4, column 1 and row 4, column 2.
  6. Row 1, column 6 has only one open neighbor, so its domino must run to row 2, column 6; the single-cell 5 region at the top-right needs a total of exactly 5 — nothing else fits. Place the 2–5 domino vertically on row 1, column 6 and row 2, column 6.
  7. The “=” region at the top-right needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 2–4 domino horizontally on row 2, column 4 and row 2, column 5.
  8. The “=” region at the center needs every pip to match (3 of its 5 cells are already filled) — nothing else fits. Place the 1–4 domino vertically on row 3, column 4 and row 4, column 4.
  9. The “=” region at the center needs every pip to match (4 of its 5 cells are already filled) — nothing else fits. Place the 1–3 domino horizontally on row 3, column 5 and row 3, column 6.
  10. Row 5, column 5 has only one open neighbor, so its domino must run to row 4, column 5; the single-cell 2 region at the bottom needs a total of exactly 2 — nothing else fits. Place the 2–2 domino vertically on row 4, column 5 and row 5, column 5.
Full answer board
65===0>362